1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Boiling Water and Energy.

  1. Oct 4, 2008 #1
    1. The problem statement, all variables and given/known data
    A kettle containing 1kg of boiling water is heated until complete evaporation. Calculate q, w, "delta"U and "delta"H for this process. Suppose that the water vapor is an ideal gas and molar "delta"H of vaporisation at 373K = 40.6kJ/mol.


    2. Relevant equations
    H = U + PV
    "delta"U = W + q


    3. The attempt at a solution

    number of moles = 1000g/(18.02g/mol) = 55.49mol

    q = (molar "delta"H of vaporisation) x number of moles = 40.6x10^3J/mol x 55.49mol = 2.25x10^6J

    U = H - PV
    "delta"U = "delta"H - [("delta"P)(V) + ("delta"V)(P)] = "delta"H - [0 + nR"delta"T] = "delta"H
    (since there is no pressure or temperature variation, "delta"U = "delta"H? I know pressure is constant, but is temperature constant too during this phase change?)

    Therefore, "delta"U = 2.25x10^6J

    W = "delta"U - q = 0J

    I feel like I'm missing something in the calculation of "delta"U. Any help is greatly appreciated. Thanks.
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?
Draft saved Draft deleted



Similar Discussions: Boiling Water and Energy.
  1. Water surface (Replies: 0)

Loading...