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9 August, 01:36

To make ice, a freezer that is a reverse Carnot engine extracts 42 kJ as heat at - 13°C during each cycle, with coefficient of performance 5.9. The room temperature is 30.9°C. How much (a) energy per cycle is delivered as heat to the room and (b) work per cycle is required to run the freezer?

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  1. 9 August, 04:03
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    (a) 49.09 kJ

    (b) 7.09 kJ

    Explanation:

    Qc = 42 kJ

    Tc = - 13 degree C = - 13 + 273 = 260 K

    TH = 30.9 degree C = 30.9 + 273 = 303.9 K

    β = 5.9

    (a) Let the heat delivered to the room per cycle is QH.

    TH / Tc = QH / Qc

    303.9 / 260 = QH / 42

    QH = 49.09 kJ

    (b) Work done per cycle = QH - Qc = 49.09 - 42 = 7.09 kJ
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