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Efficiency of refrigerator unit

  1. Apr 17, 2005 #1
    Suppose that you want to freeze 1 kg of water for a party, and only have 5 minutes to do it. Relying on nothing but your keen physical abilities, a dixie cup, duct tape and a piece of string, you manage to build a refrigerator unit which can do the job.
    The temperature inside the refrigeration unit is 273 K, and the temperature outside is 350 K. Since you are a brilliant UW student, assume that your refrigerator has the maximum possible efficiency. The Latent Heat of Fusion of water is 3.33×105 J/kg. Assume that the initial temperature of the water is 273 K.

    a. What is the change in entropy in the water per second in J/K s?

    HELP: Change in entropy = Change in energy/ Absolute Temperature.

    b. What is the heat flow ( Q_hot/time ) into the hot well per second in J/s?

    HELP: Use the entropy flux calculated in part b, 350 K, and use

    c. How much POWER must you provide to drive your refrigerator in Watts? (That is, work done per second?)

    HELP: Q(hot) = Q(cold) + Work. Remember, power is work/time.
    Entropy = Energy transfer / Temperature.
  2. jcsd
  3. Apr 17, 2005 #2


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    Gold Member

    Its 77 degrees C outside ;) woooo, party in hell?

    You really need to try to do these on your own and then ask questions. Looks like your just posting questions and hoping we'll answer them for you.
  4. Apr 17, 2005 #3

    Andrew Mason

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    Science Advisor
    Homework Helper

    What about time?

    [tex]\frac{ds}{dt} = \frac{\frac{\Delta Q}{T_c}}{\Delta t}[/tex]

    where [itex]\Delta Q = 3.33e5 J[/itex] and [itex]\Delta t = 300 sec[/tex]

    For a Carnot cycle,

    [tex]\frac{Q_c}{Q_h} = \frac{T_c}{T_h}[/tex]

    [tex]CP = Q_c/W = \frac{1}{\frac{T_h}{T_c} - 1} [/tex]

    so work out W from Q_c found in 1.

    By the way, the problem facts are quite impossible. If you place water at 273K in a reservoir at 273K there will be no freezing at all, let alone in 5 minutes.

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