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Calorimetry and Water

  1. Nov 21, 2005 #1
    When 42.1 grams of water at 80.9 degrees Celsius are added to an unknown amount of water at 20 degrees Celsius. The final temperature of the mixture is 54.8 degrees Celsius. Calculate the unknown mass of water originally at 20 degrees Celsius.
    For known mass:
    q = (42.1 g)*(54.8-80.9 C)*(4.184 J/g*C) = 4593.0258 J
    Now for unknown mass:
    m = q/C*deltaT
    4593.0258 J/(4.184 J/g*C)*(54.8-20 C) = 31.5448 = 31.5 g???
  2. jcsd
  3. Nov 22, 2005 #2
    Any volunteers?

    Thanks again.
  4. Nov 22, 2005 #3


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    (Actually I believe by trying to bump your own threads you lose the chances of a reply - i.e. most people probably look for unreplied threads rather than ones that have been addressed)

    Anyway, going through it, I think your calculations hold.
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