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591_PartUniversity Physics Solution

# 591_PartUniversity Physics Solution - 20-6 Chapter 20 QC =...

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20-6 Chapter 20 E XECUTE : ( ) ( ) 3 4 C f 0.0400 kg 334 10 J/kg 1.336 10 J. Q mL = − = − × = − × C C H H Q T Q T = − gives ( ) ( ) ( ) ( ) 4 4 H H C C 1.336 10 J 373.15 K 273.15 K 1.825 10 J. Q T T Q = − = − − × = + × 3 C H 4.89 10 J. W Q Q = + = × E VALUATE : 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. I DENTIFY : The power output is . W P t = The theoretical maximum efficiency is C Carnot H 1 . T e T = H . W e Q = S ET U P : 4 H 1.50 10 J. Q = × C 350 K. T = H 650 K. T = 1 hp 746 W. = E XECUTE : C Carnot H 350 K 1 1 0.4615. 650 K T e T = = = 4 3 H (0.4615)(1.50 10 J) 6.923 10 J; W eQ = = × = × 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 × = = = × = E VALUATE : We could also use C C H H Q T Q T = − to calculate 4 3 C C H H 350 K (1.50 10 J) 8.08 10 J. 650 K T Q Q T = − = − × = − × Then 3 C H 6.92 10 J, W Q Q = + = × the same as previously calculated. 20.24. I DENTIFY and S ET U P : C Carnot H 1 . T e T = C Carnot H C . T K T T = E XECUTE : (a) C H (1 ). T T e = H H H (1 ) 1 . (1 ) T e e K T T e e = = E VALUATE : (b) When 1, e 0. K When 0, e . K → ∞ 1 e when C H . Q Q << 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 C H Q Q and W is very small. That is bad for an engine but good for a refrigerator. 20.25. I DENTIFY : Q S T Δ = for each object, where T must be in kelvins. The temperature of each object remains constant. S ET U P : For water, 5 f 3.34 10 J/kg. L = × E XECUTE : (a) The heat flow into the ice is 5 5 f (0.350 kg)(3.34 10 J/kg) 1.17 10 J. Q mL = = × = × 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 1.17 10 J Q = − × flows out of the heat source, at 298 K. T = 5 1.17 10 J 393 J/K. 298 K Q S T × Δ = = = − Q is negative and S Δ is negative. (c) tot 429 J/K ( 393 J/K) 36 J/K. S Δ = + − = + E VALUATE : For the total isolated system, 0 S Δ > and the process is irreversible. 20.26. I DENTIFY : Apply system 0 Q = to calculate the final temperature. . Q mc T = Δ Example 20.6 shows that 2 1 ln( / ) S mc T T Δ = when an object undergoes a temperature change. S ET U P : For water 4190 J/kg K. c = Boiling water has 100.0 C 373 K. T = = ° E XECUTE : (a) The heat transfer between 100 C ° water and 30 C ° water occurs over a finite temperature difference and the process is irreversible. (b) 2 2 (270 kg) ( 30.0 C) (5.00 kg) ( 100 C) 0. c T c T + = ° ° 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 Δ = + 4730 J/K ( 4265 J/K) 470 J/K. S Δ = + − = + E VALUATE : system 0, S Δ > as it should for an irreversible process. 20.27. I DENTIFY : Both the ice and the room are at a constant temperature, so . Q S T Δ = For the melting phase transition, f . Q mL = 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 S ET U P : 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 < E XECUTE : (a)
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