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!

Entropy of fusion; two methods, different result?

  1. Mar 5, 2010 #1
    "Calculate the entropy of fusion of A at 25deg C given that its enthalpy of fusion is 32kJ/mol at its melting point 146deg C. Also given is Cp,m(liquid)= 28J/mol/K and Cp,m(s)= 19J/mol/K."

    I thought of two approaches. Both should be valid (to my knowledge), but only the first gives the correct result. The first one:
    Using Kirchhoffs Law,
    [tex] \Delta H (T2) = \Delta H(T1) + \int_{T_1}^{T_2}\Delta Cp,m dT\approx \Delta H(T1)+\Delta Cp,m (T2-T1)[/tex]
    and then applying the definition
    [tex] \Delta S(T2)= \frac{\Delta H(T2)}{T2} [/tex]
    (gives the correct result)

    But my second idea apparantly failed:
    Simply setting up the cyclic equality
    [tex] \Delta S(s\to l,T2) = \Delta S(s,T2to T1)+\Delta S(s\to l,T1)+\Delta S(l,T1\to T2) [/tex]

    where the terms are of the form
    [tex] \Delta S(i,T1\to T2) = Cp,m \ln (T2/T1) [/tex]
    Can someone tell me why this approach gives a different result, about 73J/mol/K compared with the first, and correct, result of approx. 100J/mol/K.?
    Thank you in advance.
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted