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Unformatted text preview: PHY 317K Homework 7 Solutions 1. D If two objects are in thermal equilibrium, by definition they have the same temperature. 2. E A thermometer must distinguish between different temperatures, so it is necessary to use a substance that undergoes some measurable change when its temperature changes. 3. D We are interested in the change in pressure when the temperature in this thermometer changes by Δ T = 1K. For a constant volume gas thermometer p/T = const . Therefore, p T = p + Δ p T + Δ T , so p + Δ p = p T ( T + Δ T ) = p + p T Δ T . Thus, Δ p = p T Δ T . Then: Δ p = 5 . 55 × 10 4 Pa 273 . 16K (1K) = 203Pa. 4. B For a circle (like a coin), A = πr 2 = πd 2 / 4, where d is the diameter. If d changes by Δ d , the area becomes A + Δ A = π ( d + Δ d ) 2 / 4 = πd 2 / 4 + 2 πd Δ d/ 4 + π (Δ d ) 2 / 4. Therefore, Δ A = 2 πd Δ d/ 4, if we ignore the third term since Δ d is very small compared to d . Then if we multiply the left side by d/d , we see Δ A = 2( πd 2 / 4)Δ d/d , from which it follows: Δ A/A = 2Δ d/d . Hence, a 0.17% change in d results in a 0.34% change in the area of the coin. 5. C For coefficient of linear expansion α , Δ V V = 3 α Δ T . Thus, here Δ V = 3 αV Δ T = 3(1 . 1 × 10 5 /C ◦ )(4 . 00cm) 3 (57C ◦ ) = 0 . 12cm 3 . 6. A Heat is energy transfered due to a temperature difference, by definition. 7. B Heat has the same unit as work. 8. A The heat capacity of an object is equal to the amount of heat energy needed to raise its temperature by 1C ◦ (or equivalently,1K). 1 9. D The specific heat of a substance is equal to the amount of heat energy per unit mass needed to raise the temperature of the substance by 1C ◦ (or equivalently, 1K). 10. D The heat capacity at constant volume and the heat capacity at constant pressure have different values because the system does work at constant pressure, but not at constant volume. I.e., when the volume changes, work is done. 11. D From the change in internal energy we can find the change in temperature: Δ T = Δ E int mc , where m is the mass of the cube and c is the specific heat. We are told the density ρ of aluminum is 2 . 7g / cm 3 and the specific heat is c = 0 . 217cal / g · C ◦ . If a is the edge length of the cube, m = ρa 3 = (2 . 7g / cm 3 )(15 . 9cm 3 ) = 10850g. Then: Δ T = 94000cal (10850g)(0 . 217 cal / g · C ◦ ) = 40C ◦ . 12. B If the objects are thermally isolated then the amount of heat that one object will lose is equal to the amount of heat the other object will gain: C A ( T final T A ) = C B ( T B T final ) Solving for T final we find: T final = C A T A + C B T B C A + C B . 13. B Using the formula from previous problem and C B = 2 C A we find T final = C A T A + C B T B C A + C B = C A ( T A + 2 T B ) 3 C A = T A + 2 T B 3 = 200K + 2(400K) 3 = 333K ....
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This note was uploaded on 12/08/2008 for the course PHY 317k taught by Professor Kopp during the Fall '07 term at University of Texas at Austin.
 Fall '07
 KOPP
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