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Refrigerator - Heat Energy

  1. Nov 27, 2005 #1
    100 mL of water at 288 K is placed in the freezer compartment of a refrigerator with a coefficient of performance of 4.0. How much heat energy is exhausted into the room as the water is changed to ice at 258 K?

    (The answer is 5.34 x 10^4 J)

    I tried using the Brayton-cycle refrigerator to understand this problem.

    I am assuming that the lower isobar is the part of the graph that I need to analyze, where the low-temperature heat exchanger warms the gas and cools the refrigerator. With this method, I know that [tex]Q = nC_v\Delta T[/tex]. However, I cannot think of a way to find the number of moles (n).

    I also don't know if my reasoning is correct (that is, only using the lower isobar of the graph to answer my question).

    I am also assuming that my lower temperature (258 K) is equivalent to Tc, and my higher temperature (288 K) is equivalent to Th.
    I am not sure if this is correct, however.

    I have realized that I need to use work (in) to answer the question, because K = Q / W.

    However, the work for an isobar is W = pV, and I do not know the pressure, and am unable to find it since I do not know the number of moles.

    Any advice will be greatly appreciated, thank you.
    Last edited by a moderator: Nov 27, 2005
  2. jcsd
  3. Nov 27, 2005 #2

    Andrew Mason

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    You don't know, or have to know, the details of the cycle. All you have to do is find [itex]Q_H = W + Q_C[/itex] where:
    [tex]COP = Q_C/W[/tex]

    [tex]Q_C[/tex] is the heat lost by the water (you have to take into account temperature change + heat of fusion).

    Last edited: Nov 27, 2005
  4. Nov 28, 2005 #3
    What is COP?

    And I don't think we are required to know heat of fusion, so there must be some other way to solve the problem.
  5. Nov 28, 2005 #4

    Andrew Mason

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    You have double posted this. My response will be found on the other thread:

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