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23 January, 10:20

An ideal refrigerator extracts 500 joules of heat from a reservoir at 295 K and rejects heat to a reservoir at 493 K. What is the ideal coefficient of performance and how much work is done in each cycle?

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  1. 23 January, 10:44
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    C. O. P = 1.49

    W = 335.57 joules

    Explanation:

    C. O. P = coefficient of performance = (benefit/cost) = Qc/W ... equ 1 where C. O. P is coefficient of performance, Qc is heat from cold reservoir, w is work done on refrigerator.

    Qh = Qc + W ... equ 2

    W = Qh - Qc ... equ 3 where What is heat entering hot reservoir.

    Substituting for W in equ 1

    Qh / (Qh - Qc) = 1 / ((Qh / Qc) - 1) ... equ 4

    Since the second law states that entropy dumped into hot reservoir must be already as much as entropy absorbed from cold reservoir which gives us

    (Qh/Th) > = (Qc/Tc) ... equ 5

    Cross multiple equ 5 to get

    (Qh/Qc) = (Th/Tc) ... equ 6

    Sub equ 6 into equation 4

    C. O. P = 1 / ((Th/Tc) - 1) ... equ7

    Where Th is temp of hot reservoir = 493k and Tc is temp of cold reservoir = 295k

    C. O. P = 1 / ((493/295) - 1)

    C. O. P = 1.49

    To solve for W = work done on every cycle

    We substitute C. O. P into equ 1

    Where Qc = 500 joules

    1.49 = 500/W

    W = 500/1.49

    W = 335.57 joules
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