AP7C7C~1

# AP7C7C~1 - Assignment#7 Solutions 20.2 a 9000 J 6400 J =...

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Assignment #7 Solutions! 20.2: a) 9000 J 6400 J = 2600 J. b) 2600 J 9000 J = 0.289 = 28.9%. 20.8: a) From Eq. (20.6), e = 1 r 1 γ = 1 (8.8) 0.40 = 0.58 = 58%. b) 1 (9.6) 0.40 = 60%, an increase of 2%. If more figures are kept for the efficiencies, the difference is 1.4%. 20.10: P = W Δ t = Q C K Δ t = 1 K Δ m Δ t (L f + c p Δ T ) = 1 2.8 8.0 kg 3600s (1.60 × 10 5 J kg) + (485 J kg K)(2.5K) () = 128 W. 20.17: For all cases, W = Q H Q C . a) The heat is discarded at a higher temperature, and a refrigerator is required; W = Q C ((T H T C ) 1) = (5.00 × 10 3 J) ((298.15 263.15) = 665 J. b) Again, the device is a refrigerator, and W = Q C ((273.15 /263.15) = 190 J. c) The device is an engine; the heat is taken form the hot reservoir, and the work done by the engine is W = (5.00 × 10 3 J) ((248.15 263.15) = 285 J. 20.19: The total work that must be done is W tot = mgy = (500 kg)(9.80 m s 2 )(100 m) = 4.90 × 10 5 J Q H = 250 J Find Q C so can calculate work W done each cycle: Q C Q H =− T C T H Q C (T C T H )Q H (250 J) (373.15K) (773.15K) [ ] 120.7 J W = Q C + Q H = 129.3 J The number of cycles required is W tot W = 4.09 × 10 5 J 129.3 J = 3790 cycles.

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## This note was uploaded on 03/28/2011 for the course APSC 254 taught by Professor Stephen during the Spring '11 term at UBC.

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AP7C7C~1 - Assignment#7 Solutions 20.2 a 9000 J 6400 J =...

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