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Homework Help: Thermodynamic Question (refrigerator)

  1. Apr 26, 2009 #1
    I'm having a little trouble in completing this problem. I would appreciate your answer if anyone knows how to solve it. Thanks

    An ideal refrigerator removes heat at a rate of 0.15kW from an interior temperature of 1 degree celsius. The exhaust from the refrigerator is 45 degrees celsius. How much electrical power is used.

    *Note that this is a refrigerator problem (reverse process from a carnot engine)

    I have found the efficiency using Th/Tc - 1 = .29 I do not know how to proceed after this.
  2. jcsd
  3. Apr 27, 2009 #2
    Remember efficieny is always

    what you get / what you pay for.

    so in this case its

    and for the reversible carnot this comes to:

    COP = Ql/(Qh-Ql)

    You must remember the for refrigiration cycles its not technically efficiecny, its coefficient of performance we use. This avoids the confision of having efficienys over 100%.

    Where you were going wrong was you were taking the efficieny for a carnot cycle and applying it directly.
  4. Apr 27, 2009 #3

    Andrew Mason

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    Science Advisor
    Homework Helper

    Coefficient of Performance is Qc/W. W = Qh - Qc.


    COP = (Qc /(Qh - Qc)

    1/COP = Qh/Qc - 1

    For a Carnot cycle, this becomes:

    1/COP = Th/Tc - 1

    You have to determine the COP from the temperatures. Using the value for Qc you can determine W.

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