Entropy Change in a Reversible Process multiple phase change

  • #1
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Homework Statement


(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

Homework Equations


ΔS=∫dQ/T

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 ?
 

Answers and Replies

  • #2
rude man
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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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