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Carnot refrigerator help

  1. Nov 29, 2012 #1
    1. The problem statement, all variables and given/known data
    An ideal refrigerator utilizes a carnot cycle operating between 0 and 25 degrees celsius. To turn 10 kg of water at 0 degrees celsius into 10 kg of ice at 0 degrees celsius,
    a) how much heat is exhausted into the room?
    b) how much energy must be supplied to the refrigerator?


    2. Relevant equations

    coefficient of performance of carnot refrigerator (Kcarnot): Tc/(Th-Tc)
    n (funky n for efficiency thing)=Wout/Qh [but isn't that for engine]
    K=Qc/W

    3. The attempt at a solution

    I'm guessing part A is Q so I'd have to find that somehow. And, for part B I think you use the last formula I put here and input Q to find W.
     
  2. jcsd
  3. Nov 29, 2012 #2

    haruspex

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    An ideal Carnot cycle is reversible, so W is the work, whether it's in or out.
    The only thing that changes is how efficiency is defined. Equations not involving η are unchanged.
    W = QH (1-TC/TH ) = QC (1-TH/TC )
    There are two Qs: QH and QC. Which do you want for part A?
     
  4. Nov 29, 2012 #3

    Andrew Mason

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    The COP for the refrigerator is output/input where the output is the amount of heat removed from the cold reservoir, Qc and W is the work supplied: COP = Qc/W

    Since W = Qh-Qc, and Th/Tc = Qh/Qc, you can express COP in terms of Th and Tc (your expression is correct).

    For a) what do you need to find? (hint: it is one of W, Qh or Qc). For b) what do you need to find? (same hint)

    AM
     
    Last edited: Nov 29, 2012
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