634_Physics ProblemsTechnical Physics

634_Physics ProblemsTechnical Physics - 636 P22.8 Heat...

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636 Heat Engines, Entropy, and the Second Law of Thermodynamics P22.8 COP . == 300 Q W c . Therefore, W Q c = . . The heat removed each minute is Q t C ° + × ° = × 00300 4186 220 333 10 5 . kg J kg C . C . kg . J kg 0.030 0 kg 2 090 J kg C 20.0 C 1.40 10 J min 4 bg b g af ej b g or, Q t c = 233 J s. Thus, the work done per sec = P 233 77 8 Js W . .. P22.9 (a) 10 0 1055 1 3600 1 1 293 Btu hW J 1 Btu h s W F H G I K J F H G I K J F H G I K J F H G I K J = (b) Coefficient of performance for a refrigerator: COP refrigerator (c) With EER 5, 5 10 000 Btu Btu h = : = 10 000 2000 200 5 Btu h W k W Btu . Energy purchased is t × 2 00 1 500 3 00 10 3 . kW h . kWh Cost kWh kWh = 300 10 0100 3 $ $ 3 0 0 With EER 10, 10 10 000 Btu Btu h = : = 10 000 1000 100 10 Btu Btu h k W . Energy purchased is t × 1 00 1 50 10 3 kW 1 500 h kWh Cost . kWh kWh = 150 10 3 . $ $150 Thus, the cost for air conditioning is half as much with EER 10
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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