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Power of an Ideal Refrigerator - Carnot Engines

  1. Feb 5, 2012 #1
    Power of an Ideal Refrigerator -- Carnot Engines

    1. The problem statement, all variables and given/known data

    The inside of an ideal refrigerator is at a temperature T_c, while the heating coils on the back of the refrigerator are at a temperature T_h. Owing to a malfunctioning switch, the light bulb within the refrigerator remains on when the the door is closed. The power of the light bulb is P; assume that all of the energy generated by the light bulb goes into heating the inside of the refrigerator.

    For all parts of this problem, you must assume that the refrigerator operates as an ideal Carnot engine in reverse between the respective temperatures.

    If the temperatures inside and outside of the refrigerator do not change, how much extra power P_extra does the refrigerator consume as a result of the malfunction of the switch?
    Express the extra power in terms of P, T_h, and T_c.

    Suppose the refrigerator has a 25-W light bulb, the temperature inside the refrigerator is 0^\circ {\rm C}, and the temperature of the heat dissipation coils on the back of the refrigerator is 35^\circ {\rm C}. Find the extra power P_extra consumed by the refrigerator. Keep in mind that you will need to use absolute units of temperature (i.e., kelvins).


    2. Relevant equations

    K=Q_c/W=Ht/Pt=H/P


    3. The attempt at a solution

    I'm having a hard time even beginning with this problem. I understand that heat is absorbed from the cold reservoir inside the refrigerator, and released to the hot reservoir outside. I guess I'm mostly having a hard time relating power to heat energy, specifically the relevance of the malfunctioning light bulb.

    Any help in the right direction would be greatly appreciated. Thank you!
     
  2. jcsd
  3. Feb 5, 2012 #2

    Andrew Mason

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    Re: Power of an Ideal Refrigerator -- Carnot Engines

    This is an ideal refrigerator, which means that the heat pump operates at maximum efficiency. Think of this as an ideally insulated refrigerator as well (you don't have to - it just makes it simpler) so that the refrigerator itself has no leaks. In other words, with the door closed and the lightbulb off there is no heat flow into the inside so there is no need to run the heat pump. The only reason to run the heat pump is to remove the heat added by the lightbulb.

    Does that help?

    AM
     
  4. Feb 5, 2012 #3
    Re: Power of an Ideal Refrigerator -- Carnot Engines

    Hm, I'm still not sure. If I understand you correctly, then P_extra is dependent only on the power of the lightbulb, and so I don't see how T_h and T_c are relevant. I guess I'm having a hard time conceptually relating power to the actual heat being released or to the heat processes in general for that matter.
     
  5. Feb 5, 2012 #4

    Andrew Mason

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    Re: Power of an Ideal Refrigerator -- Carnot Engines

    The efficiency (COP) of the refrigerator is what you have to determine. You determine that from Th and Tc since it is a Carnot cycle.

    Then you just assume that the rate of heat flow into the cold reservoir (that the refrigerator engine has to reverse) is P. In order to remove heat from the cold reservoir at that rate, what is the rate of work you have to supply?

    AM
     
  6. Feb 6, 2012 #5
    Re: Power of an Ideal Refrigerator -- Carnot Engines

    Oh wow, I think maybe it just clicked...

    ..so for a Carnot engine, the efficiency e is equal to 1-T_c/T_h.

    We have W = Q_h - Q_c. In this case, Q_c is P, so we have P = Q_h - W. We also know that the efficiency is equal to Q_h/W, so W = Q_h/e. And Q_h is also W+Q_c.

    So now, P=(W+Q_c) - (W+Q_c)/e

    Am I on the right track at least? Now just to get that in terms of T_c and T_h...
     
  7. Feb 7, 2012 #6

    Andrew Mason

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    Re: Power of an Ideal Refrigerator -- Carnot Engines

    For a refrigerator you want to relate heat removed, Qc to the amount of work done:

    COP = output/input = Qc/W = 1/(W/Qc) = 1/(Qh/Qc - 1)

    Since it is a Carnot cycle: 1/(Qh/Qc - 1) = 1/(Th/Tc - 1)

    You are on the right track. Express W in terms of Qc (=P) and COP. Then calculate the COP and solve for W.

    AM
     
  8. Feb 7, 2012 #7
    Re: Power of an Ideal Refrigerator -- Carnot Engines

    Great, I appreciate the help. Thank you so much!
     
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