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Homework Help: Minimum amount of work input to a refrigerator

  1. May 12, 2010 #1
    1. The problem statement, all variables and given/known data
    600Kg of water at 25C has to be frozen into ice in an ideal refrigerator. The room temperature is 20C. What is the minimum amount of work input to the refrigerator to achieve this? ( specific heat capacity of water = 4170 J/Kg/K, latent heat of melting ice = 3.33 x10^5 J/Kg

    2. Relevant equations


    [tex] C.O.P=\frac{Q_C}{W}=\frac{Q_C}{{Q_H}-{Q_C}}=\frac{{T_C}/{T_H}}{1-{T_C}/{T_H}}[/tex]

    3. The attempt at a solution
    The refrigerator has to remove heat equal to 600Kg x c x dT to get the temperature down to 273, then has to remove heat from a 273K reservoir to a 293K reservoir to keep it frozen.

    1. cooling to 273, heat removed is [tex]{Q_C}=mc\delta{T}[/tex] = [tex]{Q_C}=600*4170*25=6.255E^{6}Joules[/tex]

    2. keeping it frozen - amount of heat that needs to be removed is [tex]{Q_C}=mL=600*3.33E^{5}=1.998E^8 Joules[/tex]

    so total is the sum of these = 2.06E^8 J

    then using [tex] C.O.P=\frac{Q_C}{W}=\frac{Q_C}{{Q_H}-{Q_C}}=\frac{{T_C}/{T_H}}{1-{T_C}/{T_H}}[/tex]

    i get two different coefficients of power - one for cooling and one for freezing, what do i do with them, do i add them or subtract one from the other? or do i use the sum of the two energies and put this into the COP? or am i doing this completely wrong?
  2. jcsd
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