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Heat Transfer of ice

  1. Nov 27, 2014 #1
    1. The problem statement, all variables and given/known data
    A 6.00-kg piece of solid copper metal at an initial temperature T is placed with 2.00 kg of ice that is initially at -20.0C. The ice is in an insulated container of negligible mass and no heat is exchanged with the surroundings. After thermal equilibrium is reached, there is 1.2 kg of ice and 0.80 kg of liquid water. What was the initial temperature od the piece of copper?



    2. Relevant equations
    Q=mcΔT
    Q=mL (L in this case is the heat of fusion)
    ΣQ=0



    3. The attempt at a solution
    Qcopper= 6(390)(Tf-T)
    Qice=2(2100)(0-(-20))=8400J (heat required to raise the temperature of the ice to 0C)
    Qwater=.8(334x10^3)=26400J (heat required to melt .8 kg of ice)

    applying ΣQ=0:
    6(390)(Tf-T) + 8400+26400=0

    Tf= would have to be 0c because we still have some ice left.

    solve for T:
    T
    =118.974C
    The correct answer is 150C
    what did I do wrong?
     
    Last edited: Nov 27, 2014
  2. jcsd
  3. Nov 27, 2014 #2
    You've made some arithmetic errors.

    Chet
     
  4. Nov 27, 2014 #3
    Thanks! Is the method correct, though?
     
  5. Nov 27, 2014 #4
    Yes
     
  6. Nov 27, 2014 #5

    SteamKing

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    Staff Emeritus
    Science Advisor
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    You should check your arithmetic here throughout. For example, when I calculate 2 * 2100 * 20, I don't get 8400.
     
  7. Nov 27, 2014 #6
    Thank you very much for your help!
     
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