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1 May, 07:33

Suppose you must remove an average of 3.9*108J of thermal energy per day to keep your house cool during the summer. Part A If you upgrade from an old air conditioner with a COP of 2.3 to a new air conditioner with a COP of 6.0, by how many joules is the required mechanical work reduced each day?

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  1. 1 May, 10:47
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    The required mechanical work is required to reduce each day by 1.05*10^8 Joules.

    Explanation:

    Coefficient of Performance (COP) = Q/W

    Q is thermal energy absorbed by the air conditioner

    W is mechanical work done

    Q = 3.9*10^8 J

    COP of old air conditioner = 2.3

    W = Q/COP = 3.9*10^8/2.3 = 1.70*10^8 J

    COP of new air conditioner = 6

    W = Q/COP = 3.9*10^8/6 = 6.5*10^7 J

    Reduction in mechanical work = (1.7*10^8) - (6.5*10^7) = 1.05*10^8 J
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