1103020415 (1)

5.7.4 Escape Velocity: 5.8 Earth Satellites: The objects which revolve around the When any object is thrown vertically up, it Earth are called Earth satellites. moon is the falls back to the Earth after reaching a certain only natural satellite of the Earth. It revolves in almost a circular orbit around the Earth height. Higher the speed with which the object is with period of revolution of nearly 27.3 days. Artificial satellites have been launched by thrown up, greater will be the height. If we keep several countries including  India. These satellites have different periods of revolution on increasing the velocity, a stage will come according to their practical use like navigation, surveillance, communication, looking into when the object will reach heights so large that space and monitoring the weather. Communication Satellites: These are it will escape the gravitational field of the Earth geostationary satellites.They revolve around the Earth in equatorial plane. They have same sense and will not fall back on the Earth. This initial of rotation as that of the Earth and the same period of rotation as that of the Earth, i. e., one day velocity is called the escape velocity. or 24 hours. Due to this, they appear stationary from the Earth’s surface. Hence they are called Thus, the minimum velocity with which a geostationary satellites or geosynchronous satellites. These are used for communication, body should be thrown vertically upwards from television transmission, telephones and radiowave signal transmission, e.g., INSAT the surface of the Earth so that it escapes the group of satellites launched by India. Polar Satellites: These satellites are placed Earth’s gravitational field, is called the escape in lower polar orbits. They are at low altitude 500 km to 800 km. Polar satellites are used velocity (ve) of the body. Obviously, as the for weather forecasting and meteorological purpose. They are also used for astronomical gravitational force due to Earth becomes zero observations and study of Solar radiations. only at infinite distance, the object has to reach Period of revolution of polar satellite is nearly 85 minutes, so it can orbit the Earth16 infinite distance in order to escape. time per day. They go around the poles of the Earth in a north-south direction while the Earth Let us consider the kinetic and potential rotates in an east-west direction about its own axis. The polar satellites have cameras fixed energies of an object thrown vertically upwards on them. The camera can view small stripes of the Earth in one orbit. In entire day the whole with escape velocity ve, when it is at the surface Earth can be viewed strip by strip. Polar and of the Earth and when it reaches infinite distance. equatorial regions at close distances can be viewed by these satellites. On the surface of the Earth, 5.8.1 Projection of Satellite: K.E. = 1 mv 2 2 e For the projection of an artificial satellite, PTo.Eta.l=en-eGrgMRy m= P.E. + K.E. it is necessary for the satellite to have a certain = 1 mv 2 - GMm --- (5.33) velocity and a minimum two stage rocket. A 2 e R single stage rocket can not achieve this. When The kinetic energy of the object will go the fuel in first stage of rocket is ignited on on decreasing with time as it is pulled back by the surface of the Earth, it raises the satellite Earth’s gravitational force. It will become zero when it reaches infinity. Thus at infinite distance from the Earth K.E. = 0 Also, P.E. =  GMm = 0 f ∴ Total energy = P.E. + K.E. = 0 As energy is conserved 1 mv 2 - GMm =0 2 e R or, v = 2GM --- (5.34) e R Using the numerical values of G, M and R. the escape velocity is 11.2 km/s. 92

vertically. The velocity of projection of satellite During this elliptical path, if the satellite normal to the surface of the Earth is the vertical velocity. If this vertical velocity is less that the passes through the Earth’s atmosphere, it escape velocity (v ), the satellite returns to the experiences a nonconservative force of air e resistance. As a result it loses energy and spirals Earth’s surface. While, if the vertical velocity is greater than or equal to the escape velocity, the down to the Earth. satellite will escape from Earth’s gravitational influence and go to infinity. Hence launching Case (II) vh=vc of a satellite in an orbit round the Earth can If the horizontal velocity is exactly equal not take place by use of single stage rocket. It requires minimum two stage  rocket. to the critical velocity, the satellite moves in a With the help of first stage of rocket, stable circular orbit round the Earth. satellite can be taken to a desired height above the surface of the Earth. Then the launcher is Case (III) vc<vh<ve rotated in horizontal direction i.e. through 900 If horizontal velocity is greater than using remote control and the first stage of the rocket is detached. Then with the help of second the critical velocity and less than the escape stage of rocket, a specific horizontal velocity (v ) is given to satellite so that it can revolve velocity at that height, the satellite again moves h in an elliptical orbit round the Earth with the in a circular path round the Earth. The exact point of projection as perigee (point closest to horizontal velocity of projection that must be given to a satellite at a certain height so that it the Earth). can revolve in a circular orbit round the Earth is called the critical velocity or orbital velocity Case (IV) vh = ve (vc) If horizontal speed of projection is equal Asatellite follows different paths depending to the escape speed at that height, the satellite upon the horizontal velocity provided to it. Four different possible cases are shown in Fig. 5.10. travels along parabolic path and never returns Case (I) vh<vc: to the point of projection. Its speed will be zero If tangential velocity of projection vh is less than the critical velocity, the orbit of satellite is at infinity. an ellipse with point of projection as apogee (farthest from the Earth) and Earth at one of the Case (V) vh > ve foci. If horizontal velocity is greater than the Fig. 5.10: Various possible orbits depending escape velocity, the satellite escapes from on the value of vh. gravitational influence of Earth transversing a hyperbolic path. Expression for critical speed Consider a satellite of mass m revolving round the Earth at height h above its surface. Let M be the mass of the Earth and R be its radius. If the satellite is moving in a circular orbit of radius (R+h) = r, its speed must be the magnitude of critical velocity vc. The centripetal force necessary for circular motion of satellite is provided by gravitational force exerted by the Earth on the satellite. ∴ Centripetal force = Gravitational force mv 2 = GMm c r2 r ? vc2 = GM r ?vc = GM r ?vc = GM gh (R +h) --- (5.35) (R +h) 93

This is the expression for critical speed in ?vc = 2R GSU the orbit of radius (R + h) 3 It is clear that the critical speed of a When a satellite revolves very close to satellite is independent of the mass of the satellite. It depends upon the mass of the Earth the surface of the Earth, motion of satellite and the height at which the satellite is revolving or gravitational acceleration at that altitude. gets affected by the friction produced due to The critical speed of a satellite decreases with increase in height of satellite. resistance of air. In deriving the above expression Special case the resistance of air is not considered. When the satellite is revolving close to the surface of the Earth, the height is very small as 5.8.2 Weightlessness in a Satellite: compared to the radius of the Earth. Hence the height can be neglected and radius of the orbit is According to Newton’s second law of nearly equal to R (i.e R>>h, R+h ≈ R) motion, F = ma , where F is the net force acting on an object having acceleration a. Let us consider the example of a lift or elevator from an inertial frame of reference. ... Critical speed vc = GM Whether the lift is at rest or in motion, a R passenger in it experiences only two forces: As G is related to acceleration due to (i) Gravitational force mg directed vertically downwards (towards centre of the earth) and gravity by the relation, (ii) normal reaction force N directed vertically upwards, exerted by the floor of the lift. As these g = GM forces are oppositely directed, the net force in R2 the downward direction will be F = ma - N . ∴GM = gR2 Though the weight of a body (passenger, in ... Critical speed in terms of acceleration this case) is the gravitational force acting upon due to gravity can be obtained as it, we experience or feel our weight only due to the normal reaction force N exerted by the vc = gR2 = gR floor. This, in turn, is equal and opposite to the R relative force between the body and the lift. If you are standing on a weighing machine in a = 7.92 km/s lift, the force recorded by the weighing machine Obviously, this is the maximum possible critical is nothing but the normal reaction N. speed. This is at least 25 times the speed of the fastest passenger aeroplanes. Case I: Lift having zero acceleration Example 5.9: Show that the critical velocity of a body revolving in a circular orbit very close This happens when the lift is at rest or is to the surface of a planet of radius R and mean moving upwards or downwards with constant velocity: density ρ is 2R GSU . 3 Solution : Since the body is revolving very close to the planet, h = 0 The net force F = 0 = mg - N ∴ mg =N density U M = M Hence in this case we feel our normal V weight mg . 4 S R3 3 Case II: Lift having net upward acceleration a ?M = 4 S R3 U u 3 This happens when the lift just starts Critical Velocity moving upwards or is about to stop at a lower floor during its downward motion (remember, GM G 4 S R3 U while stopping during downward motion, the R 3 acceleration must be upwards). vc = = R 94

As the net acceleration is upwards, the not falling on the earth? The reason is that the upward force must be greater. revolving satellite is having a tangential velocity which manages to keep it moving in a circular m∴gF, h=enmcea,u = N - mg ∴ N = mg + mau, i.e., orbit at that height. we feel heavier. 5.8.3 Time Period of a Satellite: N > It should also be remembered that this is The time taken by a satellite to complete not an apparent feeling. The weighing machine one revolution round the Earth is its time period. really records a reading greater than mg. Case III (a): Lift having net downward Consider a satellite of mass m projected to acceleration a height h and provided horizontal velocity equal to the critical velocity. The satellite revolves in d a circular orbit of radius (R+h) = r. This happens when the lift just starts The distance traced by satellite in one moving downwards or is about to stop at a higher floor during its upward motion (remember, revolution is equal to the circumference of the while stopping during upward motion, the acceleration must be downwards). circular orbit within periodic time T. As the net acceleration is downwards, the ?Critical speed = Circumference of the orbit downward force must be greater. Time period ∴ F = ma = mg -N ∴ N = mg - ma , i.e., vc = 2S r N < mg, hence,dwe feel lighter. d T It should be remembered that this is not an but we have, vc = Gm apparent feeling. The weighing machine really r records a reading less than mg. ? Gm 2S r r T Case III (b): State of free fall: This will be possible if the cables of the lift are cut. In this or, GM 4S 2r2 case, the downward acceleration ad = g. r T2 If the downward acceleration becomes ? T2 4S 2r3 equal to the gravitational acceleration g, we get, GM N = mg - mad = 0. As π2, G and M are constant, T 2 ∝ r3, i.e., Thus, there will not be any feeling of the square of period of revolution of satellite is weight. This is the state of total weightlessness directly proportional to the cube of the radius and the weighing machine will record zero. of orbit. In the case of a revolving satellite, the T = 2S r3 satellite is performing a circular motion. The GM acceleration for this motion is centripetal, which is provided by the gravitational acceleration ?T 2S (R + h)3 --- (5.36) g at the location of the satellite. In this case, GM ad = g, or the satellite (along with the astronaut) is in the state of free fall. Obviously, the apparent This is an expression for period of satellite weight will be zero, giving the feeling of total revolving in a circular orbit round the Earth. weightlessness. Perhaps you might have seen Period of a satellite does not depend on its in some videos that the astronauts are floating mass. It depends on mass of the Earth, radius inside the satellite. It is really difficult for them of the Earth and the height of the satellite. If to change their position. the height of projection is increased, period of the satellite increases. Period of the satellite can In spite of free fall, why is the satellite also be obtained in terms of acceleration due to 95

gravity. density (U ) = mass (M ) volume (V ) As GM = gh (R+h)2 ?T = 2S (R +h)3 ?M = UV --- (2) gh(R + h)2 As planet is spherical in shape, volume of ?T = 2S R +h planet is given as gh V = 4 S R3 3 ?T = 2S r --- (5.37) ? M = 4 S R3U --- (3) gh 3 Special case : Substituting the values form eq. (2) and (3) When satellite revolves close to the surface in Eq. (1), we get of the Earth, R + h r≈evRoluatnidongihs≈ g. Hence the T = 2S R3 minimum period of Gu 4 S R3U 3 R (T)min = 2π --- (5.38) 3S g ?T = GU Example 5.9: Calculate the period of revolution of a polar satellite orbiting close to the surface of the Earth. Given R = 6400 km, g = 9.8 m/s2. 5.8.4 Binding Energy of an orbiting satellite: The minimum energy required by a satellite Solution : h is negligible as satellite is close to to escape from Earth’s gravitational influence is the Earth surface. the binding energy of the satellite. ... R + h ≈ R Expression for Binding Energy of satellite revolving in circular orbit round the Earth gh ≈ g Consider a satellite of mass m revolving at height h above the surface of the Earth in a R = 6400 km = 6.4×106 m. circular orbit. It possesses potential energy as T = 2S R well as kinetic energy. Let M be the mass of the g Earth, R be the Radius of the Earth, vc be critical 6.4 u106 velocity of satellite, r = (R+h) be the radius of 9.8 = 2u 3.14 the orbit. ∴Kinetic energy of satellite = 5.075u103second = 1 mvc2 2 = 85 minute (approximately) 1 GMm Example 5.10: An artificial satellite revolves = 2r --- (5.39) around a planet in circular orbit close to its surface. Obtain the formula for period of the The gravitational potential at a distance r satellite in terms of density ρ and radius R of planet. from the centre of the Earth is - GM ... Potential energy of satellite r Solution : Period of satellite is given by, = Gravitational (R +h)3 potential × mass of satellite GM T = 2π --- (1) = - GMm --- (5.40) r Here, the satellite revolves close to the The total energy of satellite is given as surface of planet, hence h is negligible, hence R+h  R T.E. = K.E. + P.E. 96

= 1 GMm - GMm influence its total energy should become non- 2 r r negative (zero or positive). Hence the minimum = - 1 GMm energy to be supplied to unbind the satellite 2r --- (5.41) is + 1 GMm This is the binding energy of a 2r satellite. Total energy of a circularly orbiting satellite is negative. Negative sign indicates Internet my friend that the satellite is bound to the Earth, due to gravitational force of attraction. For the hyperphysics.phy-astr.gsu.edu/hbase/grav. satellite to be free from the Earth’s gravitational html#grav ExercisesExercises 1. Choose the correct option. 2. Answer the following questions. i) The value of acceleration due to gravity is i) State Kepler’s law equal of area. maximum at (A) the equator of the Earth . ii) State Kepler’s law of period. (B) the centre of the Earth. (C) the pole of the Earth. iii) What are the dimensions of the universal (D) slightly above the surface of the gravitational constant? Earth. iv) Define binding energy of a satellite. ii) The weight of a particle at the centre of the v) What do you mean by geostationary Earth is satellite? (A) infinite. (B) zero. vi) State Newton’s law of gravitation. (C) same as that at other places. (D) greater than at the poles. vii) Define escape velocity of a satellite. iii) The gravitational potential due to the Earth viii)  What is the variation in acceleration due is minimum at to gravity with altitude? (A) the centre of the Earth. (B) the surface of the Earth. ix) On which factors does the escape speed (C) a points inside the Earth but not at of a body from the surface of Earth its centre. depend? (D) infinite distance. iv) The binding energy of a satellite revolving x) As we go from one planet to another planet, how will the mass and weight of around planet in a circular orbit is 3×109 J. a body change? Its kinetic energy is (A) 6×109J xi) What is periodic time of a geostationary (B) -3 ×109J satellite? (C) -6 ×10+9J (D) 3 ×10+9J xii) State Newton’s law of gravitation and express it in vector form. xiii) What do you mean by gravitational constant? State its SI units. xiv) Why is a minimum two stage rocket necessary for launching of a satellite? xv) State the conditions for various possible orbits of a satellite depending upon the tangential speed of projection. 97

2. Answer the following questions in xiv) What is critical velocity? Obtain an detail. expression for critical velocity of an orbiting satellite. On what factors does it i) Derive an expression for critical velocity depend? of a satellite. ii) State any four applications of a xv) Define escape speed. Derive an communication satellite. expression for the escape speed of an object from the surface of the each. iii) Show that acceleration due to gravity at height h above the Earth’s surface is xvi) Describe how an artificial satellite using two stage rocket is launched in an orbit gh g § R R h ·2 around the Earth. ©¨  ¸¹ iv) Drawalabelleddiagramtoshowdifferent 4. Solve the following problems. trajectories of a satellite depending upon i) At what distance below the surface of the Earth, the acceleration due to gravity the tangential projection speed. decreases by 10% of its value at the surface, given radius of Earth is 6400 v) Derive an expression for binding energy km. of a body at rest on the Earth’s surface. vi) Why do astronauts in an orbiting satellite [Ans: 640 km]. have a feeling of weightlessness? ii) If the Earth were made of wood, the mass vii) Draw a graph showing the variation of gravitational acceleration due to of wooden Earth would have been 10% the depth and altitude from the Earth’s surface. as much as it is now (without change in its diameter). Calculate escape speed viii) At which place on the Earth’s surface from the surface of this Earth. is the gravitational acceleration maximum? Why? [Ans: 3.54 km/s] ix) At which place on the Earth surface the iii) Calculate the kinetic energy, potential gravitational acceleration minimum? energy, total energy and binding energy Why? of an artificial satellite of mass 2000 kg orbiting at a height of 3600 km above the x) Define the binding energy of a satellite. surface of the Earth. Obtain an expression for binding energy of a satellite revolving around the Earth Given:- G = 6.67×10-11 Nm2/kg2 at certain attitude. R = 6400 km M = 6×1024 kg xi) Obtain the formula for acceleration due [Ans: KE = 40.02×109J, to gravity at the depth ‘d’ below the Earth’s surface. PE = -80.09 ×109J, xii) State Kepler’s three laws of planetary TE = 40.02 ×109J, motion. BE = 40.02×109J] xiii) State the formula for acceleration due iv) Two satellites A and B are revolving around a planet. Their periods of to gravity at depth ‘d’ and altitude ‘h’ revolution are 1 hour and 8 hours respectively. The radius of orbit of Hence show that their ratio is equal to satellite B is 4×104 km. find radius of § Rd · orbit of satellite A . ¨© R  2h ¸¹ by assuming that the altitude [Ans: 1×104 km] is very small as compared to the radius of the Earth. 98

v) Find the gravitational force between the x) Calculate the value of the universal Sun and the Earth. gravitational constant from the given Given Mass of the Sun = 1.99×1030 kg data. Mass of the Earth = 6×1024 kg, Radius of the Earth = 6400 km and the Mass of the Earth = 5.98×1024 kg acceleration due to gravity on the surface = 9.8 m/s2 vi) The average distance between the Earth [Ans : 6.69×10-11 N m2/kg2 ] and the Sun = 1.5×1011 m. A body weighs 5.6 kg wt on the surface [Ans: 3.5×1022 N] xi) of the Earth. How much will be its weight on a planet whose mass is 7 times Calculate the acceleration due to gravity the mass of the Earth and radius twice that of the Earth’s radius. at a height of 300 km from the surface of the Earth. (M = 5.98 ×1024 kg, R = 6400 km). [Ans :- 8.889 m/s2] [Ans: 9.8 kg-wt] vii) Calculate the speed of a satellite in an xii). What is the gravitational potential due to orbit at a height of 1000 km from the the Earth at a point which is at a height Earth’s surface. ME= 5.98×1024 kg, R = oMfa2sRs EofatbhoevEeatrhthe surface of the Earth, 6.4×106 m. is 6×1024 kg, radius of [Ans : 7.34 ×103 m/s] the Earth = 6400 km and G = 6.67×10-11 viii) Calculate the value of acceleration due Nm2 kg-2. to gravity on the surface of Mars if the [Ans: 2.08×107 J] radius of Mars = 3.4×103 km and its mass is 6.4×1023 kg. *** [Ans : 3.69 m/s2] ix) Aplanet has mass 6.4 ×1024 kg and radius 3.4×106 m. Calculate energy required to remove on object of mass 800 kg from the surface of the planet to infinity. [Ans : 5.02 ×1010J] 99

6. Mechanical Properties of Solids Can you recall? 1. Can you name a few objects which change their shape and size on application of a force and regain their original shape and size when the force is removed ? 2. Can you name objects which do not regain their original shape and size when the external force is removed? 6.1 Introduction: the larger is its deformation. Deformation Solids are made up of atoms or a group could be in the form of change in length of a wire, change in volume of an object or change of atoms placed in a definite geometric in shape of a body. arrangement. This arrangement is decided by nature so that the resultant force acting on each We know that when a deforming force constituent due to others is zero. This is the (e.g. stretching) is applied to a rubber band, it equilibrium state of a solid at room temperature. gets deformed (elongated) but when the force The given equilibrium arrangement does not is removed, it regains its original length. When change with time. It can change only when an a similar force is applied to a dough, or clay external stimulus, like compressive force from it also gets deformed but it does not regain its all sides, is applied to a solid. The constituents original shape and size after removal of the vibrate about their equilibrium positions even deforming force. These observations indicate at very low temperatures but cannot leave their that rubber and clay are different in nature. fixed positions. This fact provides the solids The property that decides this nature is called a definite shape and size (allows the solids to elasticity/plasticity. We will learn more about maintain a definite shape and size). these properties of solids in this Chapter . 6.2 Elastic Behavior of Solids: If an external force is applied to a solid the constituents are slightly displaced and restoring If a body regains its original shape and forces are developed in it. These restoring size after removal of the deforming force, it forces try to bring the constituents back to is called an elastic body and the property is their equilibrium positions so that the solid can called elasticity. Here the restoring forces are regain its shape. When the deforming forces are strong enough to bring the displaced molecules removed, the interatomic forces tend to restore to their original positions. Examples of elastic the original positions of the molecules and thus materials are metals, rubber, quartz, etc. the body regains its original shape and size. However, as we will see later, this is possible If a body regains its original shape and only within certain limits. size completely and instantaneously upon removal of the deforming force, then it is said The form of a body is decided by its size to be perfectly elastic. and shape, e.g., a tennis ball and a football both are spherical, i.e., they have the same If a body does not regain its original shape. But a tennis ball is smaller in size than shape and size and retains its altered shape a football. When a force is applied to a solid or size upon removal of the deforming force, (which is not free to move), the size or shape it is called a plastic body and the property is or both change due to changes in the relative called plasticity. Here, the restoring forces are positions of molecules. Such a force is called not strong enough to bring the molecules back deforming force. to their original positions. Examples of plastic materials are clay, putty, plasticine, thick mud, The change in shape or size or both etc. There is no solid which is perfectly elastic of a body due to an external force is called or perfectly plastic. The best example of a near deformation. ideal elastic solid is quartz fibre and that of a plastic body is putty. The larger the deforming force on a body, 100

6.3 Stress and Strain: Fig. 6.1 (a): Tensile stress. The elastic properties of a body are described in terms of stress and strain. When a body gets deformed under an applied force, restoring forces are set up internally. They oppose change in shape or size of the body. When body is in equilibrium in its altered shape or size, deforming force and restoring force are equal and opposite. The internal restoring force per unit area of a body is called stress. --- (6.1) Fig. 6.1 (b): Compressive stress. B) Tensile strain: where F is internal restoring force (external applied deforming force). SI unit of stress is The strain produced by a tensile deforming N m-2 or pascal (Pa). The dimensions of a stress force is called tensile strain or longitudinal are [ L-1 M1 T-2 ]. strain or linear strain. Strain is a measure of the deformation of a If L is the original length and ∆l is the body. When two equal and opposite forces are change in length due to the deforming force, applied to an elastic body, there is a change in then the dimensions of the body, Strain is defined as the ratio of change in dimensions of the --- (6.5) body to its original dimensions. 2 : When a deforming force acting on a body --- (6.2) produces change in its volume, the stress is called volume stress and the strain produced is called volume strain. It is the ratio of two similar quantities. A) Volume stress or hydraulic stress: Hence strain is a dimensionless physical quantity. It has no units. There are three types Let F be a force acting perpendicular to of stress and corresponding strains. the entire surface of the body. It acts normally 1: Stress produced by a deforming force acting and uniformly all over the surface area A of the along the length of a body or a rod is called body. Such a stress which produces change in tensile stress or a longitudinal stress. The size but no change in shape is called volume strain produced is called tensile strain. stress. A) Tensile stress or compressive stress: --- (6.6) Suppose a force F is applied along the length of a wire, or perpendicular to its cross section A. This produces an elongation in Volume stress produces change in size the wire and the length of the wire increases without change in shape of body, it is called accordingly, as shown in Fig. 6.1 (a). hydraulic or hydrostatic volume stress as shown in Fig. 6.2. Tensile stress = | F | --- (6.3) B) Volume strain: A A deforming force acting perpendicular to the entire surface of a body produces a volume When a rod is pushed at two ends with equal strain. Let V be the original volume and ∆V be and opposite forces, its length decreases. the change in volume due to deforming force, The restoring force per unit area is called then compressive stress as shown in Fig. 6.1 (b). --- (6.7) Compressive stress | F | --- (6.4) A 101

tangential force. Tangential force is parallel to the top and the bottom surface of the block. The restoring force per unit area developed due to the applied tangential force is called shearing stress or tangential stress. B) Shearing strain: There is a relative displacement, ∆l, of the Fig. 6.2 : Volume stress. bottom face and the top face of the cube. Such Do you know ? relative displacement of two surfaces is called When a balloon is filled with air at high shear strain. It can be calculated as follows, pressure, its walls experience a force from within. This is also volume stress. It tries whenSthheearreinlagtisvteradinisp∆llal ce=mtaenntθ∆=l θ --- (6.9) to expand the balloon and change its size is very small. without changing shape. When the volume stress exceeds the limit of bulk elasticity, the 6.4 Hooke’s Law: balloon explodes. Similarly, a gas cylinder explodes when the pressure inside it exceeds Robert Hooke (1635-1703), an English the limit of bulk elasticity of its material. physicist, studied the tension in a wire and A submarine when submerged under water is under volume stress. strain produced in it. His study led to a law now 3 : When a deforming force acting on a body known as Hooke’s law. produces change in the shape of a body, shearing stress and shearing strain are produced. Statement: Within elastic limit, stress is A) Shearing stress: directly proportional to strain. Let F be a tangential force acting on a surface area A. This force produces change in Stress = constant shape of the body without changing its size as Strain shown in Fig. 6.3. The constant is called the modulus --- (6.8) of elasticity. The modulus of elasticity of a material is the ratio of stress to the corresponding strain. It is defined as the slope of the stress-strain curve in the elastic deforming region and depends on the nature of the material. The maximum value of stress up to which stress is directly proportional to strain is called the elastic limit. The stress-strain curve within elastic limit is shown in Fig. 6.4 D Fig. 6.3 : Tangential force produces Fig 6.4: Stress versus strain graph within shearing stress. elastic limit for an elastic body. 6.5 Elastic modulus: Suppose ABCD is the front face of a cube. There are three types of stress and strain A force F is applied to the cube so that the related to change in length, change in volume bottom of the cube is fixed and only the top and change in shape. Hence, we have three surface is slightly displaced. Such force is called moduli of elasticity corresponding to each type 102

of stress and strain. Table 6.1: Young's modulus of some familiar materials 6.5.1 Young’s modulus (Y): It is the modulus of elasticity related to Material Young's modulus Y ×1010 Pa (N/m2) change in length of an object like a metal wire, rod, beam, etc., due to the applied Lead 1.5 Glass (crown) 6.0 deforming force. Hence it is also called as 7.0 Aluminium 7.6 elasticity of length. It is named after the British Silver 8.1 Gold 9.0 physicist Thomas Young (1773-1829). Brass 11.0 Copper 21.0 Consider a metal wire of length L having Steel radius r suspended from a rigid support. A load Mg is attached to the free end of the wire. Due to this, deforming force is applied at the free end of the wire in downward direction. In its Example 6.1: A brass wire of length 4.5m with crosssectional area of 3×10-5 m2 and a copper equilibrium position, wire of length 5.0 m with cross sectional area 4×10-5 m2 are stretched by the same load. The Longitudinal  stress = Applied  force same elongation is produced in both the wires. Area Find the ratio of Young’s modulus of brass and copper. = F Solution: For brass, A LB= 4.5m, AB= 3×10-5 m2 Mg --- (6.10) lB= l, FB= F Sr2 It produces a change in length of the wire. If (L+l) is the new length of wire, then l is the extension or elongation in wire. Longitudinal  strain = change in length   original  length ?YB F u 4.5  Ll - -- (6.11) 3u105 u l = For copper, LC= 5m, AC= 4×10-5 m2 Young’s modulus is the ratio of longitudinal lC= l, FC=F stress to longitudinal strain. Young cs  modulus  lloonnggiittuuddiinnaall  stress -- (6.12)  strain ?Yc F u 5.0 4 u105 u l u105 u YB F u 4.5 u 4 F u5 l YC 3u105 u MgL l S r2l Y ---(6.13) 18 u 105 1.2 15 u 105 SI unit of Young’s modulus is N/m2. Its Example 6.2: A wire of length 20 m and area dimensions are [ L-1 M1 T-2 ]. Young’s modulus indicates the of cross section 1.25×10-4 m2 is subjected to a resistance of an elastic solid to elongation or load of 2.5 kg. (1 kgwt = 9.8 N). The elongation compression. Young’s modulus of a material is produced in wire is 1×10-4 m. Calculate Young’s useful for characterization of an object subjected modulus of the material. to compression or tension. Young's modulus is Solution: Given, L = 20 m the property of solids only. A = 1.25 ×10-4 m2 103

F = mg = 2.5 × 9.8N Compressibility is the fractional decrease in volume, -∆V/V per unit increase in pressure. L = 10-4 m SI unit of compressibility is m2/ N or Pa-1 and its dimensions are [ L1 M-1 T2]. To find: Y Y = FL 2.5u9.8u20 Al 1.25u104 u104 = 3.92 × 1010 N m-2 Do you know ? 6.5.2 Bulk modulus (K): The bulk modules of water is 2.18×108 Pa and its compressibility is 45.8×10-10 Pa-1. It is the modulus of elasticity related to Materials with small bulk modulus and large compressibility are easier to compress. change in volume of an object due to applied deforming force. Hence it is also called as elasticity of volume. Bulk modulus of elasticity Example 6.3: A metal cube of side 1m is is a property of solids, liquids and gases. subjected to a force. The force acts normally If a sphere made from rubber is completely on the whole surface of cube and its volume immersed in a liquid, it will be uniformly changes by 1.5×10-5 m3. The bulk modulus of compressed from all sides. Suppose this metal is 6.6×1010 N/m2. Calculate the change in compressive force is F. Let the change in pressure. pressure on the sphere be dP and let the change Solution: Given, in its volume be dV. If the original volume of volume of cube=V = l3 = (1)3 =1m3 the sphere is V, then volume strain is defined as Change in volume = dV = 1.5×10-5 m3 Bulk modulus = K = 6.6×1010 N/m2. To find: Change in pressure dP   dV --- (6.14) K =V dP V dV The negative sign indicates that there is a dP = K dV V decrease in volume. The magnitude of the volume strain is dV dP 6.6 u1010 u1.5u105 V 1 Bulk modulus is defined as the ratio of dP = 9.9×105 N/m2. volume stress to volume strain. Table 6.2: Bulk modulus of some familiar materials K dP V  dP --- (6.17) Material Bulk modulus K § dV · dV ×1010 Pa (N/m2) ¨© V ¹¸ Lead 4.1 Brass 6.0 SI unit of bulk modulus is N/m2. Dimensions of Glass (crown) 6.0 K are [ L-1 M1 T-2 ]. Aluminium 7.5 Silver 10.0 Table 6.2 gives bulk moduli of some Copper 14.0 familiar materials Steel 16.0 Gold 18.0 Bulk modulus measures the resistance offered by gases, liquids or solids while an 6.5.3 Modulus of rigidity (η): attempt is made to change their volume. The modulus of elasticity related to The reciprocal of bulk modulus of elasticity change in shape of an object is called rigidity is called compressibility of the material. modulus. It is the property of solids only as they alone possess a definite shape. --- (6.18) 104

The block shown in Fig. 6.5 is made of Table 6.3: Rigidity modulus η of some a uniform isotropic material. It has a uniform familiar materials crosssection area A and height l. A cross section of the block is defined as any plane parallel Material Rigidity modulus η to the top and the bottom surface and cuts the ×1010 Pa (N/m2) block. Two forces of magnitude 'F' are applied along top and bottom surface as shown in Fig. Lead 0.6 (6.5). They constitute a couple. The upper Aluminium 2.5 surface is displaced relative to the lower surface Glass (crown) 2.5 by a small distance ∆l and corresponding angles 2.7 change by a small amount θ = ∆l/l. Silver 2.9 Gold 3.5 Brass 4.4 Copper 8.3 Steel Rigidity modulus indicates the resistance offered by a solid to change in its shape. Example 6.4: Calculate the modulus of rigidity of a metal, if a metal cube of side 40 cm is subjected to a shearing force of 2000 N. The upper surface is displaced through 0.5cm with Fig. 6.5: Modulus of rigidity, tangential respect to the bottom. Calculate the modulus of force F and shear strin θ. rigidity of the metal. A couple is applied by pushing the top and the bottom surfaces as shown in Fig. 6.5. Solution: Given, Similar couple would be applied if the bottom of the block is fixed and only the top is pushed. Length of side of cube = l= 40 cm = 0.40 m The forces F and - F are parallel to the Shearing force = F= 2000 N = 2×103 N cross section. This is different than the tensile stress where the force is normal to the cross Displacement of top face = ∆l = 0.5cm = 0.005m section. Area = A = l 2 = 0.16m2 As a result of the way in which the forces To find: modulus of rigidity, η are applied the block is subjected to a shear stress defined by shear stress = F/A. K F AT The SI unit of shear stress is N/m2 or Pa. The block is distorted as a result of the shear T 'l 0.005 0.0125 stress. The top and bottom surface are relatively l 0.40 displaced by a small distance ∆l. The corner angle changes by a small amount θ which is K 2.0 u103 N called shear strain and is expressed in radian. (0.16m2 ) ˜ (0.0125) Shear strain 'θ' is given by θ = ∆l/l, (for small ∆l). = 1.0 u106 N / m2 Shear modulus or modulus of rigidity: It is 6.5.4 Poisson’s ratio: defined as the ratio of shear stress to shear strain within elastic limits. Suppose a wire is fixed at one end and a force is applied at its free end so that the wire gets stretched. Length of the wire increases and at the same time, its diameter decreases, i.e., the wire becomes longer and thinner as shown in Fig. 6.6 (a). K = shear stress F/A F --- (6.17) shear strain T AT Table 6.3 gives values of rigidity Fig. 6.6 (a): When a wire is stretched its modulus η of some familiar materials. length increases and its diameter decreases. 105

Do you know ? Fig. 6.6 (b): When a wire is compressed its For most of the commonly used metals, the value of σ is between 0.25 and 0.35. length increases and its diameter increases. Many times we assume that volume is constant while stretching a wire. However, If equal and opposite forces are applied to in reality, its volume also increases. Using approximations it can be shown that σmax ≈ an object along its length inwards, the object gets 0.5 if volume is unchanged. In practice, it is much less. This shows that volume also compressed (Fig. 6.6 (b)). There is a decrease increases while stretching. 6.6 Stress-Strain Curve: in dimensions along its length and at the same Suppose a metal wire is suspended time there is an increase in its dimensions vertically from a rigid support and stretched by applying load to its lower end. The load is perpendicular to its length. When length of the gradually increased in small steps until the wire breaks. The elongation produced in the wire is wire decreases, its diameter increases. measured during each step. Stress and strain is noted for each load and a graph is drawn by The ratio of change in dimensions to taking tensile strain along x-axis and tensile stress along y-axis. It is a stress-strain curve as original dimensions in the direction of the shown in Fig. 6.7. applied force is called linear strain while Fig. 6.7 : stress-strain curve. The initial part of the graph is a straight the ratio of change in dimensions to original line OA. This is the region in which Hooke's law is obeyed and stress is directly proportional dimensions in a direction perpendicular to the to strain. The straight line portion ends at A. The stress at this point is called proportional applied force is called lateral strain. Within limit. If the load is further increased till point B is reached, stress and strain are no longer elastic limit, the ratio of lateral strain to the proportional and Hooke's law is not valid. If the load is gradually removed starting at any point linear strain is called the Poisson’s ratio. between O and B. The curve is retraced until the wire regains its original length. The change If l is the original length of wire, ∆l is is reversible. The material of the wire shows increase/decrease in length of wire, D is the elastic behaviour in the region OB. Point B is original diameter and d is corresponding change called the yield point. The corresponding point in diameter of wire then, Poisson’s ratio is given is called the elastic limit. by V Lateral strain Linear strain = d / D l / L d.L --- (6.18) D.l Poisson’s ratio has no unit. It is dimensionless. Table 6.4 gives values of Poisson ratio, σ, of some familiar materials. Table 6.4: Poisson ratio, σ, of some familiar materials Material Poisson ratio σ Glass (crown) 0.2 Steel 0.28 Aluminium 0.36 Brass 0.37 Copper 0.37 Silver 0.38 Gold 0.42 106

When the stress is increased beyond point the energy dissipated during deformation of a B, the strain continues to increase. If the load is material. removed at any point beyond B, C for example, the material does not regain its original length. It follows the line CE. Length of the wire when Fig. 6.8: Stress-stain curve for increasing there is no stress is greater than the original and decreasing load. length. The deformation is irreversible and the material has acquired a permanent set. Can you tell? Further increase in load causes a large Why does a rubber band become loose after increase in strain for relatively small increase repeated use? in stress, until a point D is reached at which fracture takes place. 6.7 Strain Energy: The material shows plastic flow or plastic The elastic potential energy gained by a deformation from point B to point D. The material does not regain its original state when wire during elongation by a stretching force the stress is removed. The deformation is called plastic deformation. is called as strain energy. The curve described above shows all Consider a wire of original length L and the possibilities for an elastic substance. In particular, many metallic wires (copper, cross sectional area A stretched by a force F aluminum, silver, etc) exhibit this type of behavior. However, majority of materials in acting along its length. The wire gets stretched every day life exhibit only some part of it. and elongation l is produced in it. The stress and Materials such as glass, ceramics, etc., break within the elastic limit. They are called the strain increase proportionately. brittle. Longitudinal stress = F Metals such as copper, aluminum, wrought A iron, etc. have large plastic range of extension. l They lengthen considerably and undergo plastic Longitudinal strain = L deformation till they break. They are called ductile. Young’s modulus = longitudinal  stress longitudinal  strain Metals such as gold, silver which can be hammered into thin sheets are called malleable. §F · ©¨ A ¹¸ Rubber has large elastic region. It can be Y §l·  FALl stretched so that its length becomes many times ¨© L ¹¸ its original length, after removal of the stress it returns to its original state but the stress strain ?F YAl --- (6.19) curve is not a straight line. A material that can L be elastically stretched to a larger value of strain is called an elastomer. The magnitude of stretching force increases In case of some materials like vulcanized from zero to F during elongation of wire. At a rubber, when the stress applied on a body decreases to zero, the strain does not return to certain stage, let ‘f ’ be the force applied and ‘x’ zero immediately. The strain lags behind the stress. This lagging of strain behind the stress is be the corresponding extension. The force at called elastic hysteresis. Figure 6.8 shows the stress-strain curve for increasing and decreasing this stage is given by Eq. (6.19) as load. It encloses a loop. Area of loop gives f = YAx L 107

For further extension dx in the wire, the work Work done per unit volume done is given by Strain e=ne12rgy(sptreersusn).i(tsvtroaliunm) e Work = (force).(displacement). dW = f dx 1 = 2 ∴ dW = YAx  dx (stress).(strain) --- (6.22) L When the wire gets stretched from x = 0 to = stress As Y strain , x = l, the total work done is given as Stress = Y. (strain) and l stress W ³dW strain = Y 0 ∴ Strain energy per unit volume ?W l YAx dx 1 ³ 0L 2 Y ˜ (strain)2 --- (6.23) ?W YA l xdx Also, strain energy per unit volume ³ L --- (6.24) 0 Thus Eq. (6.22), (6.23) and (6.24) give strain ?W YA ª x2 ºl energy per unit volume in various forms. « » 6.8 Hardness: L ¬ 2 ¼ 0 Hardness is the property of a material ?W YA ª l2  02 º which enables it to resist plastic deformation. L « 2 2 » Hard materials have little ductility and they are ¬ ¼ brittle to some extent. The term hardness also refers to stiffness or resistance to bending, W = YAl 2 scratching abrasion or cutting. It is the 2L property of a material which gives it the ability to resist permanent deformation when a load is W = 1 YAl l applied to it. The greater the hardness, greater 2L is the resistance to deformation. W = 1 Fl The most well-known example of the hard 2 materials is diamond. It is incredibly difficult to scratch a diamond. Metal with very low 1 (load).(extension) --- (6.20) hardness is aluminium. Work done = 2 Hardness of material is different from This work done by stretching force is equal its strength and toughness. to energy gained by the wire. This energy is If a force is applied to a body it produces deformation in it. Higher is the force required strain energy. for deformation, the stronger is the material, i.e., the material has more strength. Strain energy = 1 (load).(extension) --- (6.21) 2 Steel has high strength whereas plasticine Strain energy per unit volume can be obtained clay is not strong because it gets easily deformed even by a small force. by using Eq. (6.20) and various formula of Toughness is the ability of a material to stress, strain and young’s modulus. resist fracturing when a force is applied to it. Plasticine clay is relatively tough as it can be Work done per unit volume stretched and deformed due to applied force without breaking. = work done in streching wire volume of wire. = 1 F .l 2 A.L 1§ F ·§ l · 2 ©¨ A ¹¸ ©¨ L ¸¹ 108

A single material may be hard, strong and In this section we are going to study friction in tough, e.g., solids only. 1) Bulletproof glass is hard and tough but not 6.9.1 Origin of friction: strong. If smooth surfaces are observed under 2) Drill bits must be hard, strong and tough powerful microscope, many irregularities and projections are observed. Friction arises due for their work. to interlocking of these irregularities between 3) Anvils are very tough and strong but they two surfaces in contact. The surfaces can be made extremely smooth by polishing to avoid are not hard. irregularities but it is noticed that in this case 6.9 Friction in Solids: also, friction does not decrease but may increase. Hence the interlocking of irregularities is not Whenever the surface of one body slides the real cause of friction. over another, each body exerts a certain amount of force on the other body. These forces are According to modern theory, cause of tangential to the surfaces. The force on each friction is the force of attraction between body is opposite to the direction of motion molecules of two surfaces in actual contact in between the two bodies. It prevents or opposes addition to the force due to the interlocking the relative motion between the two bodies. It is between the two surfaces. When one body is in a common experience that an object placed on contact with another body, the real microscopic any surface does not move easily when a small area in contact is very small due to irregularities force is applied to it. This is because of certain in contact. Figure 6.9 shows the microscopic force of opposition acting between the surface of view of two polished surfaces in contact. the object and the surface on which it is placed. Even a rolling ball comes to rest after covering Fig. 6.9: Microscopic view of polished a finite distance on playground because of such surfaces in contact. opposing force. Our foot ware is provided with designs at the bottom of its sole so as to Due to small area, pressure at points of produce force of opposition to avoid slipping. contact is very high. Hence there is a strong force It is difficult to walk without such opposing of attraction between the surfaces in contact. force. You know what happens when you try If both the surfaces are of the same material to walk fast on polished flooring at home with the force of attraction is called cohesive force soap water spread on it. There is a possibility of while if the surfaces are of different materials slipping due to lack of force of opposition. To the force of attraction is called adhesive force. initiate any motion between a pair of surfaces, When the surfaces in contact become more and we need a certain minimum force. Also after the more smooth, the actual area of contact goes on motion begins, it is constantly opposed by some increasing. Due to this, the force of attraction natural force. This mechanical force between between the molecules increases and hence the two solid surfaces in contact with each other is friction also increases. Putting some grease or called as frictional force. The property which other lubricant (a different material) between resists the relative motion between two the two surfaces reduces the friction. surfaces in contact is called friction. 6.9.2 Types of friction: 1. Static friction: In some cases it is necessary to avoid friction, because friction causes dissipation of Suppose a wooden block is placed on energy in machines due to which efficiency a horizontal surface as shown in Fig 6.10. A of machines decreases. In such cases friction small horizontal force F is applied to it. The should be reduced by using polished surfaces, lubricants, etc. Relative motion between solids and fluids (i.e. liquids and gases) is also naturally opposed by friction, e.g., a boat on the surface of water experiences opposition to its motion. 109

block does not move with this force as it cannot to the normal reaction. Table 6.4 gives the coefficient of static fiction for some overcome the frictional force between the block materials. 2] The limiting force of friction is independent and horizontal surface. In this case, the force of the apparent area between the surfaces in contact, so long as the normal reaction of static friction is equal to F and balances it. remains the same. 3] The limiting force of friction depends upon The frictional force which balances applied materials in contact and the nature of their surfaces. force when the body is static is called force Table 6.4: Coefficient of static friction forficsttiaotnicprfreivcetniotns ,slFids.inIgn other words, static motion. If we keep increasing F, a stage will come when, for F = Fmax, the object will start moving. Material Coefficient of For F < Fmax, the force of static friction is equal Teflon on Teflon static friction µs to F. For F ≥ Fmax, the kinetic friction comes into play. Static friction opposes impending 0.4 motion i.e. the motion that would take place in absence of frictional force under the applied force. Brass on steel 0.51 Copper on steel 0.53 Aluminium on steel 0.61 Steel on steel 0.74 Fig. 6.10: Static friction. Glass on glass 0.94 Rubber on concrete (dry) 1.0 The force of static friction is self adjusting Example 6.5: The coefficient of static friction between a block of mass 0.25 kg and a horizontal force. When the applied force F is very small, the surface is 0.4. Find the horizontal force applied to it. block remains at rest. Here the force of friction Solution: Given, is also small. When F is increased by a small µs = 0.4 m = 0.25 kg value, the block remains still at rest as the force To find: Force of friction is increased to balance the applied F = µs. N = µs. (mg) F = 0.4 × 0.25 × 9.8 force. If applied force is increased, the friction F = 0.98 N also increases and reaches the maximum value. 2. Kinetic friction : Just before the body starts sliding over another Once the sliding of block on the surface starts, the force of friction decreases. The force body, the value of frictional force is maximum, required to keep the body sliding steadily is thus less than the force required to just start its it is called limiting force of ifsrircetvioenrs,eFd,L .thIef sliding. The force of friction that comes into the direction of applied force play when a body is in steady state of motion over another surface is called force of kinetic direction of static friction is also reversed, i.e., friction. it adjusts its direction also. Friction between two surfaces in contact when one body is actually sliding over the Laws of static friction: other body, is called kinetic friction or dynamic friction. 1] The limiting force of static friction is directly proportional to the normal reaction (N) between the two surfaces in contact. FL ∝ N ... FL = µs N --- (6.25) Where µs is constant of proportionality. It is called as coefficient of static friction. ?µs FL --- (6.26) N The coefficient of static friction is defined as the ratio of limiting force of friction 110

Laws of kinetic friction : friction while the force of kinetic friction is 1. The force of kinetic friction (Fk ) is directly greater than force of rolling friction. As rolling friction is the minimum, ball bearings are proportional to the normal reaction between used to reduce friction in parts of machines to two surfaces in contact. increase its efficiency. ∴ Fk α N Advantages of friction: ∴ Fk = µk N --- (6.27) Friction is necessary in our daily life. Where µk is constant of proportionality. It • We can walk due to friction between is called as coefficient of kinetic friction. ground and feet. ? Pk Fk   --- (6.28) • We can hold object in hand due to static N friction. The coefficient of kinetic friction is defined • Brakes of vehicles work due to friction; as the ratio of force of kinetic friction to the hence we can reduce speed or stop normal reaction between the two surfaces vehicles. in contact. Table 6.5 gives the co-efficient of kinetic friction for some materials. • Climbing on a tree is possible due to friction. 2. Force of kinetic friction is independent of shape and apparent area of the surfaces in Disadvantages of friction contact. • Friction opposes motion. • Friction produces heat in different parts 3. Force of kinetic friction depends upon of machines. It also produces noise. the nature and material of the surfaces in • Automobile engines consume more fuel contact. due to friction. 4. The magnitude of the force of kinetic Methods of reducing friction friction is independent of the relative • Use of lubricants, oil and grease in velocity between the object and the surface different parts of a machine. provided that the relative velocity is neither • Use of ball bearings converts kinetic too large nor too small. friction into rolling friction. Table 6.5: Coefficient of kinetic friction Material Coefficient of kinetic friction µk Can you tell? Rubber on concrete (dry) 0.25 1) It is difficult to run fast on sand. 2) It is easy to roll than pull a barrel Glass on glass 0.40 along a road. Brass on steel 0.40 3) An inflated tyre rolls easily than a Copper on steel 0.44 flat tyre. 4) Friction is a necessary evil. Aluminium on steel 0.47 Internet my friend Steel on steel 0.57 1. https://opentextbc.ca>chapter>friction. Teflon on Teflon 0.80 2. https://www.livescience.com 3. https://www.khanacdemy.org.physics 3 Rolling friction : 4. https://courses.lumenlearning.com> Motion of a body over a surface is said to be elastiscitychapter>elasticity 5. https://www.toper.com>guides>physics rolling motion if the point of contact of the body with the surface keeps changing continuously. Friction between two bodies in contact when one body is rolling over the other, is called rolling friction. For same pair of surfaces, the force of static friction is greater than the force of kinetic 111

ExercisesExercises 1. Choose the correct answer: and the softest material. i) Change in dimensions is known as….. xiii) Define friction. (A) deformation (B) formation xiv) Why force of static friction is known as ‘self-adjusting force’? (C) contraction (D) strain. xv) Name two factors on which the co- ii) The point on stress-strain curve at which efficient of friction depends. strain begins to increase even without increase in stress is called…. 3. Answer in short: (A) elastic point (B) yield point i) Distinguish between elasticity and plasticity. (C) breaking point (D) neck point ii) State any four methods to reduce friction. iii) Strain energy of a stretched wire is 18×10-3 J and strain energy per unit iii) What is rolling friction? How does it volume of the same wire and same cross arise? section is 6×10-3 J/m3. Its volume will be.... iv) Explain how lubricants help in reducing friction? (A) 3cm3 (B) 3 m3 v) State the laws of static friction. (C) 6 m3 (D) 6 cm3 vi) State the laws of kinetic friction. iv) ----- is the property of a material which enables it to resist plastic deformation. vii) State advantages of friction. (A) elasticity (B) plasticity viii) State disadvantages of friction. (C) hardness (D) ductility ix) What do you mean by a brittle substance? Give any two examples. v) The ability of a material to resist fracturing when a force is applied to it, is called…… 4. Long answer type questions: (A) toughness (B) hardness i) Distinguish between Young’s modulus, bulk modulus and modulus of rigidity. (C) elasticity (D) plasticity. ii) Define stress and strain. What are their 2. Answer in one sentence: different types? i) Define elasticity. iii) What is Young’s modulus? Describe an experiment to find out Young’s modulus ii) What do you mean by deformation? of material in the form of a long straight wire. iii) State the SI unit and dimensions of stress. iv) Derive an expression for strain energy per iv) Define strain. unit volume of the material of a wire. v) What is Young’s modulus of a rigid body? v) What is friction? Define coefficient of static friction and coefficient of kinetic vi) Why bridges are unsafe after a very long friction. Give the necessary formula for use? each. vii) How should be a force applied on a body vi) State Hooke’s law. Draw a labeled graph to produce shearing stress? of tensile stress against tensile strain for a metal wire up to the breaking point. In this viii) State the conditions under which Hooke’s graph show the region in which Hooke’s law holds good. law is obeyed. ix) Define Poisson’s ratio. x) What is an elastomer? xi) What do you mean by elastic hysteresis? xii) State the names of the hardest material 112

5. Answer the following vii) A wire of mild steel has initial length 1.5 m and diameter 0.60 mm is extended i) Calculate the coefficient of static friction by 6.3 mm when a certain force is applied for an object of mass 50 kg placed on to it. If Young’s modulus of mild steel horizontal table pulled by attaching a is 2.1 x 1011 N/m2, calculate the force spring balance. The force is increased applied. gradually it is observed that the object just moves when spring balance shows 50N. [Ans: 250 N] [Ans: µs = 0.102] viii) A composite wire is prepared by joining a tungsten wire and steel wire end to end. ii) A block of mass 37 kg rests on a rough Both the wires are of the same length and the same area of cross section. If this horizontal plane having coefficient of composite wire is suspended to a rigid support and a force is applied to its free static friction 0.3. Find out the least end, it gets extended by 3.25mm. Calculate the increase in length of tungsten wire and force required to just move the block steel wire separately. horizontally. [Ans: Fs = 108.8N] iii) A body of mass 37 kg rests on a rough horizontal surface. The minimum [Given: Ysteel = 2 × 1011N/m2, horizontal force required to just start the YTungsten = 3.40 × 108 N/m2] motion is 68.5 N. In order to keep the body moving with constant velocity, a [Ans: extension in tungsten wire = 3.244 mm, force of 43 N is needed. What is the value extension in steel wire = 0.0052 mm] of a) coefficient of static friction? and b) ix) A steel wire having cross sectional area 1.2 mm2 is stretched by a force of 120 N. coefficient of kinetic friction? If a lateral strain of 1.455 mm is produced in the wire, calculate the Poisson’s ratio. [Ans: a) µs = 0.188 b) µk = 0.118] iv) A wire gets stretched by 4mm due to a [Ans: 0.291] certain load. If the same load is applied x) A telephone wire 125m long and 1mm in radius is stretched to a length 125.25m to a wire of same material with half the when a force of 800N is applied. What is the value of Young’s modulus for material length and double the diameter of the of wire? [Ans: 1.27×10 11N/m2] first wire. What will be the change in its length? [Ans: 0.5mm] v) Calculate the work done in stretching a xi) A rubber band originally 30cm long is steel wire of length 2m and cross sectional area 0.0225mm2 when a load of 100 N is stretched to a length of 32cm by certain slowly applied to its free end. [Young’s modulus of steel= 2×1011 N/m2 ] load. What is the strain produced? [Ans: 6.667× 10-2 ] xii) What is the stress in a wire which is 50m [Ans: 2.222J] long and 0.01cm2 in cross section, if the vi) A solid metal sphere of volume 0.31m3 wire bears a load of 100kg? is dropped in an ocean where water pressure is 2×107 N/m2. Calculate change [Ans: 9.8× 108 N/m2] in volume of the sphere if bulk modulus of the metal is 6.1×1010 N/m2 xiii) What is the strain in a cable of original length 50m whose length increases by 2.5cm when a load is lifted? [Ans: 10-4 m3] [Ans: 5× 10-4 ] *** 113

7. Thermal Properties of Matter Can you recall? 1. Temperature of a body determines its 3. Solids, liquids and gases expand on hotness while heat energy is its heat heating. content. 4. Substances change their state from solid 2. Pressure is the force exerted per unit area to liquid or liquid to gas on heating up to normally on the walls of a container by the specific temperature. gas molecules due to collisions. 7.1 Introduction: quantitatively measure it. Scientific precision requires measurement of a physical quantity in In previous lessons, while describing the numerical terms. A thermometer is the device to equilibrium states of a mechanical system measure the temperature. or while studying the motion of bodies, only three fundamental physical quantities namely In this chapter , we will learn properties of length, mass and time were required. All other matter and various phenomena that are related physical quantities in mechanics or related to heat. Phenomena or properties having to do to mechanical properties can be expressed in with temperature changes and heat exchanges terms of these three fundamental quantities. are termed as thermal phenomena or thermal In this chapter, we will discuss properties or properties. You will understand why the phenomena related to heat. These require a direction of wind near a sea shore changes fourth fundamental quantity, the temperature, during day and night, why the metal lid of a as mentioned in Chapter 1. glass bottle comes out easily on heating and why two metal vessels locked together can be The sensation of hot or cold is a matter of separated by providing heat to the outer vessel. daily experience. A mother feels the temperature of her child by touching its forehead. A cook 7.2 Temperature and Heat: throws few drops of water on a frying pan to know if it is hot enough to spread the dosa Heat is energy in transit. When two bodies batter. Although not advisable, in our daily at different temperatures are brought in contact, lives, we feel hotness or coldness of a body by they exchange heat. After some time, the heat touching or we dip our fingers in water to check transfer stops and we say the two bodies are if it is hot enough for taking bath. When we say in thermal equilibrium. The property or the a body or water is hot, we actually mean that its deciding factor to determine the state of thermal temperature is more than our hand. However, equilibrium is the temperature of the two bodies. in this way, we can only compare the hotness Temperature is a physical quantity that defines or coldness of two objects qualitatively. Hot the thermodynamic state of a system. and cold are relative terms. You might recall the example given in your science textbook of You might have experienced that a glass of VIIIth standard. Lukewarm water seems colder ice-cold water when left on a table eventually than hot water but hotter than cold water to our warms up whereas a cup of hot tea on the hands. We ascribe a property ‘temperature’ to same table cools down. It means that when the an object to determine its degree of hotness. temperature of a body, ice-cold water or hot The higher the temperature, the hotter is the tea in the above examples, is different from its body. However, the precise temperature of surrounding medium, heat transfer takes place a body can be known only when we have between the body and the surrounding medium an accurate and easily reproducible way to until the body and the surrounding medium are at the same temperature. We then say that the body and its surroundings have reached 114

a state of thermal equilibrium and there is no of a gas. Hence gases neither have a definite net transfer of heat from one to the other. In volume nor shape. Interatomic spacing in fact, whenever two bodies are in contact, there solids is ~ 10-10 m while the average spacing is a transfer of heat owing to their temperature in liquids is almost twice that in solids. The difference. average inter molecular spacing in gases at normal temperature and pressure (NTP) is Matter in any state - solid, liquid or gas- ~10-9 m. consists of particles (ions, atoms or molecules). In solids, these particles are vibrating about their From the above discussion, we understand fixed equilibrium positions and possess kinetic that heat supplied to the substance increases energy due to motion at the given temperature. the kinetic energy of molecules or atoms of The particles possess potential energy due to the substance. The average kinetic energy the interatomic forces that hold the particles per particle of a substance defines the together at some mean fixed positions. Solids temperature. Temperature measures the degree therefore have definite volume and shape. When of hotness of an object and not the amount of its we heat a solid, we provide energy to the solid. thermal energy. The particles then vibrate with higher energy and we can see that the temperature of the A glass of water, a gas enclosed in a solid increases (except near its melting point). container, a block of copper metal are all Thus the energy supplied to the solid (does examples of a 'system'. We can say that heat not disappear!) becomes the internal energy in in the form of energy is transferred between the form of increased kinetic energy of atoms/ two (or more) systems or a system and its molecules and raises the temperature of the surroundings by virtue of their temperature solid. The temperature is therefore a measure difference. SI unit of heat energy is joule (J) of the average kinetic energy of the atoms/ and that of temperature is kelvin (K) or celcius molecules of the body. The greater the kinetic (°C). The CGS unit of heat energy is erg. energy is, the faster the molecules will move (1J = 107 erg). The other unit of heat energy, and higher will be the temperature of the body. that you have learnt in VIIIth standard, is calorie If we continue heating till the solid starts to melt, (cal) and the relation with J is 1 cal = 4.184 J. the heat supplied is used to weaken the bonds Heat being energy has dimension [L2M1T-2K°] between the constituent particles. The average while dimension of temperature is [L°M°T°K1]. kinetic energy of the constituent particles does not change further. The order of magnitude of 7.3 Measurement of Temperature: the average distance between the molecules of the melt remains almost the same as that of In order to isolate two liquids or gases from solid. Due to weakened bonds liquids do not each other and from the surroundings, we use possess definite shape but have definite volume. containers and partitions made of materials like The mean distance between the particles and wood, plastic, glass wool, etc. An ideal wall or hence the density of liquid is more or less the partition (not available in practice) separating same as that of the solid. On heating further, two systems is one that does not allow any flow the atoms/molecules in liquid gain kinetic or exchange of heat energy from one system to energy and temperature of the liquid increases. the other. Such a perfect thermal insulator is If we continue heating the liquid further, at the called an adiabatic wall and is generally shown boiling point, the constituents can move freely as a thick cross-shaded (slanting lines) region. overcoming the interatomic/molecular forces When we wish to allow exchange of heat energy and the mean distance between the constituents between two systems, we use a partition like a increases so that the particles are farther apart. thin sheet of copper. It is termed as a diathermic wall and is represented as a thin dark region. As per kinetic theory of gases, for an ideal gas, there are no forces between the molecules Let us consider two sections of a container separated by an adiabatic wall. Let them contain two different gases. Let us call them system A 115

and system B. We independently bring systems temperatures just as we select the standard A and B in thermal equilibrium with a system C. of length (metre) to be the distance between Now if we remove the adiabatic wall separating two fixed marks. The fact that substances systems A and B, there will be no transfer of change state from solid to liquid to gas at heat from system A to system B or vice versa. fixed temperatures is used to define reference This indicates that systems A and B are also temperature called fixed point. The two fixed in thermal equilibrium. Overall conclusion of temperatures selected for this purpose are the this activity can be summarized as follows: melting point of ice or freezing point of water If systems A and B are separately in thermal and the boiling point of water. The next step is equilibrium with a system C, then A and B are to sub-divide this standard temperature interval also mutually in thermal equilibrium. When into sub-intervals by utilizing some physical two or more systems/ bodies are in thermal property that changes with temperature and equilibrium, their temperatures are same. This call each sub-interval a degree of temperature. principle is used to measure the temperature of This procedure sets up an empirical scale for a system by using a thermometer. temperature. * The temperature at which pure water Do you know ? freezes at one standard atmospheric If TA = TB and TB = TC, then TA = TC is not pressure is called ice point/ freezing point a mathematical statement, if TX represents the of water. This is also the melting point of temperature of system X. It is the zeroth law ice. * The temperature at which pure water boils of thermodynamics and makes the science of and vaporizes into steam at one standard atmospheric pressure is called steam point/ Thermometry possible. boiling point. This is also the temperature at which steam changes to liquid water. Do you remember that to know the Having decided the fixed point phenomena, temperature of our body, doctor brings the it remains to assign numerical values to these mercury in the thermometer down to indicate fixed points and the number of divisions some low temperature. We are then asked to between them. In 1750, conventions were keep the thermometer in our mouth. We have adopted to assign (i) a temperature at which to wait for some time before the thermometer is pure ice melts at one atmosphere pressure taken out to know the temperature of our body. (the ice point) to be 0º and (ii) a temperature There is transfer of heat energy from our body at which pure water boils at one atmosphere to the thermometer since initially our body is (the steam point) to be 100º so that there are at a higher temperature. When the temperature 100 degrees between the fixed points. This was on the thermometer is same as that of our body, the centigrade scale (centi meaning hundred in thermal equilibrium is said to be attained and Latin). This was redefined as celcius scale after heat transfer stops. the Swedish scientist Anders Celcius (1701- 1744). It is a convention to express temperature As mentioned above, to precisely know as degree celcius (ºC). the thermodynamic state of any system, we need to know its temperature. The device used To measure temperature quantitatively, to measure temperature is a thermometer. generally two different scales of temperature Thermometry is the science of temperature are used. They are describe below. and its measurement. For measurement of temperature, we need to establish a temperature 1) Celsius scale:- On this scale, the ice point is scale and adopt a set of rules for assigning marked as 0 and the steam point is marked numbers (with corresponding units). as 100, both taken at normal atmospheric pressure (105 Pa or N/m2). The interval For the calibration of a thermometer, between these points is divided into 100 a standard temperature interval is selected between two easily reproducible fixed 116

equal parts. Each of these is known as Given TF = 98.4 qF, degree Celsius and is written as ºC. 100 (98.4-32) 180 2) Fahrenheit scale :- On this scale, the ice TC point is market as 32 and the steam point 100 (66.4) 180 is marked as 212, both taken at normal atmospheric pressure. The interval between = 36.89 qC these points is divided into 180 equal A device used to measure temperature, is based on the principle of thermal equilibrium. parts. Each division is known as degree To measure the temperature, we use different measurable properties of materials which Fahrenheit and is written as °F. change with temperature. Some of them are length of a rod, volume of a liquid, electrical A relationship for conversion between resistance of a metal wire, pressure of a gas at constant volume etc. Such changes in physical the two scales may be obtained from a graph properties with temperature are used to design a thermometer. Physical property that is used in of fahrenheit temperature (TF) versus celsius the thermometer for measuring the temperature temperature (TC). The graph is a straight line is called the thermometric property and the (Fig. 7.1) whose equation is material employed for the purpose is termed as the thermometric substance. Temperature TF  32 TC --- (7.1) is measured by exploiting the continuous monotonic variation of the chosen property 180 100 with temperature. A calibration, however, is required to define the temperature scale. Fig. 7.1: A plot of fahrenheit temperature There are different kinds of thermometers each type being more suitable than others for a Ex(aTmF)pvleersu7s.1c:elsAiuvsetreamgeperraotoumre (TC). certain job. In each type, the physical property temperature used to measure the temperature must vary continuously over a wide range of temperature. on a normal day is 27 °C. What is the room It must be accurately measurable with simple apparatus. temperature in °F? Solution: We have An important characteristic of a thermometer is its sensitivity, i.e., a change in the TF  32 = TC thermometric property for a very small change 180 100 in temperature. Two other characteristics are accuracy and reproducibility. Also it is ?TF 180 TC + 32 important that the system attains thermal 100 equilibrium with the thermometer quickly. Given TC = 27 qC, TF 180 u 27 + 32 100 = 48.6 + 32 If the values of a thermometric property are P1 and P2 at the ice point (0 ºC) and steam = 80.6 qF point (100 ºC) respectively and the value of this property is PT at unknown temperature T, then Example 7.2: Normal human body temperature T is given by the following equation in feherenheit is 98.4 °F. What is the body temperature in °C? 100PT  P1  Solution: We have P2  P1  TC TF  32 T --- (7.2) 100 180 ?TC 100 (TF -32) Ideally, there should be no difference 180 117

in temperatures recorded on two different 27 100 R  95.2 , thermometers. This is seen for thermometers 138.6  95.2 based on gases as thermometric substances. In a constant volume gas thermometer, the pressure ?R 27 u 138.6  95.2  95.2 of a fixed volume of gas (measured by the difference in height) is used as the thermometric 100 property. It is an accurate but bulky instrument. 11.72  95.2 106.92 : Liquid-in-glass thermometer depends Normally in research laboratories, on the change in volume of the liquid with a thermocouple is used to measure the temperature. The liquid in a glass bulb expands temperature. A thermocouple is a junction of up a capillary tube when the bulb is heated. The two different metals or alloys e.g., copper and liquid must be easily seen and must expand (or iron joined together. When two such junctions contract) rapidly and by a large amount for a at the two ends of two dissimilar metal rods small change in temperature over a wide range are kept at two different temperatures, an of temperature. Most commonly used liquids electromotive force is generated that can be are mercury and alcohol as they remain in liquid calibrated to measure the temperature. state over a wide range. Mercury freezes at -39 °C and boils at 357 °C; alcohol freezes at -115 °C Thermistor is another device used to and boils at 78 °C. Thermochromic liquids are measure temperature based on the change in ones which change colour with temperature but resistance of a semiconductor materials i.e., the have a limited range around room temperatures. resistance is the thermometric property. You For example, titanium dioxide and zinc oxide will learn more about this device in Chapter are white at room temperature but when heated 14 on Semiconductors. change to yellow. 7.4 Absolute Temperature and Ideal Gas Equation: Example 7.3: The length of a mercury column 7.4.1 Absolute zero and absolute temperature in a mercury-in-glass thermometer is 25 mm at the ice point and 180 mm at the steam point. Experiments carried out with gases at What is the temperature when the length is 60 low densities indicate that while pressure is mm? held constant, the volume of a given quantity of gas is directly proportional to temperature Solution: Here the thermometric property P is (measured in ºC). Similarly, if the volume of the length of the mercury column. Using Eq. a given quantity of gas is held constant, the (7.2), we get pressure of the gas is directly proportional to temperature (measured in ºC). These relations T 10060  25 22.58 qC are graphically shown in Fig. 7.2 (a) and (b). 180  25 Mathematically, this relationship can be written as PV ∝ TC. Thus the volume-temperature Resistance thermometer uses the change or pressure-temperature graphs for a gas are of electrical resistance of a metal wire straight lines. They show that gases expand with temperature. It measures temperature linearly with temperature on a mercury accurately in the range -2000 °C to 1200 °C but thermometer i.e., equal temperature increase it is bulky and is best for steady temperatures. causes equal volume or pressure increase. The similar thermal behavior of all gases suggests Example 7.4: A resistance thermometer has that this relationship of gases can be used to resistance 95.2 Ω at the ice point and 138.6 Ω measure temperature in a constant-volume gas at the steam point. What resistance would be thermometer in terms of pressure of the gas. obtained if the actual temperature is 27 ºC? Although actual experimental Solution: Here the thermometric property P is the resistance. Using Eq. (7.2), if R is the measurements might differ a little from the resistance at 27 ºC, we have ideal linear relationship, the linear relationship 118

holds over a wide temperature range. temperature is not possible in practice. It may be noted that the point of zero pressure or zero Fig. 7.2 (a): Graph of volume versus volume does not depend on any specific gas. temperature (in °C) at constant pressure. The two fixed point scale, described in Fig. 7.2 (b): Graph of pressure versus Section 7.3, had a practical shortcoming for temperature (in °C) at constant volume. calibrating the scale. It was difficult to precisely control the pressure and identify the fixed It may be noted that the lines do not points, especially for the boiling point as the pass through the origin i.e., have non-zero boiling temperature is very sensitive to changes intercept along the y-axis. The straight lines in pressure. Hence, a one fixed point scale have different slopes for different gases. If we was adopted in 1954 to define a temperature assume that the gases do not liquefy even if scale. This scale is called the absolute scale or we lower the temperature, we can extend the thermodynamic scale. It is named as the kelvin straight lines backwards for low temperatures. scale after Lord Kelvin (1824-1907). Is it possible to reach a temperature where the gases stop exerting any pressure i.e., pressure is It is possible for all the three phases - solid, zero? In a constant pressure thermometer, as the liquid and gas/vapour of a material - to coexist temperature is lowered, the volume decreases. in equilibrium. This is known as the triple point. Suppose the gas does not liquefy even at very To know the triple point one has to see that three low temperature, at what temperature, will phases coexist in equilibrium and no one phase its volume become zero? Practically it is not is dominating. This occurs for each substance at possible to keep a material in gaseous state for a single unique combination of temperature and very low temperature and without exerting any pressure. Thus if three phases of water - solid pressure. If we extrapolate the graph of pressure ice, liquid water and water vapour- coexist, P versus temperature TC (in ºC), the temperature the pressure and temperature are automatically at which the pressure of a gas would be zero fixed. This is termed as the triple point of water is -273.15 ºC. It is seen that all the lines for and is a single fixed point to define a temperature different gases cut the temperature axis at the scale. same point at i.e., -273.15 ºC. This point is termed as the absolute zero of temperature. The absolute scale of temperature, is so It is not possible to attain a temperature lower termed since it is based on the properties of an than this value. Even to achieve absolute zero ideal gas and does not depend on the property of any particular substance. The zero of this scale is ideally the lowest temperature possible although it has not been achieved in practice. It is termed as Kelvin scale with its zero at -273.15 °C and temperature intervals same as that on the celsius scale. It is written as K (without °). Internationally, triple point of water has been assigned as 273.16 K at pressure equal to 6.11 × 102 Pa or 6.11 × 10-3 atmosphere, as the standard fixed point for calibration of thermometers. Size of one kelvin is thus 1/273.16 of the difference between the absolute zero and triple point of water. It is same as one celcius. On celcius scale, the triple point of water is 0.01 ºC and not zero. Three identical thermometers, marked in kelvin, celcius and fahrenheit, placed in a fixed temperature bath, each thermometer showing 119

the same rise in the level of mercury for human 7.4.2 Ideal Gas Equation: body temperature, are depicted in Fig. 7.3. The relation between three properties of The relation between the three scales of a gas i.e., pressure, volume and temperature is temperature is as given in Eq. (7.3) . called ideal gas equation. You will learn more about the properties of gases in chemistry. TC TF  32 TK  273.15 --- (7.3) 100 180 100 Using absolute temperatures, the gas laws can be stated as given below. 1) Charles’ law- In Fig. 7.2 (b), the volume- temperature graph passes through the origin if temperatures are measured on the kelvin scale, that is if we take 0 K as the origin. In that case the volume V is directly proportional to the absolute temperature T. Thus V∝T Fig. 7.3: Comparison of the kelvin, or, V = constant --- (7.4) celsius and fahrenheit temperature scales Thus Charles'Tlaw can be stated as, the (Thermometer reading are not to the scale). volume of a fixed mass of gas is directly Example 7.5: Express T = 24.57 K in celsius proportional to its absolute temperature if and fahrenheit. Solution: We have the pressure is kept constant. 2) Pressure (Gay Lussac's) law- From TF  32 TC = TK - 273.15 Fig.7.2, it can be seen that the pressure- 180 100 100 temperature graph is similar to the volume- ?TC = TK -273.15 temperature graph. = 24.57-273.15 Thus P ∝ T = - 248.58qC or, P = constant --- (7.5) T TF  32 TK - 273.15 Pressure law can be stated as the pressure of 180 100 a fixed mass of gas is directly proportional ?TF 180 (TK - 273.15) + 32 to its absolute temperature if the volume is 100 9 (24.57  273.15)  32 kept constant. 5 3) Boyle’s law- For fixed mass of gas at = - 447.44 + 32 constant temperature, pressure is inversely proportional to volume. = - 415.44qF Thus P ∝ 1 Example 7.6: Calculate the temperature which V has the same value on fahernheit scale and PV = constant --- (7.6) kelvin scale. Solution: Let the required temperature be y. Combining above three equations, we get i.e., TF = TK = y then we have PV = constant --- (7.7) y  32 y  273.15 T 180 100 For one mole of a gas, the constant of or, 5y 160 9 y  2458.35 proportionality is written as R or, 4 y 160  2458.35 ∴ PV =R or PV = RT --- (7.8) T ? y 574.59 If given mass of a gas consists of n moles, Thus 574.59 °F and 574.59 K are equivalent temperatures. then Eq. (7.8) can be written as PV= nRT --- (7.9) 120

This relation is called ideal gas equation. If the substance is in the form of a long rod of length l, then for small change ∆T, in The value of constant R is same for all gases. temperature, the fractional change ∆l/l, in length (shown in Fig.7.4), is directly proportional to Therefore, it is known as universal gas constant. ∆T. Its numerical value is 8.31 J K-1 mol-1. Example 7.7: The pressure reading in a thermometer at steam point is 1.367 × 103 Pa. 'l v 'T l What is pressure reading at triple point knowing the linear relationship between temperature and or 'l D 'T --- (7.10) l pressure? Solution: We have Pprtreipsles=ur2e7s3a.1t 6te×m§¨©pTPera¹¸·twurheeoref where α is called the coefficient of linear Ptriple and P are the expansion of solid. Its value depends upon triple point (273.16 K) and T respectively. We nature of the material. Rearranging Eq. (7.10), we get are given that P = 1.367×103 Pa at steam point i.e., at 273.15 + 100 = 373.15 K. D 'l l'T § 1.367u103 · ∴ Ptriple = 273.16 × ¨ ¸ = lT  l0 = 1.000×103 Pa © 373.15 ¹ --- (7.11) l0 (T  T0 ) 7.5 Thermal Expansion: where l0 = length of rod at 0 °C lT = length of rod when heated to T °C When matter is heated, it normally expands T0 = 0 °C is initial temperature and when cooled, it normally contracts. The T = final temperature atoms in a solid vibrate about their mean positions. When heated, they vibrate faster and ∆l =lT - l0 = change in length force each other to move a little farther apart. ∆T =T - T0= rise in temperature This results into expansion. The molecules in Referring to Eq. (7.11), if l0=1 and T- T0=1 °C, a liquid or gas move with certain speed. When then heated, they move faster and force each other to move a little farther apart. This results in α = lT - l0 (numerically). expansion of liquids and gases on heating. The Coefficient of linear expansion of a solid expansion is more in liquids than in solids; gases expand even more. is thus defined as increase in the length per unit original length at 0 °C for one degree A change in the temperature of a body centigrade rise in temperature. causes change in its dimensions. The increase in the dimensions of a body due to an increase The unit of coefficient of linear expansion is in its temperature is called thermal expansion. per degree celcius or per kelvin. The magnitude There are three types of thermal expansion: of α is very small and it varies only a little with 1) Linear expansion, 2) Areal expansion, temperature. For most practical purposes, α 3) Volume expansion. can be assumed to be constant for a particular 7.5.1 Linear Expansion: material. Therefore, it is not necessary that initial temperature be taken as 0 °C. Equation The expansion in length due to thermal (7.11) can be rewritten as energy is called linear expansion. D l2  l1 l1(T2  T1) --- (7.12) Fig. 7.4: Linear expansion ∆l is exaggerated where l1 = initial length at temperature T1 °C for explanation. l2 = final length at temperature T2 °C. Table 7.1 lists average values of coefficient of linear expansion for some materials in the 121

temperature range 0°C to 100 °C. We have Table 7.1: Values of coefficient of linear D = l2  l1 l2  l1 expansion for some common materials. l1(T2  T1) l1T2  l1T1 Materials a (K-1) ? l1T2 l1T1  l2  l1 D Carbon (diamond) 0.1×10-5 Glass 0.85×10-5 T2 1 ª«l1T1  l2  l1 º Iron 1.2×10-5 l1 ¬ D » Steel 1.3×10-5 ¼ Gold 1.4×10-5 Copper 1.7×10-5 1 ¬ª«(4.256 u 27)  4.268  4.256 º Silver 1.9×10-5 4.256 1.2 u105 ¼» 2.5×10-5 Aluminium 6.1×10-5 1 «¬ª114.912  0.012 º Sulphur 6.1×10-5 4.256 1.2 u105 »¼ Mercury 6.9×10-5 Water 8.8×10-5 1 ¬ª114.912  1000º¼ 4.256 Carbon (graphite) 261.96 qC 7.5.2 Areal Expansion: The increase ∆A, in the surface area, on heating is called areal expansion or superficial Example 7.8: The length of a metal rod at 27 °C expansion. is 4 cm. The length increases to 4.02 cm when the metal rod is heated upto 387 °C. Determine 'A v 'T the coefficient of linear expansion of the metal A rod. or 'A E 'T A Solution: Given T1 = 27 °C Fig. 7.5: Areal expansion ∆A is exaggerated for explanation. T2 = 387 °C If a substance is in the form of a plate of l1 = 4 cm = 4×10-2 m l2 = 4.02 cm = 4.02×10-2 m area A, then for small change ∆T in temperature, the fractional change in area, ∆A/A (as shown We have in Fig. 7.5), is directly proportional to ∆T. D l2  l1 'A v 'T l1(T2  T1) A 4.02  4.0u102 or 'A E 'T --- (7.13) A 4 u102 (387  27) where β is called the coefficient of areal expansion of solid. It depends on the material 0.02 u102 4 u102 u 360 of the solid. Rearranging Eq. (7.13), we get 1.39u105 / qC E 'A AT  A0 --- (7.14) Example 7.9: Length of an iron rod at A'T temperature 27 °C is 4.256 m. Find the A0 (T  T0 ) temperature at which the length of the same rod increases to 4.268 m.(α for iron = 1.2×10-5 K-1) where A0= area of plate at 0 °C AT = area of plate when heated to T °C Solution: Given T0 = 0 °C is initial temperature T = final temperature T1 = 27 °C, l1 = 4.256 m, ∆A = AT - A0 = change in area l2 = 4.268m, α = 1.2×10-5 K-1 ∆T =T - T0= rise in temperature. If A0 = 1 m2 and T - T0 = 1 °C, then β = AT - A0 (numerically). 122

Therefore, coefficient of areal expansion of a If the substance is in the form of a cube solid is defined as the increase in the area per unit original area at 0°C for one degree rise of volume V, then for small change ∆T in in temperature. temperature, the fractional change, ∆V/V (as shown in Fig.7.6), in volume is directly The unit of β is per degree celcius or per proportional to ∆T. kelvin. 'V v 'T V As in the case of α, β also does not vary 'V J 'T --- (7.16) much with temperature. Hence, if A1 is the area or V of a metal plate at T1 °C and A2 is the area at higher temperature T2 °C, then where γ is called coefficient of cubical or A2  A1 volume expansion. It depends upon the nature A1(T2  T1) E --- (7.15) of the material. Its unit is per degree celcius or per kelvin. From Eq.(7.16), we can write Example 7.10: A thin aluminium plate has an J 'V VT V0 --- (7.17) area 286 cm2 at 20 °C. Find its area when it is V 'T V0 (T - T0 ) heated to 180 °C. where V0 = volume at 0 °C (β for aluminium = 4.9×10-5 /°C) VT = volume when heated to T °C Solution: Given T0 = 0 °C is initial temperature T1 = 20 °C T = final temperature T2 = 180 °C ∆V = VT - V0= change in volume A1 = 286 cm2 ∆T =T - T0= rise in temperature. β = 4.9×10-5 /°C If V0 = 1 m3, T - T0=1 °C, then We have γ =VT - V0 (numerically). The coefficient of cubical expansion of E A2  A1 a solid is therefore defined as increase in A1(T2  T1) volume per unit original volume at 0°C for one degree rise in the temperature. ∴ A2= A1 [1 + β (T2-T1)] = 286 [1 + 4.9×10-5 (180-20)] If V1 is the volume of a body at T1 °C and V2 is the volume at higher temperature T2 °C, = 286 [1 + 4.9×10-5×160] then = 286 [1 + 784.0×10-5] = 286 [1 + 0.00784] J1 V2 V1 --- (7.18) = 286 [1.00784] V1(T2  T1) ... A2 = 288.24 cm2 7.5.3 Volume expansion γ1 is the coefficient of volume expansion at temperature T1 °C. The increase in volume due to heating is called volume expansion or cubical expansion. Since fluids possess definite volume and take the shape of the container, only change in volume is significant. Equations (7.17) and (7.18) are valid for cubical or volume expansion of fluids. It is to be noted that since fluids are kept 'V v 'T in containers, when one deals with the volume V expansion of fluids, expansion of the container or 'V J 'T is also to be considered. If expansion of fluid V results in a volume greater than the volume Fig. 7.6: Volume expansion ∆V is of the container, the fluid will overflow if the exaggerated for explanation. container is open. If the container is closed, volume expansion of fluid will cause additional 123

pressure on the walls of the container. Can you Example 7.11 : A liquid at 0 °C is poured now tell why the balloon bursts sometimes on in a glass beaker of volume 600 cm3 to fill it its own on a hot day? completely. The beaker is then heated to 90 °C. How much liquid will overflow? Normally solids and liquids expand on heating. Hence their volume increases on (γliquid = 1.75×10-4 /°C, γglass = 2.75×10-5 /°C) heating. Since the mass is constant, it results in Solution: Given a decrease in the density on heating. You have learnt about the anomalous behaviour of water. V1= 600 cm3 Water expands on cooling from 4°C to 0°C. Hence its density decreases on cooling in this T1 = 0 °C temperature range. T2 = 90 °C In Table 7.2 are given typical average We have J V2 V1 values of the coefficient of volume expansion γ for some materials in the temperature range V1(T2  T1) 0°C to 100°C. ... increase is volume = V2 - V1= γ V1 (T2- T1) Table 7.2: Values of coefficient of volume expansion for some common materials. Increase in volume of beaker Materials γ (K-1) = γglass× V1 (T2- T1) = 2.75×10-5×600×(90-0) Invar 2×10-6 = 2.75×10-5×600×90 Glass (ordinary) 2.5×10-5 (3.3-3.9)×10-5 = 148500×10-5 cm3 Steel 3.55×10-5 ... increase in volume of beaker = 1.485 cm3 Iron 4.2×10-5 Gold 5.7×10-5 Increase in volume of liquid Brass 6.9×10-5 Aluminium 18.2×10-5 = γliquid × V1 (T2- T1) Mercury 20.7×10-5 = 1.75×10-4×600×(90-0) Water 58.8×10-5 Paraffin 95.0×10-5 = 1.75×10-4×600×90 Gasoline 110×10-5 Alcohol (ethyl) = 94500×10-4 cm3 ... increase in volume of liquid = 9.45 cm3 γ is also characteristic of the substance but ... volume of liquid which overflows is not strictly a constant. It depends in general on temperature as shown in Fig.7.7. It is seen = (9.45-1.485) cm3 that γ becomes constant only at very high temperatures. = 7.965 cm3 7.5.4 R elation between Coefficients of Expansion: i) Relation between β and α: Consider a square plate of side l0 at 0 °C and at T °C. ... lT lT l0 (1+αT) from Eq. (7.11). = If area of plate at 0 °C is A0, A0 = l02. If area of plate at T °C is AT, AT = lT2 = l02 (1+αT)2 or AT = A0 (1+αT)2 --- (7.19) Also from Eq. (7.14), Fig. 7.7: Coefficient of volume expansion AT = A0 (1+βT) --- (7.20) of copper as a function of temperature. 124

Using Eqs. (7.19) and (7.20), we get Solution: Given A0 (1+αT)2 = A0 (1+βT) T1 = 0 °C T2 = 100 °C or 1+ 2αT +α2T2 =1+βT A1 = 50×8 = 400 cm2 A2 = 401.57 cm2 Since the values of α are very small, the We have term α2T2 is very small and may be neglected. ∴ β = 2α --- (7.21) A2  A1 A1(T2  T1) E 2D Can you tell? (401.57  400) cm2 1. Why the metal wires for electrical 400 cm2 u (100  0) qC transmission lines sag? 1.57 0.3925u104 qC1 2. Why a railway track is not a continuous 400 u100 piece but is made up of segments separated by gaps? ?D 0.1962 u104 qC1 3. How a steel wheel is mounted on an 1.962 u105 qC1 axle to fit exactly? ∴ Coefficient of linear expansion of brass 4. Why lakes freeze first at the surface? is 1.962×10-5 /°C. The result is general because any solid can be regarded as a collection of small squares. Do you know ? ii) Relation between γ and α: * When pressure is held constant, due to change in temperature, the volume of Consider a cube of side l0 at 0 °C a liquid or solid changes very little in comparison to the volume of a gas. and lT at T °C. from Eq. (7.11). ... lT = l0 (1+αT) * The coefficient of volume expansion, γ, is generally an order of magnitude larger for If volume of the cube at 0 °C is V0, V0 = l03. liquids than for solids. If volume of the cube at T °C is VT , * Metals have high values for the coefficient VT = lT3 = l03 (1+αT)3 for linear expansion, α, than non-metals. or VT = V0 (1+αT)3 --- (7.22) * γ changes more with temperature than α and β. Also from Eq. (7.17), * We know that water expands on freezing VT = V0 (1+γT) --- (7.23) from 4 ºC to 0 ºC. Other two substances, Using Eqs. (7.22) and (7.23), we get that expnad on freezing are metals bismuth (Bi) and antimony (Sb). Thus the density V0 (1+αT)3 = V0 (1+γT) of liquid is more than corresponding solid and hence solid Bi or Sb float on their or 1+ 3αT +3α2T 2 + α3T 3 =1+γT liquids like ice floats on water. Since the values of α are very small, the 7.6 Specific Heat Capacity: terms with higher powers of α may be neglected. 7.6.1 Specific Heat Capacity of Solids and ∴ γ = 3α --- (7.24) Liquids Again the result is general because any If 1 kg of water and 1 kg of paraffin are heated in turn for the same time by the same solid can be regarded as a collection of small heater, the temperature rise of paraffin is about twice that of water. Since the heater gives cubes. equal amounts of heat energy to each liquid, it seems that different substances require different Finally, the relation between α, β and γ is D E J --- (7.25) 2 3 Example 7.12: A sheet of brass is 50 cm long and 8 cm broad at 0 °C. If the surface area at 100 °C is 401.57cm2, find the coefficient of linear expansion of brass. 125

amounts of heat to cause the same temperature C 1 'Q --- (7.27) rise of 1°C in the same mass of 1 kg. P 'T If ∆Q stands for the amount of heat absorbed or given out by a substance of mass The SI unit of molar specific heat capacity m when it undergoes a temperature change ∆T, is J/mol °C or J/mol K. Like specific heat, molar then the specific heat capacity of that substance specific heat also depends on the nature of the is given by substance and its temperature. Table 7.3 lists the values of specific heat capacity for some s 'Q common materials. m'T --- (7.26) If m = 1 kg and ∆T = 1°C then s = ∆Q. From Table 7.3, it can be seen that water Thus specific heat capacity is defined as has the highest specific heat capacity compared the amount of heat per unit mass absorbed to other substances. For this reason, water is or given out by the substance to change its used as a coolant in automobile radiators as well temperature by one unit (one degree) 1 °C as for fomentation using hot water bags. or 1K. 7.6.2 Specific Heat Capacity of Gas: Table 7.3: Specific heat capacity of some substances at room temperature and In case of a gas, slight change in atmospheric pressure. temperature is accompanied with considerable changes in both, the volume and the pressure. Substance Specific heat capacity If gas is heated at constant pressure, volume (J kg-1 K-1) changes and therefore some work is done on the surroundings during expansion requiring Steel 120 additional heat. As a result, specific heat at Lead 128 constant pressure (Sp) is greater than specific Gold 129 heat at constant volume (Sv). It is thus necessary Tungsten 134.4 to define two principal specific heat capacities Silver 234 for a gas. Copper 387 Iron 448 Principal specific heat capacities of gases: Carbon 506.5 Glass 837 a) The principal specific heat capacity of a Aluminium 903.0 Kerosene 2118 gas at constant hveoaltumabeso(Srbv)edis defined as Paraffin oil 2130 the quantity of or released Alcohol (ethyl) 2400 Ethanol 2500 for the rise or fall of temperature of unit Water 4186.0 mass of a gas through 1 K (or 1°C) when its volume is kept constant. b) The principal specific heat capacity of a The SI unit of specific heat capacity is J/ gas at constant hpereastsaubreso(rSbp)edisodrerfeinleeadseads kg °C or J/kg K and C.G.S. unit is erg/g °C or the quantity of erg/g K. The specific heat capacity is a property of the substance and weakly depends on its for the rise or fall of temperature of unit temperature. Except for very low temperatures, the specific heat capacity is almost constant for mass of a gas through 1K (1°C) when its all practical purposes. pressure is kept constant. Molar specific heat capacities of gases: If the amount of substance is specified in a) Molar specific heat capacity of a gas at terms of moles µ instead of mass m in kg, then constant volume (C ) is defined as the the specific heat is called molar specific heat (C) quantity of heat absov rbed or released for and is given by the rise or fall of temperature of one mole of the gas through 1K (or 1°C), when its volume is kept constant. 126

b) Molar specific heat capacity of a gas at = 4×10 × 300 J ... Q = 12000 J constant opfrehsesautreab(sCopr)beisd defined as the quantity or released for 7.6.4 Heat Capacity (Thermal Capacity): the rise or fall of temperature of one mole Heat capacity or thermal capacity of a body is the quantity of heat needed to raise or of the gas through 1K (or 1°C), when its lower the temperature of the whole body by 1°C (or 1K). pressure is kept constant. Relation between Principal and Molar ∴ Thermal heat capacity can be written as Specific Heat Capacities: Heat received or given out A relation between principal specific heat = mass × 1 × specific heat capacity capacity and molar specific heat capacity is given by the following expression Heat capacity = Q = m × s --- (7.29) Molar specific heat capacity = Molecular Heat capacity (thermal capacity) is measured weight × principal specific heat capacity. in J/°C. i.e. Cp= M × Sp and Cv= M × Sv Example 7.14: Find thermal capacity for a where M is the molecular weight of the gas. copper block of mass 0.2 kg, if specific heat capacity of copper is 290 J/kg °C. Table 7.4 lists values of molar specific Solution: Given heat capacity for some commonly known gases. m = 0.2 kg Table 7.4: Molar specific heat capacity of s = 290 J/kg °C some gases. Gas CP CV Thermal capacity = m × s = 0.2 kg×290 J/kg °C (J mol-1 K-1) (J mol-1 K-1) =58 J/ °C He 20.8 12.5 H2 28.8 20.4 7.7 Calorimetry: N2 29.1 20.8 O2 29.4 21.1 Calorimetry is an experimental technique CO2 37.0 28.5 for the quantitative measurement of heat exchange. To make such measurement a calorimeter is used. Figure 7.8 shows a simple 7.6.3 Heat Equation: water calorimeter. If a substance has a specific heat capacity It consists of cylindrical vessel made of of 1000 J/kg °C, it means that heat energy of 1000 J raises the temperature of 1 kg of that copper or aluminium and provided with a stirrer substance by 1°C or 6000 J will raise the temperature of 2 kg of the substance by 3 °C. If and a lid. The calorimeter is well-insulated to the temperature of 2 kg mass of the substance falls by 3 °C, the heat given out would also prohibit any transfer of heat into or out of the be 6000 J. In general we can write the heat equation as calorimeter. Heat received or given out (Q) = mass (m) × temperature change (∆t) × specific heat capacity (s). or Q = m × DT × s --- (7.28) Example 7.13: If the temperature of 4 kg Fig. 7.8: Calorimeter. mass of a material of specific heat capacity One important use of calorimeter is to 300 J/ kg °C rises from 20 °C to 30 °C. Find the determine the specific heat of a substance using heat received. Solution: Q = 4 kg × (30-20) °C × 300 J/kg °C 127

the principle of conservation of energy. s1 (m2s2  m3s3 )(T  T2 ) --- (7.31) Here we are dealing with heat energy and the system is isolated from surroundings. Therefore, m1(T1  T ) heat gained is equal to the heat lost. Also, one can find specific heat capacity In the technique known as the “method of mixtures”, a sample 'A' of the substance is of water or any liquid using the following heated to a high temperature which is accurately measured. The sample 'A' is then placed quickly expression, it the specific heat capacity of the in the calorimeter containing water. The contents are stirred constantly until the mixture attains a material of calorimeter and sample is known final common temperature. The heat lost by the sample 'A' will be gained by the water and the s3 m1s1(T1  T )  m2 s2 --- (7.32) calorimeter. The specific heat of the sample 'A' m3(T  T2 ) m3 of the substance can be calculated as under: Let Note - In the experiment, the heat from the solid sample 'A' is given to the liquid and therefore m1 = mass of the sample 'A' the sample should be denser than the liquid, so m2 = mass of the calorimeter and the stirrer that sample does not float on the liquid. m3 = mass of the water in calorimeter s1 = specific heat capacity of the substance of Example 7.15: A sphere of aluminium of 0.06 sample 'A' kg is placed for sufficient time in a vessel containing boiling water so that the sphere is s2 = specific heat capacity of the material of at 100 °C. It is then immediately transferred to calorimeter (and stirrer) 0.12 kg copper calorimeter containing 0.30 kg of water at 25 °C. The temperature of water rises s3 = specific heat capacity of water and attains a steady state at 28 °C. Calculate T1 = initial temperature of the sample 'A' the specific heat capacity of aluminium. T2 = initial temperature of the calorimeter (Specific heat capacity of water, sw = 4.18 × stirrer and water 103J kg-1 K-1, specific heat capacity of copper sCu = 0.387×103 J kg-1 K-1) T = final temperature of the combined system Solution : Given We have the data as follows: Mass of aluminium sphere = m1 = 0.06 kg Mass of copper calorimeter = m2= 0.12 kg Heat lost by the sample 'A' = m1s1 (T1- T) Mass of water in calorimeter Heat gained by the calorimeter and the stirrer = m3= 0.30 kg = m2s2 (T - T2) Specific heat capacity of copper Heat gained by the water = m3s3 (T - T2) = sCu = s2 = 0.387×103 J kg-1 K-1 Assuming no loss of heat to the Specific heat capacity of water surroundings, the heat lost by the sample goes into the calorimeter, stirrer and water. Thus = sw = s3= 4.18×103 J kg-1 K-1 writing heat equation as, Initial temperature of aluminium sphere m1s1(T1- T) = T1 =100°C = m2s 2(T - T2) + m3s3(T - T2) ---(7.30) Initial temperature of calorimeter and water = T2= 25°C Knowing the specific heat capacity of Final temperature of the mixture water (s3 = 4186 J kg-1 K-1) and copper (s2 = 387 J kg-1 K-1) being the material of the calorimeter = T = 28°C and the stirrer, one can calculate specific heat capacity (s1) of material of sample 'A', from Eq. We have (7.30) as 128

s1 (m2s2  m3s3 )(T  T2 ) (ice) to liquid (water). m1(T1  T ) = ª¬0.12 u 387  0.30u 4180º¼  28  25 Temperature T (0C) ............................ (0.06)(100  28) C D 46.44 1254u 3 3901.32 = = 4.32 (0.06) u 72 A = 903.08 J kg-1 K-1 (0,0) B Time (t) (s) ∴ Specific heat capacity of aluminium is Fig. 7.9 : Variation of temperature with time. 903.08 J kg-1 K-1. a) The change of state from solid to liquid is called melting and from liquid to solid is 7.8 Change of State: called solidification. Matter normally exists in three states: solid, b) Both the solid and liquid states of the liquid and gas. A transition from one of these substance co-exist in thermal equilibrium states to another is called a change of state. Two during the change of states from solid to common changes of states are solid to liquid and liquid or vice versa. liquid to gas (and vice versa). These changes can occur when exchange of heat takes place c) The temperature at which the solid and between the substance and its surroundings. the liquid states of the substance are in thermal equilibrium with each other is Activity called the melting point of solid (here ice) or freezing point of liquid (here water). It To understand the process of change of is characteristic of the substance and also state depends on pressure. Take some cubes of ice in a beaker. d) The melting point of a substance at one Note the temperature of ice (0 °C). Start standard atmospheric pressure is called its heating it slowly on a constant heat source. normal melting point. Note the temperature after every minute. Continuously stir the mixture of water and e) At one standard atmospheric pressure, the ice. Observe the change in temperature. freezing point of water and melting point Continue heating even after the whole of of ice is 0 °C or 32°F. The freezing point ice gets converted into water. Observe describes the liquid to solid transition the change in temperature as before till while melting point describes solid-to- vapours start coming out. Plot the graph liquid transition. of temperature (along y-axis) versus time (along x-axis). You will obtain a graph 2) From point B to D: of temperature versus time as shown in The temperature begins to rise from point Fig. 7.9. B to point C, i.e., after the whole of ice gets Analysis of observations : converted into water and we continue further heating. We see that temperature begins to 1) From point A to B: rise. The temperature keeps on rising till it reaches point C i.e., nearly 100 °C. Then it There is no change in temperature from again becomes steady. It is observed that the point A to point B, this means the temperature temperature remains constant until the entire of the ice bath does not change even though amount of the liquid is converted into vapour. heat is being continuously supplied. That is the The heat supplied is now being utilized to temperature remains constant until the entire change water from liquid state to vapour or amount of the ice melts. The heat supplied is gaseous state. being utilised in changing the state from solid a) The change of state from liquid to vapour 129

is called vapourisation while that from is more. Hence the rate of losing such molecules vapour to liquid is called condensation. to atmosphere will be larger. Thus, higher is the b) Both the liquid and vapour states of the temperature of the liquid, greater is the rate of substance coexist in thermal equilibrium evaporation. Since faster molecules are lost, the during the change of state from liquid to average kinetic energy of the liquid is reduced vapour. and hence the temperature of the liquid is c) The temperature at which the liquid and lowered. Hence the phenomenon of evaporation the vapour states of the substance coexist gives a cooling effect to the remaining liquid. is called the boiling point of liquid, Since evaporation takes place from the surface here water or steam point. This is also of a liquid, the rate of evaporation is more if the the temperature at which water vapour area exposed is more and if the temperature of condenses to form water. the liquid is higher. d) The boiling point of a substance at one standard atmospheric pressure is called its You might have seen that if your mother normal boiling point. wants her sari/clothes to dry faster, she does not fold them. More is the area exposed, faster Can you tell? is the drying because the water gets evaporated faster. The presence of wind or strong breeze 1. What after point D in graph ? Can steam and content of water vapour in the atmosphere be hotter than 100 °C ? are two other important factors determining the drying of clothes but we do not refer to them here. 2. Why steam at 100 °C causes more harm to our skin than water at 100 °C? Before giving an injection to a patient, normally a spirit swab is used to disinfect the Do you know ? region. We feel a cooling effect on our skin due to evaporation of the spirit as explained before. You must have seen that water spilled on floor dries up after some time. Where does Activity the water disappear? It is converted into water vapour and mixes with air. We say that water Activity to understand the dependence has evaporated. You also know that water can be of boiling point on pressure converted into water vapour if you heat the water till its boiling point. What is then the difference Take a round bottom flask, more than between boiling and evaporation? half filled with water. Keep it over a burner and fix a thermometer and steam outlet Both evaporation and boiling involve through the cork of the flask change of state, evaporation can occur at any as shown in figure. As water temperature but boiling takes place at a fixed in the flask gets heated, note temperature for a given pressure, unique for each that first the air, which was liquid. Evaporation takes place from the surface dissolved in the water comes of liquid while boiling occurs in the whole liquid. out as small bubbles. Later bubbles of steam form at the As you know, molecules in a liquid are bottom but as they rise to the moving about randomly. The average kinetic cooler water near the top, they energy of the molecules decides the temperature condense and disappear. Finally, as the of the liquid. However, all molecules do not temperature of the entire mass of the water move with the same speed. One with higher reaches 100 °C, bubbles of steam reach the kinetic energy may escape from the surface surface and boiling is said to occur. The region by overcoming the interatomic forces. steam in the flask may not be visible but as This process can take place at any temperature. it comes out of the flask, it condenses as tiny This is evaporation. If the temperature of the droplets of water giving a foggy appearance. liquid is higher, more is the average kinetic energy. Since the number of molecules is fixed, it If now the steam outlet is closed for a implies that the number of fast moving molecules few seconds to increase the pressure in the 130

flask, you will notice that boiling stops. arrangement of carbon atoms is different in the More heat would be required to raise the two cases. Figure 7.10 shows the phase diagram temperature (depending on the increase in of water and CO2. Let us try to understand the pressure) before boiling starts again. Thus diagram. boiling point increases with increase in pressure. Fig. 7.10 (a): Phase diagram of water (not to scale). Let us now remove the burner. Allow water to cool to about 80°C. Remove the thermometers and steam outlet. Close the flask with a air tight cork. Keep the flask turned upside down on a stand. Pour ice- cold water on the flask. Water vapours in the flask condense reducing the pressure on the water surface inside the flask. Water begins to boil again, now at a lower temperature. Thus boiling point decreases with decrease in pressure and increases with increase in pressure. Can you tell? 1. Why cooking is difficult at high altitude? 2. Why cooking is faster in pressure cooker? 7.8.1 Sublimation: Fig. 7.10 (b): Phase diagram of CO2 (not to scale). Have you seen what happens when camphor is burnt? All substances do not pass i) Vapourisation curve l - v: The curve labelled through the three states: solid-liquid-gas. l - v represents those points where the liquid and There are certain substances which normally vapour phases are in equilibrium. Thus it is a pass from the solid to the vapour state directly graph of boiling point versus pressure. Note and vice versa. The change from solid state to that the curves correctly show that at a pressure vapour state without passing through the liquid of 1 atmosphere, the boiling points of water is state is called sublimation and the substance is 100°C and that the boiling point is lowered for said to sublime. Dry ice (solid CO2) and iodine a decreased pressure. sublime. During the sublimation process, both the solid and vapour states of a substance ii) Fusion curve l - s: The curve l - s represents coexist in thermal equilibrium. Most substances the points where the solid and liquid phases sublime at very low pressures. coexist in equilibrium. Thus it is a graph of the freezing point versus pressure. At one standard 7.8.2 Phase Diagram: atmosphere pressure, the freezing point of water is 0 °C as shown in Fig. 7.10 (a). Also notice A pressure - temperature (PT) diagram that at a pressure of one standard atmosphere often called a phase diagram, is particularly water is in the liquid phase if the temperature is convenient for comparing different phases of a between 0 °C and 100 °C but is in the solid or substance. vapour phase if the temperature is below 0 °C or above 100 °C. Note that l - s curve for water A phase is a homogeneous composition of slopes upward to the left i.e., fusion curve of a material. A substance can exist in different water has a slightly negative slope. This is true phases in solid state, e.g., you are familiar with two phases of carbon- graphite and diamond. Both are solids but the regular geometric 131

only of substances that expand upon freezing. Gas and vapour can thus be defined as- However, for most materials like CO2, the 1) A substance which is in the gaseous phase l - s curve slopes upwards to the right i.e., fusion and is above its critical temperature is called curve has a positive slope. The melting point of a gas. CO2 is -56 °C at higher pressure of 5.11 atm. 2) A substance which is in the gaseous phase iii) Sublimation curve s - v: The curve and is below its critical temperature is called labelled s - v is the sublimation point versus a vapour. pressure curve. Water sublimates at pressure less than 0.0060 atmosphere, while carbon Vapour can be liquefied simply by dioxide, which in the solid state is called dry increasing the pressure, while gas cannot. ice, sublimates even at atmospheric pressure at Vapour also exerts pressure like a gas. temperature as low as -78 °C. 7.8.4 Latent Heat: iv) Triple point: The temperature and pressure at which the fusion curve, the vapourisation Whenever there is a change in the state of curve and the sublimation curve meet and all the a substance, heat is either absorbed or given out three phases of a substance coexist is called the but there is no change in the temperature of the triple point of the substance. That is, the triple substance. point of water is that point where water in solid, liquid and gaseous states coexist in equilibrium Latent heat of a substance is the quantity and this occurs only at a unique temperature and of heat required to change the state of unit pressure. The triple point of water is 273.16 K mass of the substance without changing its and 6.11×10-3 Pa and that of CO2 is -56.6 °C and temperature. 5.1×10-5 Pa. Thus if mass m of a substance undergoes 7.8.3 Gas and Vapour: a change from one state to the other then the quantity of heat absorbed or released is given by Q = mL --- (7.33) The terms gas and vapour are sometimes where L is known as latent heat and is used quite randomly. Therefore, it is important characteristic of the substance. Its SI unit is J to understand the difference between the two. kg-1. The value of L depends on the pressure and A gas cannot be liquefied by pressure alone, is usually quoted at one standard atmospheric no matter how high the pressure is. In order pressure. to liquefy a gas, it must be cooled to a certain temperature. This temperature is called critical The quantity of heat required to convert temperature. unit mass of a substance from its solid state to Critical temperatures for some common gases and water vapour are given in Table 7.5. Thus, the liquid state, at its melting point, without nitrogen must be cooled below -147 °C to liquefy it by pressure. any change in its temperature is called its latent heat of fusion (Lf ). The quantity of heat required to convert unit mass of a substance from its liquid state Table 7.5: Critical Temperatures of some to vapour state, at its boiling point without common gases and water vapour. any change in its temperature is called its Gas Critical Temperature latent heat of vteampopuerraiztuarteionve(rLsuv )s. heat energy Air (°C) (K) A plot of -190 83 for a given quantity of water is shown is Fig. 7.11. N2 -147 126 From Fig. 7.11, we see that when heat is O2 -118 155 CO2 31.1 241.9 added (or removed) during a change of state, Water vapour 374 647 the temperature remains constant. Also the slopes of the phase lines are not all the same, which indicates that specific heats of the various states are not equal. For water the latent heat 132

of fusion and 2v2a.6p×or1i0sa5Jtikogn-1arreespLef c=tiv3e.3ly3.×T1h0a5Jt iksg-31.3a3nÍd L10v 5=J of heat is needed to melt 1kg of ice at 0 °C and 22.6×105J of heat is needed to convert 1 kg of water to steam at 100 °C. Hence, steam at 100°C carries 22.6×105J kg-1 more heat than water at 100 °C. This is why burns from steam are usually more serious than those from boiling water. Melting points, boiling point and Fig. 7.11: Temperature versus heat for water at one standard atmospheric pressure latent heats for various substances are given in (not to scale). Table 7.6. Table 7.6 : Temperature of change of state and latent heats for various substances at one standard atmosphere pressure. Substance Melting point Lf Boiling point Lv (°C) (×105 Jkg-1) (°C) (×105 Jkg-1) Gold Lead 1063 0.645 2660 15.8 Water 328 0.25 1744 8.67 Ethyl alcohol 3.33 100 22.6 Mercury 0 1.0 78 8.5 Nitrogen -114 0.12 357 2.7 Oxygen -39 0.26 -196 2.0 -210 0.14 -183 2.1 -219 Example 7.16: When 0.1 kg of ice at 0 °C = mice s (T - T) is mixed with 0.32 kg of water at 35 °C in a ice  container. The resulting temperature of the mixture is 7.8 °C. Calculate the heat of fusion = 0.1 kg×4186 J×(7.8 - 0) C° of ice (swater = 4186 J kg-1 K-1). Solution: Given = 3265.08 J Head lost = Heat gained 36434.944 0.1 Lf  3265.08 mice = 0.1 kg Lf 36434.944  3265.08 = 3316.9864 0.1 mwater = 0.32 kg = 3.31698u105 J kg-1 Tice = 0 °C Twater = 35 °C Do you know ? TF = 7.8 °C The latent heat of vapourization is much larger than the latent heat of fusion. swater = 4186 kg K-1 The energy required to completely separate the molecules or atoms is greater than the Heat lost by water energy needed to break the rigidity (rigid bonds between the molecules or atoms) in = mwater swater (TF- Twater) solids. Also when the liquid is converted = 0.32 kg × 4186 J × (7.8 - 35) °C into vapour, it expands. Work has to be done against the surrounding atmosphere to allow = - 36434.944 J (here negative sign this expansion. indicates loss of heat energy) Heat required to melt ice = mice Lf = 0.1×Lf Heat required to raise temperature of water (from ice) to final temperature 133

7.9 Heat Transfer: the rod. This method of heat transfer is called conduction. Heat may be transferred from one point of body to another in three different ways- by Those solid substances which conduct heat conduction, convection and radiation. Heat easily are called good conductors of heat e.g. transferes through solids by conduction. In this silver, copper, aluminium, brass etc. All metals process, heat is passed on from one molecule are good conductors of heat. Those substances to other molecule but the molecules do not which do not conduct heat easily are called bad leave their mean positions. Liquids and gases conductors of heat e.g. wood, cloth, air, paper, are heated by convection. In this process, there etc. In general, good conductors of heat are is a bodily movement of the heated molecules. also good conductors of electricity. Similarly In order to transfer heat by conduction and bad conductors of heat are bad conductors of convection a material medium is required. electricity also. However transfer of heat by radiation does not need any medium. Radiation of heat energy 7.9.1.1 Thermal Conductivity: takes place by electromagnetic (EM) waves that travel with a speed of 3×108 ms-1 in the space/ Thermal conductivity of a solid is a vacuum . The energy from the Sun comes to us measure of the ability of the solid to conduct by radiation. It may be noted that conduction is heat through it. Thus good conductors of heat a slow process of heat transfer while convection have higher thermal conductivity than bad is a rapid process. However radiation is the conductors. fastest process because the transfer of heat takes place at the speed of light. Suppose that one end of a metal rod is 7.9.1 Conduction: heated (see Fig 7.12 (a)). The heat flows by Conduction is the process by which heat conduction from hot end to the cold end. As a flows from the hot end to the cold end of a solid body without any net bodily movement of the result the temperature of every section of the particles of the body. rod starts increasing. Under this condition, the Heat passes through solids by conduction only. When one end of a metal rod is placed in rod is said to be in a variable temperature state. a flame while the other end is held in hand, the end held in hand slowly gets hotter, although After some time the temperature at each section it itself is not in direct contact with flame. We say that heat has been conducted from the hot of the rod becomes steady i.e. does not change. end to the cold end. When one end of the rod is heated, the molecules there vibrate faster. As Note that temperature of each cross-section of they collide with their slow moving neighbours, they transfer some of their energy by collision the rod is constant but not the same. This is to these molecules which in turn transfer energy to their neighbouring molecules still farther called steady state condition. Under steady state down the length of the rod. Thus the energy of thermal motion is transferred by molecular condition, the temperature at points within the collisions down the rod. The transfer of heat continues till the two ends of the rod are at the rod decreases uniformly with distance from the same temperature in principle but this will take infinite time. Normally various sections of the hot end to the cold end. The fall of temperature rod will attain a temperature which remains constant but not same through out the length of with distance between the ends of the rod in the direction of flow of heat, is called temperature gradient. ?Temperature gradient T1  T2 x where T1= temperature of hot end T2= temperature of cold end x = length of the rod 7.9.1.2 Coefficient of Thermal Conductivity: Consider a cube of each side x and each face of cross-sectional area A. Suppose its opposite faces are maintained at temperatures T1 and T2 (T1 > T2) as shown in Fig. 7.12 (b). Experiments show that under steady state condition, the 134

quantity of heat ‘Q’ that flows from the hot face SI unit of coefficient of thermal to the cold face is conductivity k is J s-1 m-1 °C-1 or J s-1 m-1 K-1 and its dimensions are [L1 M1 T-3 K-1]. i) directly proportional to the cross-sectional area A of the face. i.e., Q ∝ A From Eq. (7.34), we also have ii) directly proportional to the temperature Q kA(T1  T2 ) --- (7.36) difference between the two faces i.e., Q ∝ (T1- T2) t x iii) directly proportional to time t (in seconds) The quantity Q/t, denoted by Pcond , is the for which heat flows i.e. Q ∝ t time rate of heat flow (i.e. heat flow per second) from the hotter face to the colder face, at right iv) inversely proportional to the perpendicular angles to the faces. Its SI unit is watt (W). SI distance x between hot and cold faces i.e., unit of k can therefore be written as W m-1 °C-1 Q ∝ 1/x or W m-1 K-1. Combining the above four factors, we Using calculus, Eq. (7.36) may be written have the quantity of heat as Q v A(T1  T2 )t dQ kA dT x dt dx ?Q kA(T1  T2 )t --- (7.34) where dT is the temperature gradient. x dx where k is a constant of proportionality and is The negative sign indicates that heat flow called coefficient of thermal conductivity. Its is in the direction of decreasing temperature. value depends upon the nature of the tm=a1te sriaanl.d=I(fnuAmer1i=cma2llayn).d ddTx 1, then dQ = k If A = l m2, T1-T2= 1 °C (or 1 K), dt x = 1 m, then from Eq. (7.34), Q = k. Hence the coefficient of thermal conductivity of a material may also be Fig 7.12 (a): defined as the rate of flow of heat per unit Section of a area per unit temperature gradient when the metal bar in the heat flow is at right angles to the faces of a steady state. thin parallel-sided slab of material. The coefficients of thermal conductivity of some materials are given in Table 7.7. Fig 7.12 (b): Section 7.9.1.3 Thermal Resistance (RT): of a cube in the Conduction rate Pcond is the amount of steady state. energy transferred per unit time through a slab Thus the coefficient of thermal of area A and thickness x, the two sides of the slab being at temperatures T1 and T2(T1 >T2), and conductivity of a material is defined as the is given by Eq. (7.36) quantity of heat that flows in one second Pcond Q kA T1 T2 --- (7.37) t x between the opposite faces of a cube of side 1 m, the faces being kept at a temperature As discussed earlier, k depends on the material of the slab. A material that readily difference of 1°C (or 1 K). transfers heat energy by conduction is a good thermal conductor and has high value of k. From Eq. (7.34), we have k Qx A(T1  T2 )t --- (7.35) 135

Table 7.7: Coefficient of thermal or °C s/J and its dimensional formula is conductivity (k). [M-1 L-2 T3 K1]. Substance Coefficient of thermal The lower the thermal conductivity k, conductivity (J s-1 m-1 K-1) the higher is the thermal resistance. RT A material with high value is a poor thermal Silver 406 conductor and is aRT good thermal insulator. Copper 385 Aluminium 205 Thermal resistivity ρ is the reciprocal of Steel 50.2 T Insulating brick 0.15 Glass 0.8 thermal conductivity k and is characteristic of a Brick and concrete 0.8 Water 0.8 material while thermal resistance is that of slab Wood 0.04-0.12 Air at 0 °C 0.024 (or of rod) and depends on the material and on the thickness of slab (or length of rod). In western countries , where the temperature Example 7.17: What is the rate of energy loss in watt per square metre through a glass falls below 0 ºC in winter season, insulating the window 5 mm thick if outside temperature is -20 ºC and inside temperature is 25 ºC? house from the surroundings is very important. (kglass = 1 W/m K) Solution : Given In our country, if we wish to carry cold drinks kglass = 1 W/m K with us for picnic or wish to bring ice-cream T1 = 25 ºC T2 = -20 ºC from the shop to our house, we need to keep x = 5 mm = 5 × 10-3 m them in containers (made up of say thermocol) that are poor thermal conductors. Hence the concept of threesrmistaalncree,sisistanincetrodRuTc, eds.imTilhaer ∴ T1 -T2 = 25 – (-20) ºC = 45 K to electrical Q T1  T2 opposition of a body, to the flow of heat through We have Pcond t kA x . it, is called thermal resistance. The greater the ∴The rate of energy loss per square metre is thermal conductivity of a material, the smaller is its thermal resistance and vice versa. Thus Pcond k T1 T2 bad thermal conductors are those which have A x high thermal resistance. = 1W m-1 K-1 × 45 K / (5 × 10-3 m) = 9 × 103 W/m2 From Eq. (7.37) 7.9.1.4Applications of Thermal Conductivity: i) Cooking utensils are made of metals (T1  T2 ) x --- (7.38) but are provided with handles of bad Pcond kA conductors. Since metals are good conductors of We know that when a current flows through a heat, heat can be easily conducted through the base of the utensils. The handles of conductor, the ratio V/I is called the electrical utensils are made of bad conductors of heat (e.g., wood, ebonite etc.) so that they resistance of the conductor where V is the can not conduct heat from the utensils to our hands. electrical potential difference between the ends ii) Thick walls are used in the construction of cold storage rooms. Brick is a bad of the conductor and I is the current or rate conductor of heat so that it reduces the flow of heat from the surroundings to of flow of charge. In Eq. (7.38), (T1-T2) is the the rooms. Still better heat insulation temperature difference between the ends of the is obtained by using hollow bricks. Air conductor and Pcond is the rate of flow of heat. Therefore in analogy with electrical resistance, (T1-T2)/ Pcond is called thermal resistance RT of the material i.e., Thermal resistance RT = x kA The SI unit of thermal resistance is °C s/ kcal 136

being a poorer conductor than a brick, it Solution : Given further avoids the conduction of heat from A = 1000 cm2 = 1000 ×10-4 m2 outside. k = 0.022 cal/ s cm °C = 0.022 ×102 cal/m °C iii) To prevent ice from melting it is wrapped x = 4 mm = 0.4 ×10-2 m in a gunny bag. A gunny bag is a poor T1 = 27 °C, T2 = -5 °C conductor of heat and reduces the heat flow from outside to ice. Moreover, the From Eq. (7.34), we have air filled in the interspaces of a gunny bag, being very bad conductor of heat, Q kA(T1  T2 )t further avoids the conduction of heat from x outside. Low thermal conductivity can also be a ? Q kA(T1  T2 ) disadvantage. When hot water is poured in t x a glass beaker the inner surface of the glass expands on heating. Since glass is a bad 0.022u102 u1000u104 u (27  (5)) conductor of heat, the heat from inside does not reach the outside surface so quickly. Hence the 0.4 u102 outer surface does not expand thereby causing a crack in the glass. 1.76 u103 cal / s 1.76 kcal / s Example 7.18: The temperature difference 7.9.2 Convection: between two sides of an iron plate, 2 cm thick, is 10 °C. Heat is transmitted through the plate at We have seen that heat is transmitted the rate of 600 kcal per minute per square metre through solids by conduction wherein energy at steady state. Find the thermal conductivity is transferred from one molecule to another but of iron. the molecules themselves vibrating with larger Solution: Given amplitude do not leave their mean positions. But in convection, heat is transmitted from one Q 600 kcal / min m2 600 10 kcal / s m2 point to another by the actual bodily movement At 60 of the heated (energised) molecules within the fluid. x 2 cm 2 u102 m In liquids and gases heat is transmitted by T1  T2 10 qC convection because their molecules are quite From Eq.(7.34), we have free to move about. The mechanism of heat transfer by convection in liquids and gases is Q kA(T1  T2 )t described below. x Consider water being heated in a vessel ?k Qx from below. The water at the bottom of the At T1  T2 vessel is heated first and consequently its density decreases i.e., water molecules at the 10 kcal / s m2 u 2u102 m bottom are separated farther apart. These hot molecules have high kinetic energy and rise 10qC upward to cold region while the molecules 0.02 kcal / m s qC from cold region come down to take their place. Thus each molecule at the bottom gets heated Example 7.19: Calculate the rate of loss of and rises then cools and descends. This action heat through a glass window of area 1000 sets up the flow of water molecules called cm2 and thickness of 4 mm, when temperature convection currents. The convection currents inside is 27 °C and outside is -5 °C. Coefficient transfer heat to the entire mass of water. Note of thermal conductivity of glass is 0.022 cal/ s that transfer of heat is by the bodily/ physical cm °C. movement of the water molecules. 137

Always remember: circulation of fresh air. This is called forced convection. Example in section 7.9.2.1 The process by which heat is transmitted are of forced convection, namely, heat through a substance from one point to convector, air conditioner, heat radiators another due to the actual bodily movement in IC engine etc. of the heated particles of the substance is called convection. 7.9.3 Radiation: 7.9.2.1 Applications of Convection: The transfer of heat energy from one place to another via emission of EM energy i) Heating and cooling of rooms (in a straight line with the speed of light) The mechanism of heating a room by without heating the intervening medium is called radiation. a heat convector or heater is entirely based on convection. The air molecules For transfer of heat by radiation, molecules in immediate contact with the heater are are not needed i.e. medium is not required. The heated up. These air molecules acquire fact that Earth receives large quantities of heat sufficient energy and rise upward. The cool form the Sun shows that heat can pass through air at the top being denser moves down to empty space (i.e., vacuum) between the Sun and take their place. This cool air in turn gets the atmosphere that surrounds the Earth . In heated and moves upward. In this way, fact, transfer of heat by radiation has the same convection currents are set up in the room properties as light (or EM wave). which transfer heat to different parts of the room. The same principle but in opposite A natural question arises as to how heat direction is used to cool a room by an air- transfer occurs is the absence of a medium conditioner. (i.e., molecules). All objects possess thermal energy due to their temperature T(T > 0 K). ii) Cooling of transformers The rapidly moving molecules of a hot body Due to current flowing in the windings of emit EM waves travelling with the velocity of light. These are called thermal radiations. the transformer, enormous heat is produced. These carry energy with them and transfer it Therefore, transformer is always kept in to the low-speed molecules of a cold body on a tank containing oil. The oil in contact which they fall. This results in an increase in with transformer body heats up, creating the molecular motion of the cold body and convection currents. The warm oil comes its temperature rises. Thus transfer of heat by in contact with the cooler tank, gives heat radiation is a two-fold process- the conversion to it and descends to the bottom. It again of thermal energy into waves and reconversion warms up to rise upward. This process is of waves into thermal energy by the body on repeated again and again. The heat of the which they fall. We will learn about EM waves transformer body is thus carried away by in Chapter 13. convection to the cooler tank. The cooler tank, in turn loses its heat by convection to 7.10 Newton’s Laws of Cooling: the surrounding air. If hot water in a vessel is kept on table, 7.9.2.2 Free and Forced Convection: it begins to cool gradually. To study how a i) When a hot body is in contact with air given body can cool on exchanging heat with its surroundings, following experiment is under ordinary conditions, like air around performed. a firewood, the air removes heat from the body by a process called free or natural A calorimeter is filled up to two third of convection. Land and sea breezes are its capacity with boiling water and is covered. also formed as a result of free convection A thermometer is fixed through a hole in the currents in air. lid and its position is adjusted so that the bulb of the thermometer is fully immersed in water. ii) The convection process can be accelerated by employing a fan to create a rapid 138

The calorimeter vessel is kept in a constant According to Newton’s law of cooling the temperature enclosure or just in open air since room temperature will not change much rate of loss of heat dT/dt of the body is directly during experiment. The temperature on the thermometer is noted at one minute interval proportional to the difference of temperature until the temperature of water decreases by about 25 °C. A graph of temperature T (along (T -T0) of the body and the surroundings y-axis) is plotted against time t (along x-axis). provided the difference in temperatures is small. This graph is called cooling curve (Fig 7.13 (a)). From this graph you can infer how the cooling Mathematically this may be expressed as of hot water depends on the difference of its temperature from that of its surroundings. You dT v (T -T0 ) will also notice that initially the rate of cooling dt is higher and it decreases as the temperature of the water falls. A tangent is drawn to the ? dT C (T -T0 ) --- (7.39) curve at suitable points on the curve. The slope dt of each tangent (dT/dt) gives the rate of fall of temperature at that temperature. Taking (0,0) where C is constant of proportionality. as the origin, if a graph of dT/dt is plotted against corresponding temperature difference Example 7.20: A metal sphere cools at the rate (T-T0), the curve is a straight line as shown in of 1.6 °C/min when its temperature is 70°C. At Fig 7.13 (b). what rate will it cool when its temperature is 40°C. The temperature of surroundings is 30°C. Fig 7.13 (a): Temperature versus Solution: Given T1 = 70° C time graph. lim 'T T2 = 40° C 'to0 't T0 = 30° C gives the slope of the § dT · tangent drawn to the ©¨ dt ¸¹1 1.6 qC / min curve at point A and indicates the rate of According to Newton’s law of cooling, if fall of temperature. C is the constant of proportionality § dT · C (T1 -T0 ) ¨© dt ¸¹1 C (70  30) or, 1.6 ?C 1.6 0.04 / min 40 Also § dT · C (T2 -T0 ) ©¨ dt ¹¸2 0.04(40  30) 0.4 qC/ min Fig 7.13 (b): Thus the rate of cooling drops by a factor of Rate of change of four when the difference in temperature of the temperature versus metal sphere and its surroundings drops by a time graph. factor of four. Internet my friend The above activity shows that a hot body 1. https://hyperphysics.phy-astr.gsu.edu/ loses heat to its surroundings in the form of heat hbase/hframe.html radiation. The rate of loss of heat depends on the difference in the temperature of the body 2. https://youtu.be/7ZKHc5J6R5Q and its surroundings. Newton was the first to 3. https://physics.info/expansion study the relation between the heat lost by a body in a given enclosure and its temperature in a systematic manner. 139

ExercisesExercises 1. Choose the correct option. iv) What is absolute zero? v) Derive the relation between three i) Range of temperature in a clinical coefficients of thermal expansion. thermometer, which measures the vi) State applications of thermal expansion. vii) Why do we generally consider two temperature of human body, is specific heats for a gas? (A)70 ºC to 100 ºC viii) Are freezing point and melting point (B) 34 ºC to 42 ºC same with respect to change of state ? Comment. (C) 0 ºF to 100 ºF ix) Define (i) Sublimation (ii) Triple point. x) Explain the term 'steady state'. (D) 34 ºF to 80 ºF xi) Define coefficient of thermal conductivity. Derive its expression. ii) A glass bottle completely filled with water xii) Give any four applications of thermal conductivity in every day life. is kept in the freezer. Why does it crack? xiii) Explain the term thermal resistance. State its SI unit and dimensions. (A) Bottle gets contracted xiv) How heat transfer occurs through radiation in absence of a medium? (B) Bottle is expanded xv) State Newton’s law of cooling and explain how it can be experimentally (C) Water expands on freezing verified. xvi) What is thermal stress? Give an example (D) Water contracts on freezing of disadvantages of thermal stress in practical use? iii) If two temperatures differ by 25 °C on xvii) Which materials can be used as thermal insulators and why? Celsius scale, the difference in temperature on Fahrenheit scale is (A) 65° (B) 45° (C) 38° (D) 25° iv) If α, β and γ are coefficients of linear, area l and volume expansion of a solid then (A) α:β:γ 1:3:2 (B) α:β:γ 1:2:3 (C) α:β:γ 2:3:1 (D) α:β:γ 3:1:2 v) Consider the following statements- (I) The coefficient of linear expansion has dimension K -1 (II) The coefficient of volume expansion has dimension K -1 3. Solve the following problems. (A) I and II are both correct i) A glass flask has volume 1×10-4 m3. (B) I is correct but II is wrong It is filled with a liquid at 30 ºC. (C) II is correct but I is wrong If the temperature of the system is raised (D) I and II are both wrong to 100 ºC, how much of the liquid vi) Water falls from a height of 200 m. What is will overflow. (Coefficient of volume the difference in temperature between the expansion of glass is 1.2×10-5 (ºC)-1 water at the top and bottom of a water fall while that of the liquid is 75×10-5 (ºC)-   1). given that specific heat of water is 4200 J kg-1 °C-1? [Ans : 516.6 × 10-8 m3] (A) 0.96°C (B) 1.02°C ii) Which will require more energy, heating (C) 0.46°C (D) 1.16°C a 2.0 kg block of lead by 30 K or heating 2. Answer the following questions. a 4.0 kg block of copper by 5 K? (slead = i) Clearly state the difference between heat 128 J kg-1 K-1, scopper = 387 J kg-1 K-1) and temperature? [Ans : copper] ii) How a thermometer is calibrated ? iii) Specific latent heat of vaporization of iii) What are different scales of temperature? water is 2.26 × 106 J/kg. Calculate the What is the relation between them? energy needed to change 5.0 g of water 140

into steam at 100 ºC. x) An aluminium rod and iron rod show 1.5 [Ans : 11.3 × 103 J] m difference in their lengths when heated iv) A metal sphere cools at the rate of at all temperature. What are their lengths 0.05 ºC/s when its temperature is 70 at 0 °C if coefficient of linear expansion ºC and at the rate of 0.025 ºC/s when for aluminium is 24.5×10-6 /°C and for its temperature is 50 ºC. Determine the iron is 11.9×10-6 /°C temperature of the surroundings and find [Ans: 1.417m, 2.917m] the rate of cooling when the temperature xi) What is the specific heat of a metal if 50 of the metal sphere is 40 ºC. cal of heat is needed to raise 6 kg of the [Ans : 30 ºC, 0.0125 ºC/s] metal from 20°C to 62 °C ? v) The volume of a gas varied linearly with [Ans: s = 0.198 cal/kg °C] absolute temperature if its pressure is xii) The rate of flow of heat through a copper held constant. Suppose the gas does not rod with temperature difference 30 °C is liquefy even at very low temperatures, at 1500 cal/s. Find the thermal resistance what temperature the volume of the gas of copper rod. will be ideally zero? [Ans: 0.02 °C s cal] [Ans : -273.15 ºC] xiii) An electric kettle takes 20 minutes to vi) In olden days, while laying the rails for heat a certain quantity of water from 0°C trains, small gaps used to be left between to its boiling point. It requires 90 minutes the rail sections to allow for thermal to turn all the water at 100°C into steam. expansion. Suppose the rails are laid at Find the latent heat of vaporisation. room temperature 27 ºC. If maximum (Specific heat of water = 1cal/g°C) temperature in the region is 45 ºC and [Ans: 450 cal/g] the length of each rail section is 10 m, xiv) Find the temperature difference between what should be the gap left given that two sides of a steel plate 4 cm thick, α = 1.2 × 10-5 K-1 for the material of the when heat is transmitted through the rail section? plate at the rate of 400 k cal per minute [Ans : 2.16 mm] per square metre at steady state. Thermal vii) A blacksmith fixes iron ring on the rim of conductivity of steel is 0.026 kcal/m s K. the wooden wheel of a bullock cart. The [Ans:10.26°C or 10.26 K] diameter of the wooden rim and the iron xv) A metal sphere cools from 80 °C to 60 ring are 1.5 m and 1.47 m respectively °C in 6 min. How much time with it take at room temperature of 27 ºC. To what to cool from 60 °C to 40 °C if the room temperature the iron ring should be temperature is 30°C? heated so that it can fit the rim of the [Ans: 10 min] wheel (αiron = 1.2×10-5 K-1). [Ans: 1727.7 °C ] *** viii) In a random temperature scale X, water boils at 200 °X and freezes at 20 °X. Find the boiling point of a liquid in this scale if it boils at 62 °C. [Ans: 131.6°X] ix) A gas at 900°C is cooled until both its pressure and volume are halved. Calculate its final temperature. [Ans: 293.29K] 141