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Thermodynamics- Refrigeration of water/ice

  1. Oct 12, 2012 #1
    1. The problem statement, all variables and given/known data
    A refrigerator is to freeze 150g of water at 0°c in one minute, the ambient temperature being 20°C. Calculate in Watts the minimum power required?

    So known stuff:
    mass of water =0.15kg
    Temp of cold reservoir= 273k
    Temp of warm reservoir= 293k
    Specific heat capacity of water = 4190 J/kg k
    Specific latent heat of fusion of water= 3.33x105 J/kg

    2. Relevant equations
    Q=mcΔt
    Q=±mLf
    COP= Q2/W
    W=Q1-Q2
    Power=W/T
    Q1/Q2 = T1/T2
    3. The attempt at a solution

    First off i thought good idea to work out how much energy would take to actually freeze all the water so which was like (0.15*4190*-20)+(0.15*-333000)= -62520 J (assume that means that 62520J was released to the system from the water as it converted to ice)

    I thought this was the Q1 value, then knowing T1 was 273 k and T2 was 293 k subbed into Q1/Q2 = T1/T2 to solve for Q2 which was (-62520)(293)/ (273) which gets Q2=-67000.

    But then Work done= Q1-Q2 so like -62520--67000= 4580J

    Then work done/time = power

    and then like 4580/60 = 76 Watts... but thats too small to freeze water that fast surely?


    Any ideas?
     
  2. jcsd
  3. Oct 13, 2012 #2
    Delete i wrote it down wrong.
     
  4. Oct 13, 2012 #3
    heres my next go at it...
    1. The problem statement, all variables and given/known data
    A refrigerator is to freeze 150g of water at 0°c in one minute, the ambient temperature being 20°C. Calculate in Watts the minimum power required?

    So known stuff:
    mass of water =0.15kg
    Temp of cold reservoir= 273k
    Temp of warm reservoir= 293k
    Specific latent heat of fusion of water= 3.33x105 J/kg

    2. Relevant equations

    Q=±mLf
    COP= QH/W
    W=QH+QC
    Power=wd/T
    Q1/Q2 = -T1/T2
    3. The attempt at a solution

    First of, i assume the water is already at 0°c or 273k, so need only need to calculate the Latent heat of fusion for the water. As the water is freezing i think its Q=-mLf so -(0.15x3.33x105) = -49950J.

    I then said this was the value for Qc.
    The question asks for the minimum power required so i though we assume its a carnot engine, so max efficiency, so
    Qc/Qh= -Tc/Th to calculate the value for Qh. which gets (-49950*293)/(273) which gets a Qh value of 53609J.

    So seeing as work done on the system is equal to Qh+Qc then work done would be 53609+(-49950) J
    =3659J work done on engine.

    then power=work done/ time
    so 3659/60 = 61 Watts.

    i know its assuming that the system is max efficient but that just feels low.
     
  5. Oct 13, 2012 #4

    rude man

    User Avatar
    Homework Helper
    Gold Member

    You were right the first time! (I did not check your computations on the total heat removed from the water though).

    EDIT: I see now you did miscalculate the heat removed from the water. It's just the latent heat of fusion that's to be removed.

    And 61W is the correct answer, assuming you computed the latent heat correctly.
     
    Last edited: Oct 13, 2012
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