buku physics11

Insulated mittens (Fig. 8.15) take advantage of the insu- Fig.8.15 Reduction of heat transfer in a mittenlative value of dead air trapped in the foam layer. Fur-lined mitten Shiny metal surfaces reflect radiant heat. The alu-minum reflector in the back of the electric arena heater in Outer waterproof shellFig. 8.12 reflects radiant heat toward the the spectators. Synthetic foam/Food wrapped in aluminum foil stays warmer longer insulation layerbecause the foil reflects heat back into the food. Fast-food Air molecules trappedrestaurants employ this technique to keep hamburgers in spaces of insulationwarm. Figure 8.16 shows how the gases in the atmosphere,such as carbon dioxide (CO2) and methane (CH4), act like a Surface of handgreenhouse for Earth. These gases cause the Sun’s radiantenergy to be re-reflected back down to Earth, resulting in anincrease in Earth’s temperature. The average levels of carbon dioxide in our atmospherehave been steadily increasing because of the increasedamount of hydrocarbons such as oil, coal, and natural gasthat have been burned since the onset of the IndustrialRevolution. An increase in global temperatures will pro-duce drastic changes in life on this planet as ecosystems arere-aligned because of the increased temperatures. Re-reflection back down to Earth Fig.8.16 The greenhouse effect Initial radiant energyGreenhouse gases1. Relate the terms “heat,” “thermal energy,” and “temperature.” gpplyin2. Why does your bare hand feel cold when you grab a metal pole as Co the a ncep opposed to a wooden one, given that both were outside in –20°C ts weather? (This principle applies when using the outhouse in the 259 winter. Instead of a painted toilet seat, a foam (polystyrene) ring can be used.)3. Compare the methods of heat transfer (conduction, convection, and radiation). How are these methods used to heat a home?4. Would a person dressed in black feel hotter or cooler than a person dressed in white? Why? c h a pt e r 8 : Thermal Energy and Heat Transfer

Fig.8.17 The robes worn by Bedouins in the Sinai desert are black. This fact seems to contradict the concepts you have learned. However, upon For image closer examination of the situation, we see that even though the black see student robe is 4–6° higher in temperature than the white robe, the black robe causes more convection under it by warming the air to a higher tem- text. perature than the white robe. The warmer air rises faster and leaves through the pores and openings of the material. This causes external air to be drawn in from the large, open end at the bottom of the robe. The increased circulation keeps the person cool.Set-back timer/thermostat for home 8.4 Specific Heat Capacityheating and air conditioning minimizesthe temperature difference between Heat transfer to any substance depends on three things:inside and outside the home, reducing Temperature difference. The greater the temperature differenceheat flow inside or out, which can saveyou money. between the hot and cold substance, the greater the heat flow. Mass of substance. The more mass a substance has, the more molecules Table 8.4 need to have thermal energy transferred to or from them. Specific Heat Capacities Type of substance. Different substances are held together by theirMaterial Specific heat own specific intermolecular forces and accept the transfer of heat to differ- capacity (J/kg°C) ent extents.Liquid nitrogen 1.1 ϫ 102 Two separate pots are heated on a stove under the same conditions, oneGold 1.3 ϫ 102 pot containing water and the other containing vegetable oil. The tempera-Lead 1.3 ϫ 102 ture of the oil rises at a faster rate than the water temperature, but the waterMercury 1.4 ϫ 102 starts to boil at a lower temperature. This phenomenom is a property ofSteam 2.0 ϫ 102 matter that describes the thermal differences between materials, called theSilver 2.3 ϫ 102 specific heat capacity.Ethyl alcohol 2.4 ϫ 102Glycerine 2.4 ϫ 102 Specific heat capacity is the amount of heat energy that is needed toMethyl alcohol 2.5 ϫ 102 increase the temperature of 1 kg of a particular substance by 1°C.Brass 3.8 ϫ 102Copper 3.9 ϫ 102 The units of heat capacity are J/kg°C and its symbol is given as c. TheIron 4.6 ϫ 102 specific heat capacities (c) for water and several other common substancesCrown glass 6.7 ϫ 102 are given in Table 8.4.Pyrex® 7.8 ϫ 102Granite 8.0 ϫ 102 The greater the value of the specific heat capacity, c, the more heatSand 8.0 ϫ 102 energy must be transferred to the substance to change its temperature byAluminium 9.1 ϫ 102 one degree Celsius.Air 1.0 ϫ 103Wood 1.8 ϫ 103 The mathematical formula for calculating heat transfer isIce 2.1 ϫ 103Concrete 2.9 ϫ 103 ⌬EH ϭ mc⌬tWater 4.2 ϫ 103 where ⌬EH is the heat energy transferred to or from a substance in joules (J),260 m is the mass of the substance in kilograms (kg), ⌬t is the temperature change of the substance in °C, and c is the specific heat capacity of the substance in J/kg°C. unit B: Work, Energy, and Power

e x a m p l e 1 Heating copper How much heat energy is required to heat a 1.0 kg piece of copper pipe from 25.0°C to 66.0°C? Solution and Connection to Theory Given m ϭ 1.0 kg t1 ϭ 25.0°C t2 ϭ 66.0°C material is copper ⌬EH ϭ ? We can find the specific heat capacity (c) of copper in Table 8.4. ccopper ϭ 3.9 ϫ 102 J/kg°C Substituting into the equation, ⌬EH ϭ mc⌬t ϭ mc(t2 – t1) ϭ 1.0 kg(3.9 ϫ 102 J/kg°C)(66.0°C Ϫ 25.0°C) ϭ 1.0 kg(3.9 ϫ 102 J/kg°C)(41.0°C) ⌬EH ϭ 1.6 ϫ 104 J Therefore, the heat energy EH required to heat a 1.0 kg piece of copper pipe from 25.0°C to 66.0°C is 1.6 ϫ 104 J. If the copper was cooled instead of heated, the value for EH would be negative because the two temperature values would be reversed, making ⌬t negative. This means that for a substance, a positive ⌬EH means it is warming and a negative ⌬EH means that it is cooling.e x a m p l e 2 Cooling ironA 0.50 kg block of iron at 80.0°C is cooled by removing 2.28 ϫ 104 J ofheat energy. What will the final temperature of the metal be?Solution and Connection to TheoryGivenm ϭ 0.50 kg t1 ϭ 80.0°C ⌬EH ϭ Ϫ2.28 ϫ 104 J material is iron t2 ϭ ?The word “removing” indicates that EH is negative. From Table 8.4, thespecific heat capacity for iron is 4.6 ϫ 102 J/kg°C.⌬EH ϭ mc⌬t ϭ mc(t2 Ϫ t1)t2 ϭ ᎏEᎏH ϩ t1 mct2 ϭ ᎏ(0.50Ϫkg2ᎏ).2΂48.6ϫϫ101ᎏ04 2J ᎏkg·Jᎏ°C΃ ϩ 80.0°Ct2 ϭ Ϫ99.13°C ϩ 80.0°C ϭ Ϫ19°CTherefore, the final temperature for this piece of iron is Ϫ19°C. c h a pt e r 8 : Thermal Energy and Heat Transfer 261

g pplyin 1. What three things does heat transfer depend on?Co the 2. Answer Question 2 at the end of Section 8.3 (holding onto a metala ncep ts pole in the cold as opposed to a wooden one) by using the values for262 specific heat capacity in Table 8.4. 3. Compare the heat required to heat 1.5 kg pieces of gold (we wish), iron, and silver from 12°C to 40°C. How does the solution change when these materials are cooled? 8.5 Heat Exchange — The Law of Conservation of Heat Energy In the previous section, we examined what happens to the temperature of a specific substance when heat is either added or taken away. According to the law of conservation of energy, the total amount of heat energy must be constant as long as none of it is lost to the surroundings. If we think of heat energy as moving from a hot object to a cold object, then we can simplify the law of conservation of heat energy to be ԽEHlostԽ ϭ ԽEHgainedԽ or ⌬EH1 ϭ Ϫ⌬EH2 Because this application of the law of conservation of energy specifically involves the exchange of heat energy, this principle is also called the principle of heat exchange. The negative sign represents the direction in which heat is flowing. It shows that the values of EH are the same, but one material is losing heat while the other one is gaining heat. e x a m p l e 3 A spot of tea? A 0.500 kg pot of hot water for tea has cooled to 40.0°C. How much freshly boiled water must be added (at 100.0°C, of course) to raise the temperature of the tea water to a respectable 65.0°C? Solution and Connection to Theory Given cool water hot water mc ϭ 0.500 kg mh ϭ ? cw ϭ 4.2 ϫ 103 J/kg°C cw ϭ 4.2 ϫ 103 J/kg°C t1 ϭ 40.0°C, t2 ϭ 65.0°C t1 ϭ 100.0°C, t2 ϭ 65.0°C ⌬tc ϭ (65.0°C Ϫ 40.0°C) ϭ 25.0°C ⌬th ϭ (65.0°C Ϫ 100.0°C) ϭ Ϫ35.0°C unit B: Work, Energy, and Power

The fact that one ⌬t value is negative and the other is positive shows thatone material is losing heat while the other is gaining heat.According to the law of conservation of heat energy,EHc ϭ ϪEHhmccw⌬tc ϭ Ϫmhcw⌬thAs both materials are water, cw from each side can cancel.0.500 kg(25.0°C) ϭ Ϫmh(Ϫ35.0°C)mh (35.0°C) ϭ 12.5 kg°Cmh ϭ 0.357 kgTherefore, 0.357 kg of water is required to increase the temperature ofthe tea water to 65.0°C.e x a m p l e 4 Cooling an eggA hard-boiled egg with a heat capacity of 2.4 ϫ 103 J/kg°C and a mass of50.0 g is cooled from 100.0°C by 1.0 L of water at 5.0°C. What will be thefinal temperature of the water and egg after they have been allowed to sitfor a few moments? Metric conversions can easily be donemegg ϭ 50.0 g ϭ 50.0 ϫ 10Ϫ3 kg ϭ 5.00 ϫ 10Ϫ2 kg mwater ϭ 1.0 kg using the exponent multiplier. In thisSolution and Connection to Theory case, a gram is ᎏ1010ᎏ0 of a kilogram, so the multiplier “ϫ10Ϫ3” can be added to make a quick conversion. See Appendix C for other multipliers.Given Recall that by definition, 1 mL of water was assigned a mass of 1 g.egg Therefore, 1.0 L has a mass of 1.0 kgcegg ϭ 2.4 ϫ 103 J/kg°C (see Fig. 8.18).t1egg ϭ 100.0°Ct2egg ϭ ? °C Fig.8.18 The relationship betweenwater mass and volume of watercwater ϭ 4.2 ϫ 103 J/kg°C (see Table 8.4) 1 mL Mass of 1 gramt1water ϭ 5.0°C 1 mL has a mass of 1 gramt2water ϭ ? °C 1 L (1000 mL) has a mass of 1 kilogramThe heat lost from the egg will be used to warm up the water aslong as we assume that none of the heat is lost to the surround- 1 litre 1 kilogramings. Both the water and the egg will end up at the same temper-ature, which is our missing quantity (t2). c h a pt e r 8 : Thermal Energy and Heat Transfer 263

Therefore, ⌬tegg ϭ t2egg Ϫ t1egg and ⌬twater ϭ t2water Ϫ t1water EH1 ϭ ϪEH2 meggcegg⌬tegg ϭ Ϫ mwatercwater⌬twater 50.0 ϫ 10Ϫ3 kg(2.4 ϫ 103 J/kg°C)(t2 Ϫ 100°C) ϭ Ϫ1.0 kg(4.2 ϫ 103 J/kg°C)(t2 Ϫ 5°C) 120 J/°C(t2) Ϫ 12000 J ϭ Ϫ4.2 ϫ 103 J/°C(t2) Ϫ (Ϫ21000 J) 4320 J/°C(t2) ϭ 33000 J t2 ϭ 7.6°C Both the water and the egg end up at an intermediate temperature (between 5°C and 100°C) of about 8°C. g pplyin 1. State examples of the law of conservation of energy that you haveCo seen throughout the various units you have studied.a the ts ncep 2. In western movies, the cowboy or cowgirl takes a bath by adding hot water to a partially filled tub or barrel to warm it up. If the per-Fig.8.19 Convection in a son is sitting in 100 kg of lukewarm water (15°C), how much boil- ing water must be added to the bath water in order to raise thecooling lake temperature to a modest 35°C? Note: If you wish to calculate the volume of water needed, use the equation ␳ ϭ ᎏmVᎏ, where ␳ is the Warm Cold density of water (1000 kg/m3), m is the mass, and V is the volume. Autumn264 Seasonal Turnover of Lakes For fishing enthusiasts, the end of summer and the beginning of fall is ushered in when the lake “turns over.” In this process, the colder tem- peratures cool the water at the surface of the lake, causing it to become more dense and sink under the warm water layer under the surface. This sinking causes convection currents, which mix the different tem- perature layers. As a result, the water at the bottom of the lake comes to the surface and the water at the surface of the lake moves to the bottom. This process continues until the water reaches its maximum den- sity at a temperature of 4°C. Below this temperature, the water at the surface cools further, eventually turning to ice. Because water is at its maximum density in its liquid state, the ice, which is less dense than water, floats on the surface and acts as an insulating blanket for the lake water and inhabitants below. unit B: Work, Energy, and Power

Fig.8.20 For image For image see student see student text. text. 3. Summarize the conduction of heat in the water system during each season. Use the kinetic molecular theory to explain how heat is transferred. 4. Explain the conduction of heat in water in terms of the conserva- tion of energy. 5. How does this process benefit the lake and its ecosystem? 6. Water’s unusual property of density allows life in the water to sur- vive the winter. Explain. 8.6 Changes of State and Latent HeatIn Section 8.2, we learned that when heat is added or removed from a sub- Fig.8.21 Changes of statestance, the change in thermal energy affects the temperature of the substance.If the change in thermal energy is great enough, it will cause a change of state for waterin the substance. Figure 8.21 shows a cooling-warming curve for water as aconsistent amount of heat is added or removed. As the water is heated orcooled, it goes through all three states of matter: vapour/gas, liquid, and solid. As steam cools, its temperature drops, as expected. But at 100ºC, the tem-perature remains constant. Thermal energy is still being removed from thevapour, but it doesn’t register on the thermometer as a drop in temperature. 140 Warming curve for H2O Cooling curve for H2O 140 Time 120 Gas/vapour 120Temperature (°C)100 Liquid 100Temperature (°C) Solid 80 80 60 60 40 40 20 20 M t 0 0 Ϫ20 Ϫ20 Time c h a pt e r 8 : Thermal Energy and Heat Transfer 265

This extra or stored amount of energy is associated with a change of state andis called latent heat. From Fig. 8.21, we see that latent heat seems to bereleased during condensation and freezing, and stored when heat is addedduring melting and vapourization. The two types of latent heat are the latentheat of fusion and the latent heat of vapourization.Latent Heats of Fusion and VaporizationThe latent heat of fusion (Lf) is the amount of energy needed to melt 1.0 kgof ice. The word “latent” means “hidden.” It refers to the fact that duringmelting, the added heat does not cause a rise in temperature and will layhidden until it is released in the freezing process. The greater the mass ofmelting material, the more energy is required. Latent heat is defined as theamount of heat required per unit of mass to change state. The latent heat offusion (melting) is given by the equation Lf ϭ ᎏEᎏH mwhere EH is the heat in joules and m is the mass in kilograms.Lf is measured in joules per kilogram (J/kg).The latent heat of vapourization, Lv, is the heat required per unit massto vapourize a liquid. Lv ϭ ᎏEᎏH mLike Lf, Lv is measured in joules per kilogram (J/kg).Table 8.5 lists the latent heats of fusion and vapourization for some com-mon substances. Table 8.5 Latent Heats of Fusion and VapourizationSubstance Lf (J/kg) Melting/ Lv (J/kg) Boiling/ freezing condensationAluminum 9.0 ϫ 104 point (°C) 1.1 ϫ 107Ethyl alcohol 1.1 ϫ 105 8.6 ϫ 105 point (°C)Methyl alcohol 6.8 ϫ 104 659 1.1 ϫ 106Iron 2.5 ϫ 105 Ϫ130 6.3 ϫ 106 1509Lead 2.3 ϫ 104 Ϫ97.8 8.7 ϫ 105 78Nitrogen 2.5 ϫ 104 1530 2.0 ϫ 105 64.7Oxygen 1.4 ϫ 104 2.1 ϫ 105Silver 1.1 ϫ 105 327 2.3 ϫ 106 1820Water 3.3 ϫ 105 Ϫ209.9 2.3 ϫ 106 1780 Ϫ218.9 Ϫ196.8 Ϫ183 960 1950 0 100266 u n i t B : Wo r k , E n e rg y, a n d Powe r

These values were found experimentally by measuring the amount of heatrequired to melt or vapourize a certain amount of mass of material, asdescribed in the following example.e x a m p l e 5 Calculating the latent heat of fusionA 0.27 kg sample of a material requires 8.91 ϫ 104 J of energy to meltit without raising its temperature. What is the latent heat of fusion forthis material?Given EH ϭ 8.91 ϫ 104 J Lf ϭ ?m ϭ 0.27 kgSolution and Connection to TheoryLf ϭ ᎏEᎏH m ϭ ᎏ8.91 ϫᎏ104J 0.27 kg ϭ 3.3 ϫ 105J/kgTherefore, the latent heat of fusion is 3.3 ϫ 105 J/kg.Referring to Table 8.5, we see that this substance is most likely water.e x a m p l e 6 Calculating the heat of vapourizationOne danger of leaving a tea kettle unattended on the stove is having itboil dry. How much heat is required to change 0.20 kg of water (about theamount for a good cup of tea) from a liquid state to vapour?Given Lv ϭ 2.3 ϫ 106 J/kg EH ϭ ?m ϭ 0.20 kgSolution and Connection to TheoryLv ϭ ᎏEᎏH Rearrange the equation. mEH ϭ (m)Lv ΂ ΃ϭ (0.20 kg) 2.3 ϫ 106 ᎏJᎏ kg ϭ 4.6 ϫ 105 JTherefore, the amount of heat required to vapourize 0.20 kg of water is4.6 ϫ 105 J. c h a pt e r 8 : Thermal Energy and Heat Transfer 267

Note: The addition or subtraction of heat will change the temperature of asubstance up until it reaches the critical temperature at which a change instate occurs. At this temperature, any heat transfer is used to facilitate thephase change.e x a m p l e 7 The heat required to carry a substance through a change of stateWhat would be the total amount of heat required to warm a 0.100 kgsolid sample of water from Ϫ30ºC, through melting at 0ºC, to a final liq-uid temperature of 80ºC?Given t2 (solid) ϭ 0ºC t1 (liquid) ϭ 0ºCt1 (solid) ϭ Ϫ30ºC m ϭ 0.100 kgt2 (liquid) ϭ 80ºCLf (water) ϭ 3.3 ϫ 105 J/kg cice ϭ 2.1 ϫ 103 J/kgºCcwater ϭ 4.2 ϫ 103 J/kgºCET ϭ ?Solution and Connection to TheoryThe total heat would be the sum of the heat required to warm the solidto the melting point, cause the melting, and then the final warming of theliquid from the melting to the final temperature.E ϭ E ϩ E ϩ ET warming of ice melting of ice warming of water ϭ m⌬twarming1 cice ϩ Lf (m) ϩ m⌬twarming2 cwaterϭ (0.100 kg)(0ºC Ϫ (Ϫ30ºC))2.1 ϫ 103 J/kgºC ϩ (3.3 ϫ 105 J/kg)(0.100 kg) ϩ (0.100 kg)(80ºC Ϫ 0ºC)4.2 ϫ 103 J/kgºCϭ 6.3 ϫ 103 J ϩ 3.3 ϫ 104 J ϩ 3.4 ϫ 104 Jϭ 7.3 ϫ 104 JTherefore, the total heat energy required to heat 0.100 kg of water fromϪ30ºC to 80ºC is 7.3 ϫ 104 J.Effects of Latent HeatAny heat that is added during a “warming cycle” to raise the temperature of asubstance can be recovered later when the substance cools. The heat used toaffect the phase changes of melting and vapourization is stored as latent heat.It can be recovered only when the substance undergoes the reverse phase268 u n i t B : Wo r k , E n e rg y, a n d Powe r

changes of condensation and freezing. Our formulas from the above examples gpplyincan also be used to calculate the heat that must be removed from the system. Co a the The “extra” amount of heat required to cause a change in state is tsncepresponsible for the moderate climates near large bodies of water. In thespring, melting the ice on these bodies of water draws some of the heat fromthe atmosphere, which makes the air temperature lower than normal. Thisheat that we miss during the cool spring months is stored as the latent heatof fusion. During the winter freeze, this latent heat of fusion is released asheat into the atmosphere and the result is higher temperatures. In many Canadian backyards, tomato plants are protected from frost inspring and fall by spraying the foliage with water. During a frost, the wateron the plants freezes, releasing the latent heat of fusion. This extra energycan protect the plants from temperatures as low as Ϫ2ºC. Orange growers in Florida spray their crops with water if there is a threat of freezing. Explain how this procedure could provide enough heat to save their crops. 1. a) Which latent heat value, fusion or vapourization, is larger for any given substance? b) Use the kinetic molecular theory to explain what is going on at the molecular level during these two changes of state. 2. How much heat does a freezer remove from 0.25 kg of water in an ice cube tray if the temperature of the water is 0°C? 3. A 0.54 kg sample of material requires 1.782 ϫ 105 J of energy to melt it. What is the latent heat of fusion for this material? 4. A tank containing 740 kg of water cools from 15°C to 0°C and com- pletely freezes. How many hours would a 1.2 kW electric heater have to work to provide the equivalent heat the water gives off? (The heater provides 1.2 ϫ 103 J/s of heat.)8.7 Calorimetry—Some Practical ApplicationsOne practical application of the physics of heat energy is the careful and The old unit for the measure of heatprecise measurement of heat transfer in the process of calorimetry. energy was the calorie. We now use the joule as the unit of measure for During calorimetry, a calorimeter measures the heat flow into a supply heat energy, but the term “calorime-of water by tracking the temperature increase as the process continues. try,” or “measuring heat,” is stillKnowing the specific heat capacity, mass, and the initial and final used today. The conversion factortemperatures of the water allows us to calculate the heat flow by applying between these units is 4.18 J/C,the formula EH ϭ mc⌬t. Calorimetry may be used to find specific heat where C represents calories.capacities of materials or to find the energy content of foods.c h a pt e r 8 : Thermal Energy and Heat Transfer 269

Calorimetry and Specific Heat CapacityFig.8.22 A simple calorimeter The specific heat capacities of some selected substances are given in Table 8.4. These are all found experimentally using a simple calorimeter, such asHot test the one shown in Fig. 8.22, and the law of conservation of heat energy (heat object lost by a material is equal to the heat gained by the calorimeter). Lid The specific heat capacity is found by heating a material in question to a Inner known high temperature and placing it into the calorimeter. Since heat energy vessel is conserved, the specific heat capacity of the sample may be found using Water Outer mscs(t2 – t1s) ϭ Ϫmwcw(t2 – t1w) vessel orInsulation ᎏmw ϫᎏcw ᎏt2 Ϫ ᎏ(t1w) layer ms t2 Ϫ (t1s) ΂ ΃cs ϭ Ϫ The items in the brackets are all known quantities, so the key measurement that is required is the final temperature of the calorimeter. e x a m p l e 8 Specific heat capacity of aluminum A 0.700 kg piece of aluminum is heated to 100°C in boiling water and quickly placed into a calorimeter that contains 0.200 kg of water at 20.0°C. The final temperature of the calorimeter was determined to be 54.4°C. What is the specific heat capacity, cAl, for aluminum? Solution and Connection to Theory Given water aluminum mw ϭ 0.200 kg cw ϭ 4.2 ϫ 103 J/kg°C mAl ϭ 0.700 kg t1w ϭ 20.0°C cAl ϭ ? t2 ϭ 54.4°C t1Al ϭ 100°C t2Al ϭ 54.4°C mAlcAl⌬tAl ϭ Ϫmwcw⌬tw Rearranged, the equation becomes ᎏmwᎏcw ᎏtt22wAl ϪϪᎏ(΂tt11wA)l΃ mAl Ϫ΂ ΃cAl ϭ ᎏ0.200 kᎏg ϫ 4.2ᎏϫ 103 ᎏJ/kg°C ᎏ5544..44°°CC ϪϪᎏ2100.00°°CC 0.700 kg ΂ ΃cAlϭ Ϫ cAl ϭ 9.05 ϫ 102 J/kg°C The specific heat capacity for aluminium is 9.05 ϫ 102 J/kg°C, which may be compared to the value in Table 8.4.270 u n i t B : Wo r k , E n e rg y, a n d Powe r

Calorimetry and Food Energy Fig.8.23 A bomb calorimeterIn food calorimetry, a bomb calorimeter like the one in Fig. Ignition Thermometer8.23 is used to burn electrically ignited food. The heat that terminalsis liberated is transferred to the water in the calorimeter, Stirrerwhich registers a temperature rise. The entire apparatus isinsulated from the surroundings and the water is stirred to Waterensure precise results. The energy content of the food is Insulationthen determined by the equation Sealed reaction chamber containing ⌬EH ϭ mwcw⌬tw Ignition wire food item and oxygen Potato chipwhere mw is the mass of the water in the calorimeter, cw is thespecific heat capacity of water (4.2 ϫ 103 J/kg°C), and ⌬tw isthe temperature change of the water in the calorimeter.e x a m p l e 9 How well balanced is that breakfast?A 30 g sample of Kellogg’s Corn Pops cereal is burned inside a bombcalorimeter and the 2.00 kg of water warms from 24.0°C to 81.1°C. Whatis the amount of energy that is stored in the cereal?Solution and Connection to TheoryGivenmc ϭ 30.0 g mw ϭ 2.00 kg cw ϭ 4.2 ϫ 103 J/kg°C⌬tw ϭ t2 Ϫ t1 ϭ 81.1°C Ϫ 24.0°C ϭ 57.1°C⌬E ϭ mc⌬t⌬E ϭ 2.00 kg(4.2 ϫ 103 J/kg°C)(57.1°C)⌬E ϭ 480 000 J or 480 kJ for 30.0 g of cereal.The Corn Pops cereal has 480 kJ of energy in every 30.0 g, or 16 kJ/g.Fig.8.24 Methods of Heat Transfer Direction uttin g of heat flow it all gethTo Conduction p er Hot object Convection Cold object Heat lost Radiation Heat gained1. How is calorimetry related to the conservation of energy? gpplyin2. A 1.200 kg piece of iron is heated to 95°C. It is placed into a Co the a ncep calorimeter that contains 0.430 kg of water at 10.0°C. Find the ts final temperature of the calorimeter. 271 c h a pt e r 8 : Thermal Energy and Heat Transfer

S T S c i e n c e — Te c h n o l o g y — S o c i ety —S E Environmental Interrelationships Global Warming: Heating Ourselves to DeathFig.STSE.8.1 When we think about global warming, we might initially think that we Atmospheric Concentration would be better off without cold winters. The negative impact of global of Carbon Dioxide (1744–1992) warming far outweighs the benefits of warmer winters. Global warm-CO2 concentration 375 ing refers to the consistent increase in the average ambient temperature (parts per million by volumes) Siple Station ice core on Earth. The increase is believed to be caused by an increase in the amount of greenhouse gases in the atmosphere (see Fig. STSE.8.1 and 350 Mauna Loa Table STSE.8.1). These gases, which include carbon dioxide and 325 300 methane, act like the glass in a greenhouse. As shown in Fig. STSE.8.2, radiant heat energy from the Sun reaches Earth, but much of the 275 reflected heat is kept inside by the atmosphere, which acts like a giant 250 solar blanket. Over the last century, the increase in global population as 1700 1750 1800 1850 1900 1950 2000 well as in industrialization has also meant that more energy is used and Year more waste heat is produced. This increase has resulted inFig.STSE.8.2 a change in global temperature (see Fig. STSE.8.3), which Reflected by Some Some has far-reaching effects that will upset the delicate balance of Earth’s ecosystems. Earth’s surface longwave is longwave is Parts of Earth covered with ice and snow act as mirrors, to space lost to space lost to space reflecting heat away from Earth. As global warming melts Greenhouse Gases these areas, Earth will warm at an even faster rate. Global warming is responsible for flooding from melted ice and for rapid environmental changes in many ecosys-Solar Converted Surface gains Surface gains tems that negatively affect plant and animal species. Recentenergy into heat, more heat more heat efforts to curb the emission of greenhouse gases should beabsorbed causing the and longwave and longwave applauded, but it may be a case of too little too late.at surface emission of radiation is radiation is longwave emitted again emitted again radiation Fig.STSE.8.3 Global Temperature Changes (1880–1999) 1.4 Departure from long-term mean (°F) 1.2 1 0.8 0.6 0.4 0.2 0 Ϫ0.2 Ϫ0.4 Ϫ0.6 Ϫ0.8 1880 1890 1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 Year272 u n i t B : Wo r k , E n e rg y, a n d Powe r

Table STSE.8.1 Gases Involved in the Greenhouse Effect: Past and Present Concentrations and SourcesGreenhouse gas Concentration in 1750 Present concentration Percent change Natural and anthropogenic sourcesCarbon dioxide 228 ppm 330 ppm 29% Organic decay; Forest fires; Volcanoes; 0.70 ppm Burning fossil fuels; Deforestation;Methane 1.70 ppm Land-use change 280 ppbNitrous oxide 0 310 ppb 143% Wetlands; Organic decay; Termites; Natural gas & oil extraction; Biomass burning;Chlorofluorocarbons Unknown 900 ppt Rice cultivation; Cattle; Refuse landfills(CFCs) Varies with latitudeOzone and altitude in the 11% Forests; Grasslands; Oceans; Soils; atmosphere Soil cultivation; Fertilizers; Biomass burning; Burning of fossil fuels Not applicable Refrigerators; Aerosol spray propellants; Cleaning solvents Global levels have Created naturally by the action of sunlight generally decreased on molecular oxygen and artificially in the stratosphere through photochemical smog production and increased near Earth's surfaceDesign a Study of Societal Impact Since the late 1980s, several international agreements have been signed by the world’s major industrial countries to reduce emissions of green- house gases. Have these agreements worked? How does the level of green- house gases emitted this year compare to 2, 5, and 10 years ago? How drastic does the decrease in greenhouse gases need to be for the world to avoid the predictions of devastation associated with global warming?Design an Activity to Evaluate Evaluate the performance of several different commercially available travel coffee mugs for their ability to keep coffee hot. Evaluate whether a thermostat set-back timer saves energy during a simulated 24-hour period. By plotting a cooling curve for a warm liquid left in a freezer, calculate the net heat loss and compare it with the heat energy required to warm the liquid back to its original temperature.Build a Structure 273 Design and construct an insulated box that will maintain the temper- ature of a heated container of water. Monitor the temperature of the container over time, using a computer interface and a temperature probe. If you hold a competition, the contestant with the smallest drop in temperature after a set amount of time is the winner. c h a pt e r 8 : Thermal Energy and Heat Transfer

S U M M A RY S P E C I F I C E X P E C TAT I O N S You should be able to Understand Basic Concepts: Define and describe the concepts and units related to thermal energy and kinetic molecular theory as they relate to heat transfer. Apply the kinetic molecular theory to explain how thermal energy is transferred in matter by the processes of conduction, convection, and radiation. Relate the principle of heat exchange to the law of conservation of energy. Use the law of conservation of energy to quantitatively determine the amount of heat transferred between two bodies. Define “closed” and “open” systems and relate them to the efficiency of heat transfer. Develop Skills of Inquiry and Communication: Design and carry out an experiment to measure the efficiency of transfer- ring electrical energy to thermal energy in an electric heating appliance. Demonstrate the safe and appropriate handling of electrical appliances by checking for damaged or hazardous equipment before conducting experiments. Analyze and interpret tabulated experimental data to determine the characteristic physical property of specific heat capacity. Relate Science to Technology, Society, and the Environment: Analyze the environmental and social impact of global warming. Relate the processes of reflection and transmission of radiant solar energy to explain global warming. Describe how the knowledge of the processes of conduction, convection, and radiation are used in the design of systems that reduce or enhance heat transfer. Relate the natural phenomena of sea and land breezes and seasonal turnover to the process of heat convection in fluids. Equations EH ϭ mc⌬t m1c1⌬t ϭ Ϫ(m2c2⌬t) Lf ϭ ᎏEᎏH m Lv ϭ ᎏEᎏH m274 u n i t B : Wo r k , E n e rg y, a n d Powe r

EXERCISESConceptual Questions 9. As you will read in later chapters, water from lakes such as Lake Ontario and Lake Huron is1. State the principle of heat exchange and used to cool steam turbines in many of our discuss how it applies to the law of nuclear power plants. Lake water is also used conservation of energy. by fossil-fuel-burning power plants as well as other types of industrial complexes. After use,2. Liquids are used to transfer heat in car radia- this warm water is returned to the lakes. Why tors. Why do we mix water and ethylene gly- would industries use water to cool their col in the radiator given that water has a machinery? Suggest some of the societal and higher heat capacity than ethylene glycol? environmental implications of warming our lakes artificially.3. Explain why it is relatively easy to remove baked potatoes from a hot barbecue without a 10. Should fireplaces be used as part of support hot pad if they are wrapped in aluminum foil. walls in houses? Explain.4. Two equal mass samples of copper and iron 11. Why does running hot water over a metal lid heated with the same amount of heat energy on a glass jar make it easier to open? will achieve different temperatures. Which sample will reach the higher temperature 12. What happens to the size of a hole in the cen- and why? tre of a metal ring when the ring is heated? Explain.5. To save energy, many home owners have installed temperature set-back thermostats 13. Why are railway track sections separated that set the inside house temperature lower by a small air gap? (Hint: They are called at night during the winter. Some critics have expansion slots.) suggested that the furnace has to work harder in the morning to warm the house 14. The background universe temperature is con- up again. Write a brief paragraph that will stantly cooling. How does this temperature explain to any sceptics that the overall heat trend help explain the Big Bang Theory? flow from the inside to the outside of the house is definitely less. 15. On a cold day, you grab a shiny, metal fence gate with your bare hand. The post feels very cold.6. Use the kinetic molecular theory to explain The next gate is made of wood and does not feel the temperature change that occurs when a cold. Both gates are at the same temperature cold and a hot liquid are mixed. because they are close together outside. Explain why one gate feels colder than the other.7. Why is copper used to cover the bottom of many commercial pots and pans for the 16. Why does the temperature scale have a lower stove top? limit but not an upper limit?8. You are spending Labour Day Weekend at 17. Where is the energy going in a process where the beach and it turns out to be a clear, hot there is a state change yet no temperature day. Using the concept of heat capacity, change? describe which warms up faster, the water or the sand. Which of the two will cool 18. Describe the heating/cooling system of a car in down faster at night? terms of heat transfer, temperature differences, state changes, heat capacities, and efficiency. c h a pt e r 8 : Thermal Energy and Heat Transfer 275

19. Thermograms are pictures which are colour- Energy added (kJ)27. The graph in Fig. 8.25 illustrates the findings coded according to the amount of thermal of an investigation in which a 1.0 kg mass of energy radiated from the object. Running from three different substances is heated. coldest to hottest, the colours are black, blue, pink, red, and white. Describe the colours you Fig.8.25 would see on a thermogram of a limousine that has been running for a long time. The 200 passenger compartment is separated by a parti- tion and has a curtain on one window. The A window on the driver’s side is half open. 15020. In what other areas of study would thermo- 100 B grams be useful diagnostic tools? C21. How does the greenhouse effect work in 50 terms of types of radiation? 0 20 40 60 8022. Which freezes first, hot water in an ice cube Temperature (°C) tray or cold water in the same-sized tray? a) Find the slope (with units) of each of the Problems three lines. 8.2 Thermal Energy and Temperature b) Which substance needed the most heat for the same temperature change as the others?23. Convert the following: a) 100°C ϭ ______ K c) Which substance has the highest heat b) –25°C ϭ ______ K capacity? c) –273°C ϭ ______ K d) 0°C ϭ ______ K d) How is the slope related to heat capacity? e) 57 K ϭ ______ °C f) 300 K ϭ ______ °C 28. What temperature change would occur in each of the following circumstances? 8.4 Specific Heat Capacity a) 250 g of mercury gains 1.93 kJ of heat energyNote: Use the specific heat capacities in Table 8.4 b) 5.0 kg of water gains 100 kJ of heat energywhen required. 29. A 25.0 g piece of iron, originally at 500°C,24. How much heat energy is gained per kilogram of has its temperature fall to 100°C when it is water when it is heated from 10.0°C to 90.0°C? placed in a cold water bath. How much energy was lost by the piece of nickel during25. A 400.0 g aluminum cooking pot is heated the cooling process? from 25.0°C to 99.0°C. What amount of heat energy does it absorb? 8.5 Heat Exchange — The Law of Conservation of Heat Energy26. What was the initial temperature of a 1.50 kg piece of copper that gains 2.47 ϫ 104 J of 30. A 100.0 g aluminum coffee cup at 15.0°C energy when it is heated to a final tempera- is filled with 250.0 g of piping hot coffee ture of 150°C? (about the same heat capacity as water) at 95.0°C. After all the heat transfer between the coffee and the cup is completed, what is the final temperature of the coffee (and the cup)?276 u n i t B : Wo r k , E n e rg y, a n d Powe r

31. Native Canadians used to boil water from 35. What mass of oxygen would freeze by the maple tree sap by heating rocks and placing removal of 7.4 ϫ 104 J of heat energy? them into vats of sap sitting in hollowed-out tree logs. For this problem, assume that the 36. How much heat must be removed by a freezer syrup and the rocks have the heat capacities of in order to change twenty 60 g sections of water and sand, respectively. What mass of water at 0ºC to ice at 0ºC? stone, heated to 1000°C, would be needed to increase 20.0 kg of sap from 60.0°C to 85.0°C? 37. Brandy is made by distilling ethyl alcohol, which is done by boiling the alcohol until it32. A duck wants to take a bath and fills the tub turns into a gas. How much heat is required with 50.0 L of water at a temperature of to completely distil 0.750 kg of ethyl alcohol? 38.0°C. After taking a phone call from his friend Bob, a lemming, the bath water had 38. A 0.200 kg block of ice at Ϫ15ºC is placed cooled by 10.0°C. How much more hot water into a pan on a stove, heated to a liquid, and at 80.0°C must the duck add to return the then to vapour with a final temperature of bath to the desired temperature? 115ºC. Calculate the total amount of heat required for this process.33. A 200 W heater is used to heat 0.10 kg of a liquid from 20.0°C to 80.0°C, which takes 8.7 Calorimetry — Some Practical one minute. Applications a) What is the heat given off by the heater to the liquid? 39. The Forensic Sciences Lab in Toronto wants b) What is the heat capacity of the liquid? to find out what materials the Romans used in their water pipe systems. A 97.5 g sample 8.6 Changes of State and Latent Heat of the metal pipe at 20.0°C is placed in a water calorimeter containing 0.10 kg of water34. a) What is the latent heat of fusion of a 1.5 kg at 53.2°C. What is the specific heat capacity substance that requires 3.75 ϫ 104 J to of the metal if the calorimeter’s final tempera- melt it? ture was 52.2°C? Referring to Table 8.4, what are the possibilities for the type of metal used b) When the substance is cooled to its freez- in these Roman pipes? Which do you think ing point, how much heat energy is given was the one that was actually used? off by 1.0 kg of the substance when it freezes into a solid? 40. What mass of copper at 87.0°C, when added to 300 g of water at 17.0°C in a calorimeter, would yield a final temperature of 26.0°C?c h a pt e r 8 : Thermal Energy and Heat Transfer 277

LABORATORY EXERCISES 8.1 Efficiency of an Electric Appliance Purpose Data To measure the percent efficiency of an electric 1. This lab requires the mass of an amount of appliance such as a tea kettle, a hot plate, or an water in the heater, the time that the electricity immersion heater. was on, and the initial and final temperatures of the water. Equipment 2. Record the power rating of the heater in An electric water heating appliance (immersion watts as it is written on the electrical infor- heater with polystyrene cup, electric tea kettle, mation label. or hot plate and beaker) Thermometer Uncertainty Stopwatch Instrumental uncertainties of Ϯ.1 of the small- Procedure est division should be assigned to all tempera- tures and the mass taken. The time should be 1. Set up the lab apparatus, as shown in assigned an appropriate uncertainty based on Fig. Lab.8.1. your reflexes. 2. Measure 200 mL of water (0.200 kg) and pour Analysis it into a clean, dry heating container (poly- styrene cup, electric tea kettle, or hot plate). 1. Calculate the heat gained by the water using the relationship ⌬EH ϭ m⌬tc 3. Place the thermometer into the water and measure the temperature after it as been 2. Find the maximum heat energy that was pro- allowed to sit for about two minutes to reach duced by the heater by using the relationship thermal equilibrium. EH ϭ P⌬t, where P is the power rating marked on the heating device and t is the time. 4. Turn on the heating apparatus (immersion heater for the polystyrene cup, electric tea kettle, 3. Determine the percent efficiency of the or the hot plate if the container is the beaker) energy transfer in the heating apparatus. and start the stopwatch at the same time. Discussion 5. Keep heating until the water temperature rises a considerable amount (up to about 1. Look at the percent efficiency values found by 50°C–60°C) other students in the class. Rank the three heating appliances (kettle, hot plate, or immer- 6. Turn off the heating system. Stop the stop- sion heater) from most to least efficient. watch and note the time, but continue to stir the water gently with the thermometer, being 2. This experiment could be considered repre- careful to record the highest temperature that sentative of the amount of heat required to is reached. make one cup of tea. On average, the school cafeteria heats water for 75 cups of tea in one Fig.Lab.8.1 Stopwatch day for an entire year. How much money would be wasted due to inefficiency if the cost Thermometer 00:00 of electrical energy is 1¢ for every 4.5 ϫ 105 J? Stopwatch 00:00 Beaker Conclusion Water Electric hot plate Summarize your results for your value of the percent efficiency of your heating apparatus.278 u n i t B : Wo r k , E n e rg y, a n d Powe r

8.2 Specific Heat Capacity LABORATORY EXERCISESPurpose 4. Measure the mass of one of the metal samples and record it in your data table.To measure the heat capacity of different metalsby using the law of conservation of heat energy. 5. Attach a thread to the sample and suspend it in the warming water with the glass rod, asEquipment shown in Fig. Lab.8.2. Leave the sample sus- pended in the water until the water has beenSafety glasses boiling for at least five minutes so that you can250 mL polystyrene coffee cup be sure that the metal reaches the same tem-Length of thread perature as the water.Glass rodThermometer 6. Using the balance, find the mass of the poly-100 mL graduated cylinder styrene coffee cup calorimeter. Fill theHot plate calorimeter with 100 mL of tap water. Find theBalance combined mass of the cup and water to find theSamples of metals (aluminum, copper, zinc, mass of the water added. Measure the tempera-lead, iron) ture of the water.400 mL beaker 7. Transfer the metal very quickly to theProcedure calorimeter by lifting the glass rod with both hands and sliding the thread and metal off1. Prepare a data table using the sample table the rod and into the water. provided. 8. Stir the water gently with the thermometer2. Prepare the lab equipment as shown in Fig. and record the highest temperature that the Lab.8.2. metal and water reach.3. Fill the 400 mL beaker half full with water 9. Repeat Steps 4–8 for as many other metals as and begin heating it on the hot plate, to be you can in the time that you have. used later in the lab.Fig.Lab.8.2 Move metal sample Glass rod Glass rod Beaker Thermometer Thread Polystyrene cup Water boiling Metal sampleElectric hot plate c h a pt e r 8 : Thermal Energy and Heat Transfer 279

LABORATORY EXERCISES Data Data Table Sample Mass of Initial temperature Initial temperature Final temperature ⌬t of Specific heat material sample (kg) water (°C) capacity of of sample/ of cool water (°C) of sample/ sample (J/kg°C) hot water (°C) water (°C) Uncertainty 2. Compare the value that you determined in the lab with the accepted values that you Assign both instrumental and procedural uncer- found in Question 1 above. Did your value tainties for each time you take a measurement. agree with the accepted value to within your The balance and thermometer should be meas- experimental uncertainty? ured to Ϯ.1 of the smallest division. 3. If your value did not agree with the accepted Analysis value, give possible reasons from your lab technique that could explain the difference. Use the principle of heat exchange to calculate the Be careful that your reasons make sense. specific heat capacity of each metal (⌬EH ϭ m⌬tc). Spilling water would affect a difference in Remember that the heat lost by the metal is the calculations that could quite easily be dis- heat gained by the water. missed by your final results. Discussion Conclusion 1. Look up the specific heat capacities of all the Summarize your experimental results by refer- metals you used in Table 8.4 and list them in ring to your original purpose. your notebook, including their units.280 u n i t B : Wo r k , E n e rg y, a n d Powe r

Special Relativity 9and Rest Energy Chapter Outline For image see student 9.1 It’s All In Your Point of View 9.2 Relative Motion and the text. Speed of Light 9.3 Implications of Special Relativity 9.4 Strange Effects of Special Relativity 9.5 Mass and Energy ST S E The Higgs BosonBy the end of this chapter, you will be able to• describe various types of motion using different frames of reference• describe and calculate the variations in time, length, and mass of objects moving at speeds close to the speed of light 281• outline the relationship between mass and energy

Fig.9.1 Relative motiong 9.1 It’s All In Your Point of ViewCo pplyina In 1905, Einstein’s theory of relativity shocked the world of physics by over- thets turning some long-held beliefs about the universe. Although it involves some ncep complex math, the theory of relativity is based on a very simple set of prin- ciples, leading to some interesting situations that challenge our imagination.282 Einstein understood that our understanding of the universe depended on our point of view, the frame of reference that we use to observe it. A funny thing happened on my way to school today that could be used to illustrate how important your “point of view” can be. While waiting for the light to turn green at an intersection, the car began to roll back- ward. At least that is what I thought when I rammed the brake even further into the floor. It turns out that I was not rolling backward at all, in fact I had not even moved a centimetre. The big bus in the lane to my right began to move forward with traffic and for an instant, I thought I was moving backward relative to the bus, even though it was the bus that was moving relative to me (see Fig. 9.1). The point of view from which we observe motion is called a frame of ref- erence. This is the stationary “platform” from which we judge or measure all other motion. Einstein noted that all the laws of mechanics apply the same way, whether you observe them from rest or from a frame of reference that is mov- ing at a constant speed in a straight line. Imagine that you are driving into the city with friends and your buddy passes you a soft drink. This can be done just as safely in the car moving at a constant speed as it could when the car is at rest. If the reference frame has a constant velocity, then it is in a state that can be described by Newton’s first law. Since that law is sometimes called the law of inertia, Einstein called this point of view an inertial frame of reference. Einstein’s first principle of relativity The laws of physics that describe changing circumstances are the same for all inertial frames of reference. 1. a) Inertial reference frames are those in which Newton’s laws of motion hold true. Consider a person sitting in a car with shiny, slippery seats. To a stationary observer, why would the person sit- ting in the car remain in the same spot if the car is still or moving forward or backward with a constant speed? What will happen to the person when the car now speeds up, slows down, or turns? b) To the person in the car, turning a corner causes him or her to accelerate sideways, with no apparent force being applied. Using the concept of frames of reference, explain why this effect is not a violation of Newton’s second law of motion. Note: In the above example, the force of friction between the tires and the road exerts an inward force on the car toward the centre of the curve. unit B: Work, Energy, and Power

This inward force is called the centripetal force. The slippery seatdoesn’t transfer this force to the person. 9.2 Relative Motion and the Speed of Light s៬vcWe opened this chapter with an example of how things can appear to be dif- Object Frame offerent depending on the frame of reference from which we choose toobserve them. In the soft drink example on the previous page, our car, its moving forward referenceoccupants, and their drinks are all moving relative to the ground. But rela-tive to the car, the drinks would not be moving. From the point of view of (soft drink can) (car)the drinks, the ground is moving at a constant speed toward them. In thenext example, illustrated in Fig. 9.2, using relative velocity, we can showhow things appear to move in different frames of reference. If one of yourmisguided soft drink swilling friends decides to throw his half-empty canout of the car at a velocity of 5 m/s (sv→c) forward with respect to the car,which is moving at 25 m/s (cv→g), then its velocity is added to the velocity ofthe car, 30 m/s (sv→g) relative to the ground. Fig.9.2 Relative velocities of typical objects A duck standing beside the road would observe the can moving forward 283at 30 m/s relative to the ground. We arrive at this answer by simply addingthe two velocity vectors, as we learned in Chapter 3. The same methodapplies to the relative velocities of all ordinary objects. According to Einstein’s second principle of relativity, one exception to thismethod of calculating relative velocities is the speed of light, c ϭ 3.0 ϫ 108 m/s. Figure 9.3 shows that if your friend in the car shines a flashlight forward(Lvc) as the car moves, the duck still sees the light travelling at velocity c, notc ϩ 25 m/s, as you might expect. No matter how fast the car moves, anobserver on the ground will see light from the car moving at 3.0 ϫ 108 m/s. Einstein’s second principle of relativity The speed of light, c, has the same fixed value (3.0 ϫ 108 m/s) for all observers. chapter 9: Special Relativity and Rest Energy

Fig.9.3 Relative velocity of light This principle of relativity has some startling implications. In 1853, Jean Foucault discovered that light waves travel at a constant rate of about 3.0 ϫ 108 m/s, which improved previous estimates from scientists such as James Clerk Maxwell and Galileo. The material or medium that acted as the frame of reference in which light waves were believed to travel was some- what of a mystery. Today, our cars have their speed measured with respect to the ground, boats with respect to the water, and air planes with respect to the air in which they move. The frame of reference in which the speed of light was to be measured (the light’s medium), the ether, was thought to be highly elastic so light could travel at such great speeds, but of low density so it wouldn’t impede the motion of the planets. The ether was thought to be the universal medium or frame of reference against which all other motions could be measured. In the 1800s, two American physicists, A.A. Michelson and E.W. Morley, designed an experiment to measure the speed of Earth with respect to the ether. This experiment used an extremely sensitive light- measuring apparatus, shown in Figs. 9.4 and 9.5, called an interferometer. Using the interferometer, scientists split a beam of light into two per- pendicular paths similar to the two-boat analogy shown in Fig. 9.6(b).Fig.9.4 The Michelson interferometer Fig.9.5 The optics of an interferometer Light source For image Mirrorsee student Telescope text. Half-silvered mirrors Mirror284 u n i t B : Wo r k , E n e rg y, a n d Powe r

Fig.9.6 Comparing Earth’s Source Mirrora lamotion in ether to motion din water Half-silvered mirror Detector vEarth lb(a) Mirrorb (b) If one of the paths was parallel to Earth’s motion and the other was per-pendicular to Earth’s motion, then the two beams would take differenttimes to travel equal distances. Take our beloved couple, Tarzan andTarzana, from Chapter 3. They take two separate but equidistant trips at thesame speed with respect to the water (Fig. 9.6(b)); one directly across andback, and the other downstream and back. We would fully expect them toarrive back at different times. But this is not what happens with light.Michelson and Morley couldn’t measure any time difference between thetwo paths of light, even when using a light interferometer that takes advan-tage of the interference of two light beams (covered in Chapter 12). Thissensitive device would have noticed a time difference as small as 2.0 ϫ 10Ϫ17 s.But no matter how they did their experiment (direction of device, day ortime of year), they always got the same result. This result is often referredto as the null result, which explains that an “unsuccessful” experiment inscience can be just as informative as a “successful” one. One explanation ofthe null result was that there was no ether or absolute frame of reference.Another explanation was that Earth was a passive passenger being pushedalong with the “ether wind.” Today, Einstein’s answer is accepted—there isno ether! The speed of light is constant in all directions for all observers.This concept is quite simple, but it has some bizarre implications!1. Einstein is looking in a mir- Fig.9.7 c gpplyin ror while sitting in a rocket Co the that’s moving at the speed a ncep of light. Explain in your ts own words what appears in 285 the mirror and the rationale behind your answer. chapter 9: Special Relativity and Rest Energy

9.3 Implications of Special Relativity Simultaneity — “Seeing is Believing” Normally, we judge that an event has occurred by observing the light from the event with our eyes. It sounds obvious. But, if light always moves at the finite speed, c, for all observers, this judgement is not as obvious as we thought. At a baseball game, we see the ball being hit, but we do not hear the “crack” of the bat until some time later. Even though light travels incred- ibly fast, the hit actually occurred before we saw it. The hit and the viewing of it don’t occur at the same time because it takes time for the light from the event to reach us. If we had been moving, especially at a high speed, our judgement of the event would be even less accurate.Fig.9.8 Exactly whendo events occur? A BA B O O In Fig. 9.8, the warning lights at A and B flash simultaneously. When the flash occurs, observers O and O′ are exactly halfway between A and B. Observer O is on the ground and observer O′ is moving at very high speed to the right. The light from A and B travels the same distance to observer O, and he observes the flashes from both A and B as occurring at the same time. Observer O′, however, is moving at high speed toward B. But light trav- els at speed c for all observers. Since observer O′ is approaching B and reced- ing from A, he sees the flash of light from B before the flash from A. Because of these discrepancies, one person’s observations and judgements about simultaneous events can’t be considered more correct than another’s. What you see depends on your frame of reference. Simultaneity If any two observers are moving with respect to one another, then their judgement of simultaneous events may not necessarily agree.286 u n i t B : Wo r k , E n e rg y, a n d Powe r

Fig.9.9 Simultaneity connecti ts Event vϽc ngthe occurs Concep MovingStationary No agreement observer observer between observers 00:03s 00:01sTime DilationThe fact that light has the same speed for all observers even affects theresults of simple kinematics situations when the behaviour of light isobserved by fast-moving observers. Consider what happens to light from aflashlight inside a boxcar that is moving at a speed close to the speed of light,as shown in Fig. 9.10. Fig.9.10 How a stationary observer sees light from a fast-moving train OЈ D ϭ 1/2ct L 1/2vt t(a) (c) OЈ OЈ OЈ 287 v ⌬t (b) OThe fact that two observers may see the same event occur at different times sug-gests that time is not absolute. Could time pass at a different rate in differentreference frames? This possibility is exactly what Einstein’s theory predicts. Consider observer O′ in Fig. 9.10(a) shining a flashlight towards themirror on the ceiling of the boxcar. He sees light from the flashlight hit themirror and reflect back, travelling a distance of 2L in time t0 (t0 ϭ ᎏ2cLᎏ).Consider observer O on the ground in Fig. 9.10(b). If the boxcar is passing chapter 9: Special Relativity and Rest Energy

DEDUCTION OF TIME DILATION by him at near the speed of light, then the light from the flashlight appears to travel diagonally across the car to the mirror and back, a larger distanceFrom Fig. 9.10(c), at the same speed, c. The time measured by observer O is longer than that measured by observer O′ in the boxcar. This effect is a general result of the D ϭ 1/2ct L theory of relativity and is called time dilation. 1/2vt t Time Dilation For a stationary observer, moving clocks appear to runD2 ϭ L2 ϩ (2ᎏ1ᎏvt)2 slower than they do for someone moving with the clock. For the moving observer, however, it is the clock of the stationary person that appears towhere v is the speed of the be running slower. The moving and the stationary observers each believeboxcar. that the other’s clock is the one that is running slow. ᎏcᎏt 2 ϭ L2 ϩ ᎏvᎏt 2 Analyzing Fig. 9.10 for relative distances and solving for t, the station- ary observer’s time, we get the following relativistic equation.΂ ΃ ΂ ΃2 2 t ϭ ᎏtoᎏc2ᎏt2 ϭ L2 ϩ ᎏv2ᎏt2 44 Ί๶1 Ϫ ᎏvcᎏ22c2t2 ϭ 4L2 ϩ v2t2 where t is the relativistic time, which is the time measured by the station- ary observer.t2(c2 Ϫ v2) ϭ 4L2 to is the time measured by the moving observer.ϭ ᎏ2L ϭ ᎏc 12ϪᎏL ᎏcvᎏ22 v is the speed of the object and c is the speed of light, 3.0 ϫ 108 m/s. ͙cෆ2 Ϫ vෆ2 e x a m p l e 1 Time dilation at speeds close to the speed of lightΊ๶tRecall that for the observer O′in the car, to ϭ ᎏ2ᎏL c ᎏt0ᎏ 1 Ϫ ᎏcvᎏ22Ί๶๶Іt ϭ A student is late for school and decides to drive to school in her new hi-tech car of rest mass 1.50 ϫ 103 kg and length 4.2 m at a speed of 2.5 ϫ 108 m/s. If the entire trip at that speed took 2.0 s according to the student, how long is the trip according to the stationary school principal? Solution and Connection to Theory Given Lo ϭ 4.2 m v ϭ 2.5 ϫ 108 m/s to ϭ 2.0 s mo ϭ 1.50 ϫ 103 kg c ϭ 3.0 ϫ 108 m/s t ϭ ? ϭ ᎏto ϭ ᎏᎏ2.0 s ᎏ ϭ ᎏ2.0 s 1 Ϫ ᎏvcᎏ22 1 Ϫ ᎏ((32..50 ϫϫ 11ᎏ0088 mm//ss))22 ͙ෆ0.306 Ί๶ Ί๶๶๶๶t t ϭ ᎏ02..50ᎏ5s3 ϭ 3.6 s The principal would observe the student’s trip to have taken 3.6 s.288 u n i t B : Wo r k , E n e rg y, a n d Powe r