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Minimum Minimum Power of a Heat Engine to freeze 50kg water in one hour

  1. Feb 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Fifty kg of water at 0oC must be frozen into ice in a refrigerator. The room temperature is 20oC. The latent heat of fusion of water is 3.33x105 J kg-1. What is the minimum power required if the freezing is to take place in one hour?

    m = 50 kg
    Lf = 3.33x105 J kg-1.
    Twater = 0oC
    Troom = 20oC
    [itex]\Delta[/itex]t = 1 hour = 3600s


    2. Relevant equations

    [itex]\Delta[/itex]U = Q1 + Q2 - W
    W = Q1 + Q2
    (Q1/Q2) = -(T1/T2)

    3. The attempt at a solution

    Since there are 50kg of water and Lf = 3.33x105 J kg-1, the total amount of energy that is required to turn the water into ice is 1.665x107 J. For the freezing to occur in one hour, at least 4625 J must be taken from the water per second. This led me to believe that the minimum power of the refrigerator must be 4625 watts. However, according to the selected answers in the back of the book, the minimum power should come out to be 340 watts.

    After reading this, I assumed that 1.665x107 J (energy required to turn all water into ice) corresponds to Q1, or the amount of heat extracted from the system. I then assumed that T1=20oC (room temp), but without being given the value of T2, Q2, efficiency of the engine, etc, I am unsure how to proceed with the problem, perhaps I am simply overlooking something or do not have one of the necessary equations.

    Any help is appreciated, Thanks!
     
  2. jcsd
  3. Feb 22, 2012 #2

    gulfcoastfella

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    If you assume an ideal refrigeration cycle (which is based on a reversed Carnot Cycle), then the COP (coefficient of performance) is given as

    COP = T[itex]_{L}[/itex] / (T[itex]_{H}[/itex] - T[itex]_{L}[/itex] )

    The COP is also defined as

    COP = Q[itex]_{L}[/itex] / W

    Take it from there and see what you get; let me know if it works...
     
  4. Feb 22, 2012 #3

    gulfcoastfella

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    by the way, in thermo all temps should be Kelvin or Rankine unless otherwise specified for a particular equation...
     
  5. Feb 22, 2012 #4
    I'm still stuck, what would TH be? Shouldn't it be the temperature of the output reservoir, which isn't specified by the problem, and I don't think enough is known to use the second version of the equation either...

    Also, as for the temperatures, I hadn't bothered converting them to Kelvin because up to that point, I hadn't needed to find anything other than difference, which is the same for both scales.
     
  6. Feb 23, 2012 #5

    gulfcoastfella

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    A refrigeration cycle takes heat from a low temperature reservoir (T[itex]_{L}[/itex] , the inside of the refrigerator) and dumps it to a high temperature reservoir (T[itex]_{H}[/itex] , the air in the room). The problem as you wrote it doesn't provide the temp inside the refrigerator, but for the purposes of this problem, it's 0°C.

    You already calculated Q[itex]_{L}[/itex] as shown in your first post. You're set to get your answer from the second equation.
     
  7. Feb 23, 2012 #6
    Finally got it, I was completely over thinking the entire process! Thanks for the help!
     
  8. Feb 23, 2012 #7

    gulfcoastfella

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    Glad to be of help!
     
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