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A household refrigerator with a COP of 1.2 removes heat from the refrigerated space at a rate of 60kJ/min. Determine: (a) The electric power consumed by the refrigerator, and (b) The rate of heat transfer to the kitchen air.

Respuesta :

Answer:

a) Power =50 KJ/min

b)Rate of heat transfer = 110 KJ/min

Explanation:

Given that

COP = 1.2

Heat removed from space Q = 60 KJ/min

As we know that COP of refrigerator is the ratio of heat removed to work input.

Lets take power consume by refrigerator is W

So

COP= Q/W

1.2=60/W

W=50 KJ/min

So the power consume is 50 KJ/min.

From first law of thermodynamic

Heat removed from the kitchen = 50 + 60 KJ/min

Heat removed from the kitchen =110 KJ/min

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