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Homework Help: Mc(delta) t (What am I doing wrong?)

  1. Feb 2, 2009 #1
    1. The problem statement, all variables and given/known data

    A 19.2 g sample of ice at -10.0°C is mixed with 100.0 g of water at 74.9°C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of H2O(s) and H2O(l) are 2.08 and 4.18 J g-1 °C-1, respectively, and the enthalpy of fusion for ice is 6.02 kJ/mol.
    __________ degrees Celsius

    2. Relevant equations

    q = mass * specific heat * (delta) temp

    3. The attempt at a solution

    I thought I knew how to do it but keep getting it wrong. =(

    19.2 g ice at -10.0 deg going to 0 deg:
    19.2g x 2.08 J/g deg x 10 deg = 399.96 J
    Melting 19.0 g ice at 0 deg requires-
    19.2g x (1/18.02 g/mol) x 6.02kJ/mol x 1000 J/kJ x = 6350.77 J
    399.96 + 6350.77 = 6750 J
    That energy will come from cooling 100.0 g H20:
    6750 = (100)(4.18)(T-74.9)
    T(final) = 91.04

    thats wrong on webassign though...

    I also tried doing 74.9 - T to see if that works and I get 58.8 but thats wrong too.

    Thanks for any help!!! =)
     
  2. jcsd
  3. Feb 2, 2009 #2

    Borek

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    Staff: Mentor

    Once the ice gets melted you have a mixture of 19.2g water at 0 deg C and 100g water at some other temperature. Neither of these temperatures is final.

    Also note that 91.04 is obviously wrong - you can't take a water at 74.9 deg C, add ice to it and end with water that is hotter than it was before.
     
  4. Feb 2, 2009 #3
    Hmmm...then how would I relate the temps if they are not final? What else would I need to do?? :-/
     
  5. Feb 2, 2009 #4

    Borek

    User Avatar

    Staff: Mentor

    You know that cold water has to get warmer and the cold one has to get colder. Heat gained = heat lost, they have both identical final temperature.
     
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