A refrigerator is to remove heat from the cooled space at a rate of 300 kj/min to maintain its temperature at - 8 Â°c. if the air surrounding the refrigerator is at 25 Â°c, determine the minimum power input required for this refrigerator.

Question

Skyler Carlson

Physics

29 October, 17:45

A refrigerator is to remove heat from the cooled space at a rate of 300 kj/min to maintain its temperature at - 8 Â°c. if the air surrounding the refrigerator is at 25 Â°c, determine the minimum power input required for this refrigerator.

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Emily Alvarado · Answer 1

The solution for this problem is:

Coefficient of performance (COP) for a refrigerator = Qc/W where QC = 300 in this case and W is the energy input requirement per minute. The best attainable COP is for a reversible heat cycle and = Tc / (Th-Tc) = 265 / (33) = 8.03

Therefore W = 300/8.03 = 37.36 kJ/min = 622.66667 J/s = 622.66667 watts = 0.622 kW.