Thermodynamics is the study of heat. It s what comes into play when you drop an ice cube
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1 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 wait 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 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 (formerly 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= 5 ^F-32h 9 F= 9 C+32 5 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 Celsius, which means that you can convert between Celsius and Kelvin this way: K = C C = K To convert from Kelvin to Fahrenheit degrees, use this formula: F= 9 K ^ - h+ = K Technically, you don t say degrees Kelvin but rather Kelvins, as in 53 Kelvins. However, people persist in using degrees Kelvin, so you may see that usage in this book as well.
2 218 Part IV: Obeying the Laws of Thermodynamics Q. What is 54 Fahrenheit in Celsius? A. The correct answer is 12 C. 1. Use this equation: C= 5 ^F-32h 9 2. Plug in the numbers: C= 5 ^F- 32h= ^0. 55h = 12cC 9 $ ^ h 1. What is 23 Fahrenheit in Celsius? 2. What is 89 Fahrenheit in Celsius? 3. What is 18 Celsius in Fahrenheit? 4. What is 18 Celsius in Kelvin?
3 Chapter 12: You re Getting Warm: Thermodynamics What is 18 Kelvin in Celsius? 6. What is 57 Kelvin in Fahrenheit? 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 thermal 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. A Temperature = T Length = L Figure 12-1: Linear expansion. B Temperature = T + T Length = L + L 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 f = T o + T
4 220 Part IV: Obeying the Laws of Thermodynamics Linear expansion results in an expansion in any linear direction of the following: L f = L o + L If the temperature goes down a small amount, this equation applies: T f = T o T You get a contraction instead of an expansion: L f = L o L Like the coefficient of friction, a coefficient is in play here the coefficient of linear expansion, which is given the symbol α. So you can write this: L = α T L This equation is usually written in this form: o L = α L o T Here, α is usually measured in 1/ C, or C 1. Q. You re heating a 1.0 m steel bar, coefficient of linear expansion C 1, raising its temperature by 5 C. What is the final length of the bar? A. The correct answer is m. 1. Use this equation: 2. Plug in the numbers: L = α L o T = ( ) (1.0) (5) = m 3. The final length is L f = L o + L = m L = α L o T 7. You re heating a 1.0 m aluminum bar, coefficient of linear expansion C 1, raising its temperature by 100 C. What is the final length of the bar? 8. You re heating a 2.0 m gold bar, coefficient of linear expansion C 1, raising its temperature by 200 C. What is the final length of the bar?
5 Chapter 12: You re Getting Warm: Thermodynamics You re heating a 1.5 m copper bar, coefficient of linear expansion C 1, raising its temperature by 300 C. What is the final length of the bar? 10. You re heating a 2.5 m lead bar, coefficient of linear expansion C 1, raising its temperature by 40 C. What is the final length of the bar? 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: V fraction the solid exp ands is proportionalto T change in temperature V _ i _ i o The constant involved in volume expansion is called the coefficient of volume expansion. This constant is given by the symbol β, and like α, it s measured in C 1. Using β, here s how you can express the relationship shown in the preceding equation: V = β T V When you multiply both sides by V o, you re left with the following: Which means that: o V = β V o T V f = V o + β V o T
6 222 Part IV: Obeying the Laws of Thermodynamics Q. You re heating a 1.0 m 3 steel block, coefficient of volume expansion C 1, raising its temperature by 45 C. What is the final volume of the bar? A. The correct answer is m Use this equation: 2. Plug in the numbers: V = β V o T = ( ) (1.0) (45) = m 3 3. The final volume is V f = V o + V = m 3 V = β V o T 11. You re heating a 2.0 m 3 aluminum block, coefficient of volume expansion C 1, raising its temperature by 30 C. What is the final volume of the block? 12. You re heating a 2.0 m 3 copper block, coefficient of volume expansion C 1, raising its temperature by 20 C. What is the final volume of the block?
7 Chapter 12: You re Getting Warm: Thermodynamics You re heating a 1.0 m 3 glass block, coefficient of volume expansion C 1, raising its temperature by 27 C. What is the final volume of the block? 14. You re heating a 3.0 m 3 gold block, coefficient of volume expansion C 1, raising its temperature by 18 C. What is the final volume of the block? Getting Specific with Heat Capacity It s a fact of physics that it takes 4186 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 1 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 = m c T 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, T 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. So it takes 4186 J of heat energy 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 J. Nutritionists use the food energy term Calorie (capital C) to stand for 1000 calories, 1.0 kcal, so 1.0 Calorie equals 4186 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. To convert Btus to Joules, use the relation that 1 Btu equals 1055 J. If you add heat to an object, raising its temperature from T o to T f, the amount of heat you need is expressed as: Q = m c (T f T o )
8 224 Part IV: Obeying the Laws of Thermodynamics Q. You re heating a 1.0 kg copper block, specific heat capacity of 387 J/(kg- C), raising its temperature by 45 C. What amount of heat do you have to apply? A. The correct answer is 17,400 J. 1. Use this equation: Q = m c T 2. Plug in the numbers: Q = m c T = (387) (1.0) (45) = 17,400 J 15. You re heating a 15.0 kg copper block, specific heat capacity of 387 J/(kg- C), raising its temperature by 100 C. What heat do you have to apply? 16. You re heating a 10.0 kg steel block, specific heat capacity of 562 J/(kg- C), raising its temperature by 170 C. What heat do you have to apply?
9 Chapter 12: You re Getting Warm: Thermodynamics You re heating a 3.0 kg glass block, specific heat capacity of 840 J/(kg- C), raising its temperature by 60 C. What heat do you have to apply? 18. You re heating a 5.0 kg lead block, specific heat capacity of 128 J/(kg- C), raising its temperature by 19 C. What heat do you have to apply? 19. You re cooling a 10.0 kg lead block, specific heat capacity of 128 J/(kg- C), lowering its temperature by 60 C. What heat do you have to extract? 20. You re cooling a 80.0 kg glass block, specific heat capacity of 840 J/(kg- C), lowering its temperature by 16 C. What heat do you have to extract?
10 226 Part IV: Obeying the Laws of Thermodynamics 21. You put 7600 J into a 14 kg block of silver, specific heat capacity of 235 J/(kg- C). How much have you raised its temperature? 22. You add 10,000 J into a 8.0 kg block of copper, specific heat capacity of 387 J/(kg- C). How much have you raised its temperature? 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 liquefies, 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 solid to liquid. 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 f, is the heat per kilogram needed to make the change between the solid and liquid phases (such as when water turns to ice). Liquid to gas: The latent heat of vaporization, L v, 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 s, 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 2 ) to the vapor state). The latent heat of fusion of water is about J/kg. That means it takes J of energy to melt 1 kg of ice.
11 Chapter 12: You re Getting Warm: Thermodynamics 227 Q. You have a glass of 50.0 g of water at room temperature, 25 C, but you d prefer ice water at 0 C. How much ice at 0.0 C do you need to add? A. The correct answer is 15.6 g. 1. The heat absorbed by the melting ice must equal the heat lost by the water you want to cool. Here s the heat lost by the water you re cooling: Q water = m c T = m c (T f T o ) 2. Plug in the numbers: Q water = m c T = m c (T f T o ) = (0.050) (4186) (0 25) = J 3. So the water needs to lose J. How much ice would that melt? That looks like this, where L m is the latent heat of melting: 4. You know that for water, L m is J/kg, so you get this: Q = m ice L m = m ice You know that equation has to be equal to the heat lost by the water, so you can set it to: In other words: m ice Q ice = Q water 3 Q water = = 523. # 10 J 5 L m 335. # 10 Jkg / 6. You know that the latent heat of melting for water is L = 3.35 x 10 5 J/kg, which means that: m 3 = 523. # 10 = 156. # # 10 ice 5-2 kg So you need kg, or 15.6 g of ice. Q ice = m ice L m 23. You have g of coffee in your mug at 80 C. How much ice at 0.0 C would it take to cool g of coffee at 80 C to 65 C? 24. You have g of cocoa at 90 C. How much ice at 0.0 C do you have to add to the cocoa (assuming that it has the specific heat capacity of water) to cool it down to 60 C?
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