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Thermodynamics problem - need help

  1. Oct 30, 2006 #1
    You take a jug of water from the refrigerator, and pour 240mL of it into a glass. The temperature of the beverage is 10.5ºC. You then add one 45g ice cube at 0ºC.

    Determine the final temperature and the amount of ice remaining if any.

    I've tried to work it out below, but my I must've made an error because the temperature has to be between 0 and 10.5. Where have I gone wrong?

    4.184*240(x-10.5) = -[45*333+4.184*45(x-0)]
    1004.16x - 10543.68 = -[29970+376.56x]
    1004.16x - 10543.68 = -29970-376.56x
    1,380.72x - 10543.68 = -29970
    1,380.72x = -19,426.32
    x = -19,426.32/1,380.72
    x = -14.06970276

    thanks
     
  2. jcsd
  3. Oct 30, 2006 #2

    marcusl

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    It is most convenient to work in calories rather than converting to joules.

    1. First find the amount of heat available in the water.
    240g * 10.5 C = 2520 cal

    2. find the amoutn of ice this will melt.
    Heat of fusion is 80 cal/g.
    Amount of melt = 2520 cal / (80 cal/g) = 31.5 g

    3. Find remaining ice
    45g - 31.5g = 13.5g

    4. Since we've pulled water temp to 0C, and there's ice remaining at 0C, final temp=0C
     
  4. Oct 30, 2006 #3
    thanks marcusl :)
     
  5. Oct 30, 2006 #4
    I'm not all that sure about that. I havent done a problem in calories before.
     
  6. Oct 30, 2006 #5

    marcusl

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    It's a small difference, doesn't really matter either way. 4.18 Joules of heat energy raise the temperature of 1 gram of water by 1 degree C. By definition, one calorie of heat energy raises the temperature of 1 gram of water by 1 degree C. The calorie unit is a little simpler or more direct in this problem. End result is the same :smile:
     
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