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Refrigeration: The Equivalent Rate of Transport of Heat in Watts?

  1. Sep 12, 2009 #1
    A 1.2 ton air conditioner has the capacity to freeze 2400 lb of ice at 0°C in a day. What is the equivalent rate of transport of heat in watts?

    QH(out)= QL(in) + Wi
    Coefficient of Performance= QL/ (QH-QL)
    Coefficient of Performance of an Ideal System: TL/ (TH-TL)
    QL/QH= TL/TH

    I really don't know where to begin with this problem. I know 1.2 tons is equal to 2400lbs. So, the air conditioner and the amount of meat it freezes is a 1 to 1 ratio, but what does any of that have to do with the problem? I'm not given the heat-out or heat-in so how could I possibly calculate the transport of heat in watts? The solution to a similar problem (1 ton air conditioner and 2000lbs of meat is 3.5kW), but I need to know how to solve these type of problems.
    Any help would be greatly appreciated!
  2. jcsd
  3. Sep 12, 2009 #2
    Take a good look at the definition of "1 ton of Refrigeration". Find out how many kJ/s or (Watts) that one ton of refrigeration is equivalent to.

    Last edited: Sep 12, 2009
  4. Sep 12, 2009 #3
    Thanks! That was a lot simpler than I expected it to be.
  5. Sep 12, 2009 #4
    Glad that I could help.

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