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Thermodynamics-reversible carnot engine

  1. Mar 2, 2015 #1
    1. The problem statement, all variables and given/known data
    a refrigerator is used to convert 1 kg of liquid water at 30 degree celsius to ice at -18 degree celsius.the refrigerator rejects heat to surroundings which are at 30 degree celsius.the specific heat values of liquid water and ice are 4.18 kj/kgk and 1.9 kj/kgk respectively and latent heat of fusion of water is 335 kj/kg.the minimum work in kj that is to be supplied to refrigerator to accomplish ice making is?
    2. Relevant equations
    efficiency= Q2/W = (T2/T1-T2)
    H = mCpT
    Q2=heat rejected by refrigerater
    T2= temperature of ice (-18 degree celsius)
    T1= temperature of surroundings

    3. The attempt at a solution
    don't know how to find mass of ice??the correct answer is 48.2kj
    suggest me how to proceed
     
  2. jcsd
  3. Mar 2, 2015 #2

    Bystander

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    Step 1: read the problem statement.
     
  4. Mar 2, 2015 #3
    how shall i convert that,please explain in detail
     
  5. Mar 3, 2015 #4

    rude man

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    Start as follows:
    For each of 3 separate steps,
    what is the heat lost by the 1kg water/ice?
     
  6. Jun 1, 2015 #5
    How you got 48.2 kJ? please give answer in step by step?
     
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