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Carnot freezer engine problem

  1. May 8, 2009 #1
    1. The problem statement, all variables and given/known data
    A Carnot freezer in a kitchen has constant temperature of 260k, while the air in the kitchen has a constant temperature of 300K. Suppose the insulation for the freezer is not perfect and energy is conducted into the freezer at a rate of .15 Watts. Determine the average power required for the freezer's motor to maintain the constant temperature in the freezer.


    2. Relevant equations

    Qc/Qh=Tc/Th for a Carnot engine

    efficiency = (Th - Tc)/(Th)

    Work = Qh-Qc

    3. The attempt at a solution

    I'm not really sure where to begin. It doesn't seem like I have enough formulas to tackle this problem.
     
  2. jcsd
  3. May 8, 2009 #2
    The definition of power might be useful
     
  4. May 8, 2009 #3
    I know power = work/time. So .15J of heat per second are entering the freezer. For a carnot freezer, Work done = Th-Tc = 40J.
     
    Last edited: May 8, 2009
  5. May 11, 2009 #4
    No one? :(
     
  6. May 11, 2009 #5
    Don't confuse T with Q.

    If .15J are leaking into the freezer each second, what heat (Qc) needs to be
    pumped out per second?
    Now use your equations to find work done by the motor per second.
     
  7. May 12, 2009 #6

    Andrew Mason

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    Science Advisor
    Homework Helper

    Consider the amount of work required to remove a certain amount of heat from the cold reservoir to the hot reservoir. What is the relationship between W, Th and Tc for a Carnot refrigerator?

    In this case, .15 Joules/sec have to be moved. The question asks you how much work (per second) you have to do in order to move that amount of heat.

    AM
     
  8. May 12, 2009 #7
    Your missing the Thermal Coefficient. Plug it in and you're there.
     
  9. May 12, 2009 #8
    We've effectively told you Qc.
    You are given Tc and Th.

    Use your first equation to find Qh,
    then your second equation to find work.

    This ain't rocket science, although it might be relevant to it:)
     
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