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Thermodynamics help

  1. Feb 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A household refrigerator with a coefficient of performance of 2.1 removes energy from the refrigerated space at a rate of 600 Btu/h. Evaluating electricity at $ 0.08 per kW · h determine the cost of electricity in a month when the refrigerator operates for 360 hours.

    2. Relevant equations

    coefficient of performance = Q_in / W_cycle
    W_cycle = Q_out - Q_in

    3. The attempt at a solution

    I assumed that Q_out was 600 Btu/hr, since it said that was the energy removed. The coefficient of performance is obviously 2.1, and I tried to find Q_in using the above equations but had some trouble. I also know that you need to find the work and then you can just use the given "$0.08 per kW" to find the money needed, which is the answer. But I am really confused with this one.
     
  2. jcsd
  3. Feb 9, 2009 #2
    I didn't think Q/W could be greater than 1. Since 1 represents 100% effeciency. If an engine had a Q/W value of 2.1, wouldn't it be working at 210% efficiency??
     
  4. Feb 9, 2009 #3
    thermal efficiency is different than coefficient of performance
     
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