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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 . = = 3 00 Q W c . Therefore, W Q c = 3 00 . . The heat removed each minute is Q t C = ° ° + × + ° ° = × 0 030 0 4 186 22 0 0 030 0 3 33 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 b gb ga f b g e j b gb ga f or, Q t c = 233 J s. Thus, the work done per sec = = = P 233 3 00 77 8 J s W . . . P22.9 (a) 10 0 1055 1 3 600 1 1 2 93 . . Btu h W J 1 Btu h s W J s 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 a f (c) With EER 5, 5 10 000 Btu h W Btu h = P : P = = = 10 000 2 000 2 00 5 Btu h W kW Btu h W . Energy purchased is P t = = × 2 00 1 500 3 00 10 3 . kW h . kWh a fb g Cost kWh kWh = × = 3 00 10 0 100 3 . . \$ \$300 e j b g With EER 10, 10 10 000 Btu h W Btu h = P : P = = = 10 000 1 000 1 00 10 Btu Btu h W kW h W . Energy purchased is P t = = × 1 00 1 50 10 3 . . kW 1 500 h kWh a fb g Cost . kWh kWh = × = 1 50 10
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