Question

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:________________.

Answer 1

50%

Step-by-step 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%