SSC JE Mechanical Engineering 2014 Set 1 Paper

Question : 136 of 200

Marks: +1, -0

The food compartment of a refrigerator is maintained at 4°C by removing heat from it at a rate of 360 kJ/min. If the required power input to the refrigerator is 2 kW, the COP of the refrigerator is

Validate

Solution:

Cop of refrigerator

={ \Heat\extracted}/{\Work \input}

360/{2×60}

kJ/min = 3
∵ w = 2 Kw = 2 kJ/s = 2× KJ/min

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