Chapter 12 You're Getting Warm: Thermodynamics In This Chapter Converting between temperature scales Working with linear expansion Calculating volume expansion Using heat capacities Understanding latent heat Thermodynamics is the study of heat, It's what comes into play when you drop an ice cube Into a cup of hot tea and walt to see what happens - If the ice cube or the tea wins out. In physics, you often run across questions that Involve thermodynamics in all sorts of situations. This chapter refreshes your understanding of the topic and lets you put it to use with practice problems that address thermodynamics from all angles. Converting between Temperature Scales EMBER You start working with questions of heat by establishing a scale for measuring temperature. The temperature scales that you work with in physics are Fahrenheit, Celsius (sometimes called centigrade), and Kelvin. Fahrenheit temperatures range from 32" for freezing water to 212" for boiling water. Celsius goes from 0" for freezing water to 100' for boiling water. Following are the equations you use to convert from Fahrenheit (F) temperatures to Celsius (C) and back again: C=2(F-320) F =&C+32 The Kelvin (K) scale is a little different: Its 0 corresponds to absolute zero, the temperature at which all molecular motion stops. Absolute zero is at a temperature of -273.15' Celsius, which means that you can convert between Celsius and Kelvin this way. K = C+ 273.15 C= K-273.15 To convert from Kelvin to Fahrenheit degrees, use this formula: F = (K-273.15)+32 = K -459.67Part IV: Obeying the Laws of Thermodynamics REMEMBER Technically, you don't say "degrees Kelvin" but rather "Kelvins," as in 53 Kelvins. However, people persist In using "degrees Kelvin." Q. What is 54' Fahrenheit in Celsius? 2. Plug In the numbers: A. The correct answer is 12"C. C=(F-320)=(0.56)(54-329) =12.C 1. Use this equation: C= 3(F-320) 7. What is 23' Fahrenheit in Celsius? 2. What is 89 Fahrenheit in Celsius? Solve It Solve it 3. What Is 18 Celsius in Fahrenheit?. 4. What is 18 Celsius in Kelvin? Solve it Solve It 5. What is 18 Kelvins in Celsius? 6. What is 57 Kelvins In Fahrenheit? Solve It Solve ItChapter 12: You're Getting Warm: Thermodynamics 275 Getting Bigger: Linear Expansion Ever try to open a screw-top jar by running hot water over It? That hot water makes the lid of the jar expand, making It easier to turn. This simple solution is physics on the Job - it's all about thennal expansion. You can see an example of thermal expansion In Figure 12-1, where a bar is undergoing expansion in one direction, called linear expansion. Temperature = T Length = L Figure 12-1: Temperature = T+AT Linear B expansion. Length = L+AL D John Miley a Song, Inc. When you talk about the expansion of a solid in any one dimension under the Influence of heat, you're talking about linear expansion. When you raise the temperature a small amount, this equation applies: T,= T. + AT Linear expansion results in an expansion in any linear direction of the following: Ly - L. + AL If the temperature goes down a small amount, this equation applies: T,=T.-AT You get a contraction instead of an expansion: Ly =L. -AL Like the coefficient of friction, a coefficient is in play here - the coefficient of linear expansion, which is given the symbol a. So you can write this: AL -GAT16 Part IV: Obeying the Laws of Thermodynamics This equation Is usually written In this form: AL = aLAT Which means: Ly= L. + aLAT Here, a is usually measured in 1/"C, or "C-1. 0. You're heating a 1.0 m steel bar, coeffi- 2. Plug In the numbers: clent of linear expansion 1.2 x 10-5"C-1, raising its temperature by 5*C. What is AL = a L AT = (1.2 x 10-5)(1.0)(5) the final length of the bar? = 6.0 x 10-5 m = 0.00006 m 3. The final length is A. The correct answer is 1.00006 m. 1. Use this equation: Ly= L. + AL = 1.00006 m AL = a L AT 7. You're heating a 1.0 m aluminum bar, coeffi- 8. You're heating a 2.0 m gold bar, coefficient cient of linear expansion 2.3 x 10- C-, of linear expansion 1.4 x 10-5.C-1, raising Its raising its temperature by 100*C. What Is temperature by 200"C. What Is the final the final length of the bar? length of the bar? Solve it Solve ItChapter 12: You're Getting Warm: Thermodynamics 2 17 9. You're heating a 1.5 m copper bar, coeffi- 10. You're heating a 2.5 m lead bar, coefficient clent of linear expansion 1.7 x 10-"C-1, rals- of linear expansion 2.9 x 10-5.C-1, raising its Ing its temperature by 300"C. What Is the temperature by 40"C. What is the final final length of the bar? length of the bar? Solve It Solve It Plumping It Up: Volume Expansion In addition to linear expansion, physics problems can ask you to find volume expansion. Linear expansion takes place in only one dimension, but volume expansion happens in all three dimensions. In other words, you have this: (traction the solid expands) is proportional to AT (change In temperature) The constant Involved in volume expansion Is called the coefficient of volume expansion. This constant is given by the symbol f, and like a, it's measured in "C-. Using f, here's how you can express the relationship shown in the preceding equation: AV - BAT When you multiply both sides by V., you're left with the following: AV -BV AT Which means: V,- V.+BV.AT18 Part IV: Obeying the Laws of Thermodynamics JAPLE 2. Plug in the numbers: Q. You're heating a 1.0 m' steel block, coeffi- clent of volume expansion 3.6 x 10-5"C-!, AV = BV. AT - (3.6 x 10-5)(1.0)(45) raising Its temperature by 45*C. What Is the final volume of the bar? = 1.6 x 10-3 m' = 0.0016 m3 A. The correct answer Is 1.0016 m'. 3. The final volume is 1. Use this equation: V,= V. + BV. AT = 1.0016 m3 AV = BV AT 1 1. You're heating a 2.0 m' aluminum block, 12. You're heating a 2.0 m' copper block, coel- coefficient of volume expansion 6.9 x ficlent of volume expansion 5.1 x 10-5 C-. 10-5"C-1, raising its temperature by 30"C. raising Its temperature by 20"C. What Is the What is the final volume of the block? final volume of the block? Solve It Solve it 13. You're heating a 1.0 m' glass block, coeffi- clent of volume expansion 1.0 x 10-5 C-1, 14. You're heating a 3.0 m' gold block, coeffi- raising its temperature by 27"C. What is the cient of volume expansion 4.2 x 10-5"C-1, final volume of the block? raising Its temperature by 18"C. What is the Solve It final volume of the block? Solve itChapter 12: You're Getting Warm: Thermodynamics 219 Getting Specific with Heat Capacity It's a fact of physics that it takes 4,186 J to raise the temperature of 1.0 kg of water by 1"C. But it takes only 840 J to raise the temperature of 1.0 kg of glass by I"C. You can relate the amount of heat, Q, it takes to raise the temperature of an object to the change in temperature and the amount of mass involved. Use this equation: Q = mcAT In this equation, Q is the amount of heat energy involved (measured in Joules if you're using the MKS system), m is the amount of mass, AT is the change in temperature, and c is a constant called the specific heat capacity, which is measured in J/(kg-"C) in the MKS system. BEA So It takes 4,186 J of heat to warm up 1.0 kg of water 1.0'C. One calorie is defined as the amount of heat needed to heat 1.0 g of water 1.0"C, so 1 calorie equals 4.186 J. Nutritionists use the food energy term Calorie (capital C) to stand for 1,000 calories, 1.0 kcal, so 1.0 Calorie equals 4,186 J. And when you're speaking in terms of heat, you" have another unit of measurement to deal with: the British thermal unit (BTU). 1.0 BTU is the amount of heat needed to raise one pound of water 1.0'F and is a unit often used to measure energy in the HVAC industry. To convert BTUs to Joules, use the relation that 1 Btu equals 1,055 J. If you heat an object, raising its temperature from T, to T,, the amount of heat you need is expressed as: AQ = mc(T,-T.) Q. You're heating a 1.0 kg copper block, 2. Plug in the numbers: specific heat capacity of 387 J/(kg-"C), raising its temperature by 45*C. What Q = mcAT = (387 J/(kg-"C))(1.0 kg)(45*c) amount of heat do you have to apply? = 17,400 J A. The correct answer is 17,400 J. 1. Use this equation: Q = mcAT20 Part IV: Obeying the Laws of Thermodynamics 16. You're heating a 10.0 kg steel block, spe- 15. You're heating a 15.0 kg copper block, spe- cific heat capacity of 562 J/(kg-"C), raising clfic heat capacity of 387 J/(kg-"C), raising Its temperature by 170"C. What heat do Its temperature by 100*C. What heat do you have to apply? you have to apply? Solve It Solve It 17. You're heating a 3.0 kg glass block, specific 18. You're heating a 5.0 kg lead block, specific heat capacity of 840 J/(kg-"C), raising Its heat capacity of 128 J/(kg-"C), raising Its temperature by 60"C. What heat do you temperature by 19"C. What heat do you have to apply? have to apply? Solve it Solve it 19. You're cooling a 10.0 kg lead block, specific 20. You're cooling an 80.0 kg glass block, spe- heat capacity of 128 J/(kg-"C), lowering its cific heat capacity of 840 J/(kg-"C), lowering temperature by 60*C. What heat do you its temperature by 16"C. What heat do you have to extract? have to extract? Solve It Solve It 21. You put 7,600 J into a 14 kg block of silver, specific heat capacity of 235 J/(kg-"C). How 22. You add 10,000 J Into a 8.0 kg block of much have you raised its temperature? copper, specific heat capacity of 387 J/(kg-"C). How much have you raised Solve it its temperature? Solve ItChapter 12: You're Getting Warm: Thermodynamics 227 Changes of Phase: Latent Heat Heating blocks of lead is fine, but if you heat that lead enough, sooner or later It's going to melt. When It melts, Its temperature stays the same until it liquefles, and then the temperature of the lead Increases again as you add heat. So why does its temperature stay constant as it melts? Because the heat you applied went into melting the lead. There's a latent heat of melting that means that so many Joules must be applied per kilogram to make lead change phase from solld to liquid. ABER The units of latent heat are J/kg. There are three phase changes that matter can go through - solid, liquid, and gas - and each transition has a latent heat: Solid to liquid: The latent heat of melting (or heat of fusion), L. or Ly, Is the heat per kilogram needed to make the change between the solid and liquid phases (such as when ice turns to water). Liquid to gas: The latent heat of vaporization, L, Is the heat per kilogram needed to make the change between the liquid and gas stages (such as when water boils). Solid to gas: The latent heat of sublimation, L, Is the heat per kilogram needed to make the change between the solid and gas phases (such as the direct sublimation of dry ice (CO?) to the vapor state). The latent heat of fusion of water is about 3.35 x 105 J/kg. That means it takes 3.35 x 105 J of energy to melt 1 kg of ice. You have a glass of 50.0 g of water at 3. So the water needs to lose 5.23 x 10" J. room temperature, 25*C, but you'd prefer How much ice would that melt? That ice water at 0'C. How much Ice at 0.0"C looks like this, where &, is the latent do you need to add? heat of melting: A. The correct answer is 15.6 g. 1. The heat absorbed by the melting ice 4. You know that for water, L is must equal the heat lost by the water 3.35 x 105 J/kg, so you get this: you want to cool. Here's the heat lost by the water you're cooling: AQke " malm = mix(3.35 x 105 J/kg) AQ water " mcAT = mc(T,-Ta 5. You know that equation has to be equal to the heat lost by the water, so 2. Plug In the numbers: you can set It to: AQ water = mcAT AQke = AQ eater = mc(T, -T.) =(0.050 kg) ( 4,186 J)(09C-25.C) In other words: =-5.23x10' J mice = gen - 5.23x10' J 3.35 x 10' J/kg