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Temperature Fusion (heat gain = heat loss)

  1. Nov 29, 2009 #1
    1. The problem statement, all variables and given/known data
    If 10 kilogram of ice at zero degrees celsius is added to 20 kilogram of steam at 100 degrees celsius, what is the temperature of the resulting mixture?


    2. Relevant equations
    uhm... perhaps how many solutions are required to arrive at the answer?


    3. The attempt at a solution
    I and some of my classmates arrived at 16.11 degrees celsius but my instructor said its wrong. He said the answer is around 40 degrees celsius. Now, he gave it as an assignment to us to find the correct solution.
     
    Last edited: Nov 29, 2009
  2. jcsd
  3. Nov 29, 2009 #2

    Borek

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    Staff: Mentor

    In no way that's advanced physics.

    Show how you got 16 deg answer. What equations have you used?
     
    Last edited: Nov 29, 2009
  4. Nov 29, 2009 #3
    QE = QA
    Heat Evolve = Heat Absorb
    MECEdelta TE = MACAdelta TA
     
  5. Nov 29, 2009 #4

    Borek

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    Staff: Mentor

    So far so good.

    That's not all.

    I have asked you to show how you got the wrong answer.
     
  6. Nov 29, 2009 #5
    MECEdelta TE = MACAdelta TA
    (10 kg)(0.50 cal/kg*C)(tf - 0C) = (2 kg)(0.48 cal/kg*C)(100C - tf)
    5 cal/C * tf = 96 cal - 0.96 cal/C * tf
    5.96 cal/C * tf = 96 cal
    tf = (96 cal)/(0.96 cal/C)
    tf = 16.11 C
     
  7. Nov 29, 2009 #6

    Borek

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    As I told you - you are missing part of the equation. Two parts, to be precise. What you did would be OK for a mixture of water at 0 deg C and water at 100 deg C - but you start with ice and steam.
     
  8. Nov 29, 2009 #7

    fluidistic

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    Gold Member

    I agree. A bit more help, heat from hot water has to be transfered to the ice to melt it, even though the temperature does not change. Quantify this heat.
     
  9. Nov 29, 2009 #8
    I think I got the solution to arrive at 40 degrees celsius. Please verify if my answer is correct.

    Q out of steam = Q into ice (for c I use kcal/kgC but You can just as easily use J/kgC)

    Q out = m*Lv + m*c*deltaT = m*(Lv + c*deltaT) = 2.0kg*(540kcal/kg + 1.0kcal/kg/C*(100-Tf))

    Q in = m*Lf +m*c*deltaT = m*(Lf =c*deltaT) = 10kg*(80kcal/kg + 1.0kcal/kgC*(Tf -0))

    Setting these equal gives 2.0*540 +2.0*100 -2.0*Tf = 10*80 + 10*Tf

    Solving (10+2)*Tf = 2*540 +2*100-10*80 = 480 so Tf = 480/12 = 40degC
     
  10. Nov 29, 2009 #9

    Borek

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    Staff: Mentor

    I have just skimmed, looks reasonably.

    --
     
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