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Entropy Change in a Reversible Process multiple phase change

  1. Jul 29, 2016 #1
    1. The problem statement, all variables and given/known data
    (a)How much heat must be added to a block of 0.120kg of frozen ammonia initially at 100oC to convert it to a gas at 80oC given the following information?
    (b) assuming this could be done using a reversible process what would be the total entropy change associated with this operation given that ΔS=∫dQ/T (from b to a where b= Ti and a=Tf
    T melt = -78 C
    Tvap = -33 C
    c solid = 2030
    Lf = 332000
    C liquid = 4750
    Lv=1370000
    c, gas = 28
    M = 17.0 g/mol

    2. Relevant equations
    ΔS=∫dQ/T

    3. The attempt at a solution
    I've figured out the first part the heat added is simply 2.36 x 105J

    But I can't seem to get the second part analyzing the integral:
    $$ ΔS= \int_{173.15K}^{353.15K}\frac {dQ} T \ $$
    where $$ dQ = mcdT $$

    by integrating the function I get:

    $$ ΔS =0.120kg * 4750J/kgK *( ln(353.15) - ln(173.15)) $$

    I get ~ 400 J/K as an answer and the actual answer is 865 J/K ... I dont get what i'm doing wrong, is this the right path to take or am I actually suppose to take the integral from init Temp to melting point temp then from melting point temp to vaporization temp then to 80 oC ?
     
  2. jcsd
  3. Jul 29, 2016 #2

    rude man

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    Gold Member

    You must compute the entropy change separately for the different stages:
    1. 173K to 195K
    2. heat of fusion at 195K
    3. 195K to 240K
    4. heat of vaporization at 240K
    5. 240K to 353K

    That's 3 different integrals and two "gimmes" to compute the entire change in S.
     
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