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A refrigerator has a cooling load of 50 kW. It has a COP of 2. It is run by a heat engine which consumes 50 kW of heat to supply the necessary power to the refrigerator. The efficiency of the heat engine is:________________.

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Answer:

50%

Explanation:

Given information

Cooling load=50 kW

COP=2

Consumption=50 kW

Calculations

Revised input is given by cooling load/COP=50/2=25 kW

Efficiency= Work output/ Revised input=25/50=0.5

Efficiency=0.5*100=50%

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