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Latent Heat with kcals

  1. Jun 21, 2007 #1
    1. The problem statement, all variables and given/known data

    How much heat is necessary to change 395 g of ice at -10°C to water at 20°C?

    2. Relevant equations

    mLf + mc(delta)T

    3. The attempt at a solution

    (.395kg)(80kcal/kg)+(.395kg)(1.00kcal/kg x 1 deg C)(20--10)
    (31.6kcal)+(.395kcal degC)(30)
    31.6kcal + 11.85 kcal=43.45 kcal

    I got the answer wrong obviously, the correct answer should be 41.5 kcal. Please help me, my teacher isn't so great and an explaination as to where I went astray would be appreciated. Thanks
    navy_bison: zzz:
  2. jcsd
  3. Jun 21, 2007 #2


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    i. Heating 395g ice at -10C to 0C
    0.395[kg]*10[k]*2100[J/kg*K] = 8295J

    ii. Melting 395g ice
    0.395 * 333000 = 131535 J

    iii. heating 0C water to 20C
    0.395 * 20 * 4190 = 33101J

    total = 172931 J
    1kcal = 4 184 J

    172931J = 41.34 kcal [depends on what values on the specific heat constants you use, and so on]
  4. Jun 21, 2007 #3
    RE: last post

    What is the general formula for figuring the latent heat using J? The example in my book (it's been 10 years since a science class for me) only uses the kcals to calculate the number. I'm not understanding exaclty where the numbers you've used are coming from. Is there a good website that you know of that could go more indepth?
  5. Jun 21, 2007 #4


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    I used the number from my text book.. It doesent matter what units you have, you can always convert them.

    There is only one general forumula for calculating latent heat:
    mass * coefficient[energy/units of mass] = Energy
  6. Jun 21, 2007 #5
    Thanks for the help, I've found a table that represents these specific heat vaules. I understand know that the ice must be heated, then melted (latent), then heated again. Add these up and get the answer. Thank you so much for the help. Recommending this site to all the people in my class!
  7. Jun 21, 2007 #6


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    good luck!

    hehe but remember to contribute with help yourselves... and also to search a bit for old topics considering similar problems.
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