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20 April, 23:04

A Carnot air conditioner takes energy from the thermal energy of a room at 70°F and transfers it as heat to the outdoors, which is at 96°F. For each joule of electric energy required to operate the air conditioner, how many joules are removed from the room? (Halliday, 05/2018, p. 606) Halliday, D., Resnick, R., Walker, J. (05/2018). Fundamentals of Physics, 11th Edition [VitalSource Bookshelf version]. Retrieved from vbk://9781119306856 Always check citation for accuracy before use.

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  1. 21 April, 01:39
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    Given that,

    Hot temperature

    T_H = 96°F

    From Fahrenheit to kelvin

    °K = (°F - 32) * 5/9 + 273

    °K = (96 - 32) * 5/9 + 273

    K = 64 * 5/9 + 273 = 35.56 + 273

    K = 308.56 K

    T_H = 308.56 K

    Low temperature

    T_L = 70°F

    Same procedure to Levine

    T_L = (70-32) * 5/9 + 273

    T_L = 294.11 K

    A carnot refrigerator working between a hot reservoir and at temperature T_H and a cold reservoir and at temperature T_L has a coefficient of performance K given by

    K = T_L / (T_H - T_L)

    K = 294.11 / (308.56 - 294.11)

    K = 294.11 / 14.45

    K = 20.36

    Then, the coefficient of performance is the energy Q_L drawn from the cold reservoir as heat divided by work done,

    So, for each joules W = 1J

    K = Q_L / W

    Then,

    Q_L = K•W

    Q_L = 20.36 * 1

    Q_L = 20.36 J

    Q_L ≈ 20J

    So, approximately 20J of heats are removed from the room
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