Arefrigerator 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.

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.