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Carnot Engine refrigerator temp

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data

    A Carnot engine is operated as a refrigerator, taking in 18 J of heat every second and expelling 20 J into a 23oC room. What is the temperature inside the refrigerator?

    a. 15.4oC
    b. 3.33oC
    c. -6.62 oC
    d. -15.4 oC

    2. Relevant equations

    e=1-T(cold)/T(hot)
    U=Q-W

    3. The attempt at a solution

    I have no idea how to approach this problem :s
     
  2. jcsd
  3. Dec 7, 2011 #2

    vela

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    First determine how much heat is transferred to or from the hot and cold reservoirs and how much work is needed to do that.
     
  4. Dec 7, 2011 #3
    isnt the heat just 18 and the work 20?
     
  5. Dec 7, 2011 #4

    vela

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    Nope. For one thing, there are two heat transfers, heat absorbed and heat expelled by the engine/refrigerator. There's probably a picture in your textbook illustrating the flow of energy through a Carnot refrigerator.
     
  6. Dec 7, 2011 #5
    how can the efficiency be more than one?? because its receiving 18 and giving out 20?
     
  7. Dec 7, 2011 #6
    how can i get the heat?
     
  8. Dec 7, 2011 #7
    does it make sense to say that Q(absorbed)-20J=Q(given out)+18J??
     
  9. Dec 7, 2011 #8

    vela

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    No, that doesn't make sense.

    Forget the refrigerator for a second. How does a heat engine work? I'm not interested in details. I'm looking for a general description.
     
  10. Dec 7, 2011 #9
    Ok...i dont understand them very well but i know that they turn heat into work, that some heat is lost, and that the efficiency measures how much heat actually becomes work.
     
  11. Dec 7, 2011 #10

    vela

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    Yes, that's perfect. A heat engine absorbs heat QH from a hot reservoir. Some of it is turned into work W. The rest of it QC gets expelled to a cold reservoir. This fact is expressed in the equation QH=W+QC.

    A Carnot refrigerator is a heat engine that runs backward. Instead of absorbing heat from the hot reservoir, it expels heat into the hot reservoir. Instead of expelling heat to the cold reservoir, it absorbs heat from the cold reservoir. And instead of producing work, it requires work to be done on it to move the heat from the cold to hot reservoirs.

    The efficiency of a system is always output divided by input. For a heat engine, the output is work W, and the input is QH. Therefore, the efficiency is given by W/QH. Because W will always be smaller than QH, the efficiency of a heat engine will always be less than 1. Because the goal of a refrigerator is to cool off the cold reservoir, QC is the measure of its output. The input to the refrigerator is the work required to do the cooling, so the efficiency of a refrigerator is QC/W. This number is called the coefficient of performance. It's not necessarily less than 1.

    So given this basic picture, can you identify which variables are given by the 18 J and 20 J figures in the problem statement?
     
    Last edited: Dec 7, 2011
  12. Dec 7, 2011 #11
    So in this case 18J are absorbed from the cold and 20J are expelled to the hot so it needs 2J of work to run?
     
  13. Dec 7, 2011 #12

    vela

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    Exactly!

    For the refrigerator, you have, again, QH=QC+W, so the coefficient of performance is
    [tex]\text{COP} = \frac{Q_C}{W} = \frac{Q_C}{Q_H-Q_C}[/tex]This relationship holds for any refrigerator, but for a Carnot refrigerator, you can replace the Qs with Ts to get
    [tex]\text{COP} = \frac{T_C}{T_H-T_C}[/tex]Can you take it from here?
     
  14. Dec 7, 2011 #13
    Ohh so i could make them equal to eachother and get 18J/2J=Tc/(296.15-Tc) to solve for Tc
     
  15. Dec 7, 2011 #14

    vela

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    Right!
     
  16. Dec 7, 2011 #15
    :D thank you!!! I was so confused and now I get it :)
     
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