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Changes of State ice to steam problem

  1. Feb 27, 2006 #1
    Ok, I am having some trouble with this problem:

    How much energy does it take to convert .500 kg ice at -20C to steam at 250 C? Specific heat capacities: ice: 2.1 j/g C, liquid: 4.2 j/g C, steam: 2.0 j/g C. Hvap = 40.7 kj/mol, and Hfus = 6.02 Kj/mol.


    What I have done so far is use q=mst for all 3 phases to obtain the energy in 3 phases.

    q= (.500)(2.1)(20) = 21000 J, this is for the ice from -20C to 0C
    q= (.500)(4.2)(100) = 210000 J, this is from 0 C to 100 C because of liquid water to boiling pt (steam).
    q= (.500)(2.0)(150) = 150000 J, this is from 100C to 250C

    Am I on the right track? Or am I doing this all wrong? I don't have a clue what to do next. However I think it must use the Clausius-Clapeyron equation in some context. I'm not sure what to do with the Hfus either. Please HELP!!

    Thanks,
    Jkotha
     
  2. jcsd
  3. Feb 27, 2006 #2

    Astronuc

    User Avatar

    Staff: Mentor

    One also has to account for the heat of fusion (ice absorbs energy (heat) at constant temperature while changing to liquid) and heat of vaporization (steam condenses (by releasing heat) to liquid at constant temperature).
     
  4. Feb 28, 2006 #3
    I Got It!

    Hey, thanks. YOU ARE THE MAN. I can't believe I forgot about those.
     
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