1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: How do I find the entropy variation?

  1. Dec 13, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the ∆S per mol between liquid water at -5 ºC and ice at -5ºC at 1020hPa
    ∆CP,m fusion = 37,3 J K-1 mol-1
    ∆fusH = 6,01 kJ mol-1
    The answer is 21.3 J/K mol

    2. Relevant equations
    Usually I solve these problems by steps when they are at P=1 atm but since its at P=1020 hPa i don't really know how to proceed.
    I tried using the following equation:
    ∆S= Q/T + nCpm ln T2/T1
    Since I am given the Cpm I interpret that the pression is constant throughout the process

    3. The attempt at a solution
    I first tried to solve it like I normally would, without taking into account that the press ion a was not 1 atm.
    I want to try and find the temperature at which water soldifies at 1020 hPa but I don't know how with the data I am given. I thought that since Qp is the enthalpy maybe I could find it using Q = n∆T specific heat of water
    but even thought I know what the specific heat of water is (41814 J/K kg) I'm not given this data so I'm unsure if this is what I have to do. 25383443_1748247851875564_819929200_o.jpg 25383256_1748250738541942_717490943_o.jpg
    This is what I first tried doing but after I realized my mistake I did this:
    I didn't obtain the value for entropy I should have with neither of the methods I tried. Where am I wrong?
  2. jcsd
  3. Dec 13, 2017 #2
    You are correct about thinking of this as a multi-step process.

    ice at -5C and 1020 hPa to ice at 0 C and 1020 hPa
    ice at 0 C and 1020 hPa to ice at 0 C and 1 atm
    ice at 0 C and 1 atm to water at 0 C and 1 atm
    water at 0 C and 1 atm to water at 0C and 1020 hPa
    water at 0C and 1020 hPa to water at 5 C and 1020 hPa
  4. Dec 14, 2017 #3
    fisica1.jpg fisica2.jpg
    I tried doing it this way and it really does make more sense but I can't still obtain the 21.3 J/K I should and I really don't know why. I thought maybe it's because of the Cpm I have been given , but it turns out it doesnt really differ from the one I was given, so I thought I might change the sign since one it's fusion and the other one its for solidification. Still, I haven't obtained the result I need and I don't know why. Could anyone please help me?
  5. Dec 14, 2017 #4
    You did this mostly correctly in your first attempt. The first thing to recognize is that 1020 hPa is essentially 1 atm, so there is no need to account for a pressure effect. So the change in entropy in going from 1 mole of ice at -5 C to water at - 5 C is given by:
    $$\Delta S=C_i\ln(273/268)+\frac{\Delta H(0)}{273}+C_w\ln(268/273)=(C_W-C_i)\ln(268/273)+22.0$$
    We know that ##C_w-C_i=37.3\ J/mol.K##
  6. Dec 14, 2017 #5
    I don't quite understand how you're using logarithmic properties here, could you explain it to me please?
  7. Dec 14, 2017 #6
    Are you saying that you are not sure how to determine the change in entropy of a solid or liquid when it is heated or cooled from an initial temperature to a final temperature (and why this should involve natural logs of temperatures)?
  8. Dec 31, 2017 #7
    I do, sorry. I had to revisit the basics. Thank you for your help.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted