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Internal energy

  1. Feb 13, 2004 #1
    Seem to have ran into a brick wall with this problem
    Determine the change in internal energy of 1 kg of water at 100 degrees C when it is fully boiled. Once boiled this volume of water changes to 1671 Liters of steam at 100 degrees C Assume the pressure remains constantat 1 atm

    things i know
    1 L =1E-3 m3
    1atm = 1.013E5 N/m2
    1 L = 1000 cm3 = 1E-3 m3
    formulsa used
    Q=mLv = 1kg*2.26E6 J/k Q=2260000 J
    W = -P(Vsteam-Vwater) = (1.013E5)*[(1671000 - 1000)*.001
    W = -169171E3 J

    change in U = Q+W
    U = -166911E3 J
    If i did make a mistake i think it is where W is .
    Thanks Joe
  2. jcsd
  3. Feb 14, 2004 #2


    User Avatar
    Science Advisor

    Hmm... I can't remember if your calc. of Q is
    correct but as far as W goes, if memory serves right,
    I believe the conversion should be 1 Atm*Liter = 101 Joule,
    and then you get:
    W = - P * dV = - 1 (Atm) * (1671-1) (Liter) =
    = -1670 * 101 J = -168670 J

    Live long and prosper.
  4. Feb 14, 2004 #3

    Doc Al

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    Staff: Mentor

    In your equation, you have the volumes in cubic cm instead of liters. You should have:
    W = -P(Vsteam-Vwater) = (1.013E5)*[(1671 - 1)*.001]

    (This is consistent with drag's calculation.)

    Also, 3 significant figures in your answer is plenty.
    Last edited: Feb 14, 2004
  5. Feb 14, 2004 #4
    Now if my Q is right and I use drags and Doc Als conversion then U is equal to 2.09E6 J
    is that correct

    these were some of the hw questions i didn't finish

    Thanks alot

    Joe :smile:
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