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Homework Help: Another problem (Phase Change Heat Calculation)

  1. Dec 12, 2017 #1
    1. The problem statement, all variables and given/known data

    upload_2017-12-13_0-57-4.png
    2. Relevant equations
    q = mcat
    mols*hvap

    3. The attempt at a solution

    To decrease temp from 70 to 56 degrees (bp), q=mcat = 1195.5
    to convert this gas into liquid , must do mols * hvap = 29129 joules
    to decrease temp from 56 to 0 thought, i get q= mcat = -7047 J (which is negative- is this wrong?)

    it seems like the correct answer is when 7047 is positive after adding these up,. i get 37.37 kj. which seams like the answer (cant see it cause my professor erased it...)

    Why wouldn't i subtract the last number?
     
  2. jcsd
  3. Dec 13, 2017 #2

    Borek

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    Staff: Mentor

    No idea where you got the minus from - cooling the liquid is no different from cooling the gas (for which you got the correct answer).
     
  4. Dec 13, 2017 #3
    from mcat , i got the negative from change in temperature (final mius initial) so (0-56)
     
  5. Dec 13, 2017 #4

    Borek

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    Staff: Mentor

    So, for colling of liquid ΔT = final - initial = 0 - 56 = -56, yes?

    And what was the change in temperature for the cooling of gas?
     
  6. Dec 13, 2017 #5
    Oh I see. We are looking for any heat removed from cooling the gas so we should add heat to cool the liquid since it was used to cool the gas as well
     
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