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3 July, 22:54

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.

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  1. 4 July, 02:39
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    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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