591_PartUniversity Physics Solution

591_PartUniversity - 20-6 Chapter 20 QC = mLf = 0.0400 kg 334 103 J/kg = 1.336 104 J EXECUTE QC T = C gives QH TH QH =(TH TC QC = 1.336 104 J

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20-6 Chapter 20 EXECUTE: () ( ) 34 Cf 0.0400 kg 334 10 J/kg 1.336 10 J. Qm L =− × × CC HH QT = − gives ( ) 44 C C 1.336 10 J 373.15 K 273.15 K 1.825 10 J. T Q =− − × =+ × ⎡⎤ ⎣⎦ 3 CH 4.89 10 J. WQ Q =+= × EVALUATE: For a heat engine, C Q is negative and H Q is positive. The heat that comes out of the engine (0 ) Q < goes into the ice ( 0 Q > ). 20.23. IDENTIFY: The power output is . W P t = The theoretical maximum efficiency is C Carnot H 1. T e T H . W e Q = SET UP: 4 H 1.50 10 J. Q C 350 K. T = H 650 K. T = 1 hp 746 W. = EXECUTE: C Carnot H 350 K 1 1 0.4615. 650 K T e T = 43 H (0.4615)(1.50 10 J) 6.923 10 J; We Q == × = × this is the work output in one cycle. 3 4 (240)(6.923 10 J) 2.77 10 W 37.1 hp. 60.0 s W P t × = × = EVALUATE: We could also use = − to calculate C H 350 K (1.50 10 J) 8.08 10 J. 650 K T QQ T ⎛⎞ × × ⎜⎟ ⎝⎠ Then 3 6.92 10 J, the same as previously calculated. 20.24. IDENTIFY and SET UP: C Carnot H T e T C Carnot HC . T K TT = EXECUTE: (a) (1 ). TT e H ) 1 . ) Te e K TT e e −− EVALUATE: (b) When 1, e 0. K When 0, e . K →∞ 1 e when . << C Q is small in this limit. That is good for an engine since C Q is wasted. But it is bad for a refrigerator since C Q is what is useful. 0 e when and W is very small. That is bad for an engine but good for a refrigerator. 20.25. IDENTIFY: Q S T Δ= for each object, where T must be in kelvins. The temperature of each object remains constant. SET UP: For water, 5 f 3.34 10 J/kg. L EXECUTE: (a) The heat flow into the ice is 55 f (0.350 kg)(3.34 10 J/kg) 1.17 10 J. L × The heat flow occurs at 273 K, T = so 5 1.17 10 J 429 J/K. 273 K Q S T × Δ= = = Q is positive and S Δ is positive. (b) 5 Q × flows out of the heat source, at 298 K. T = 5 393 J/K. 298 K Q S T −× = Q is negative and S Δ is negative. (c) tot 429 J/K ( 393 J/K) 36 J/K. S + = + EVALUATE: For the total isolated system, 0 S Δ > and the process is irreversible. 20.26. IDENTIFY: Apply system 0 Q = to calculate the final temperature. . cT = Δ Example 20.6 shows that 21 ln( / ) Sm cTT when an object undergoes a temperature change. SET UP: For water 4190 J/kg K. c =⋅ Boiling water has 100.0 C 373 K. T ° EXECUTE: (a) The heat transfer between 100 C ° water and 30 C ° water occurs over a finite temperature difference and the process is irreversible. (b) 22 (270 kg) ( 30.0 C) (5.00 kg) ( 100 C) 0. −+ = °° 2 31.27 C 304.42 K. T ° (c) 304.42 K 304.42 K (270 kg)(4190 J/kg K)ln (5.00 kg)(4190 J/kg K)ln . 303.15 K 373.15 K S + 4730 J/K ( 4265 J/K) 470 J/K. S +− = + EVALUATE: system 0, S Δ> as it should for an irreversible process. 20.27. IDENTIFY: Both the ice and the room are at a constant temperature, so . Q S T For the melting phase transition, f . L = Conservation of energy requires that the quantity of heat that goes into the ice is the amount of heat that comes out of the room.
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The Second Law of Thermodynamics 20-7 SET UP: For ice, 3 f 334 10 J/kg. L When heat flows into an object, 0, Q > and when heat flows out of an object, 0. Q < EXECUTE: (a) Irreversible because heat will not spontaneously flow out of 15 kg of water into a warm room to freeze the water.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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591_PartUniversity - 20-6 Chapter 20 QC = mLf = 0.0400 kg 334 103 J/kg = 1.336 104 J EXECUTE QC T = C gives QH TH QH =(TH TC QC = 1.336 104 J

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