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Homework Help: What is final temperature?

  1. Apr 29, 2009 #1
    1. The problem statement, all variables and given/known data
    A 81-g ice cube at 0°C is placed in 878 g of water at 25°C. What is the final temperature of the mixture?

    2. Relevant equations
    No heat is loss therefore m*Cp*deltaT+m*Cp*deltaT=0
    Cp ice=2.09
    Cp water=4.19

    3. The attempt at a solution

    The answer is wrong...don't know why...
  2. jcsd
  3. Apr 29, 2009 #2


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    What about the latent heat of fusion for ice?
  4. Apr 29, 2009 #3
    So you think you'll have ice at 23 degrees?
    The ice will melt (at least partially). This will take energy (heat) from the water so teh final temperature will be lower.
    You have to take into account the melting heat Q=m_ice*l_melting
    where m_ice is the mass of ice that melts (may be equal to the initial mass) and l_melting is the latent heat (heat of fusion) for ice.

    In order to see if the ice melts completely (or not) you can compare the heat necessary to melt all th e ice with the heat released by the water when it cools to zero degrees.
  5. Apr 29, 2009 #4

    Which is still wrong...what am I doing wrong now?
  6. Apr 29, 2009 #5
    You don't have ice heating up. It's already at 0 degrees. There is no term like 81*2.09*(Tf-25)
    It's the water from melting that will heat up.
  7. Apr 29, 2009 #6
    So then it should be...


    That's not correct either...
  8. Apr 29, 2009 #7


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    After melting completely, the ice water takes heat from water to reach the final temperature.
  9. Apr 29, 2009 #8
    Ok...if that is so...then...
    81g*4.190j/kg*Tf + 878g*4.19j/kg*Tf -91970.5 + 81*333 = 0
    so Tf = 16.1757
  10. Apr 29, 2009 #9


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    Appears to be correct.
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