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Calculating final temperature in insulated container

  1. Feb 6, 2015 #1
    1. The problem statement, all variables and given/known data
    0.60kg of ice at O degrees C, 2.0 kg of water at 0 degrees 0 degrees, 3kg of iron at 325 degrees C are place in a sealed and insulated container. The specific heat for iron is c=400J/kg degree C, for water is c=4200J/kgdegree, for water is c=4200J/kgdegree celcius, for ice is c=2000 and the latent heat for ice is 3.3x10^5J/kg.

    What is the final temperature inside the insulated container

    2. Relevant equations
    Q=mcΔT

    3. The attempt at a solution
    Q=0.6x2000(Tf)+2(4200)(Tf)+3x400(Tf-325) = 0
    Q=1200Tf + 8400Tf + 1200Tf - 390000=0
    I solved for Tf and got the wrong answer. Help me please.
     
  2. jcsd
  3. Feb 6, 2015 #2

    SteamKing

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    It would help you and us if you would detail your calculations more carefully.

    That said, what happens to the ice in the container after the hot iron is put in? It looks like you have taken care of the temp. change of the iron and the liquid water there initially, but the ice appears to be missing from your calculations. :L
     
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