Entropy Change in a Reversible Process multiple phase change

AI Thread Summary
The discussion focuses on calculating the total entropy change for a reversible process involving the phase change of frozen ammonia. The initial heat required to convert the ammonia from solid to gas is determined to be 2.36 x 10^5 J. For the entropy change, the user initially attempts to integrate using a single formula but realizes that multiple stages must be considered. The correct approach requires separate calculations for each phase transition: solid to liquid, liquid to gas, and the temperature changes within each phase. The final entropy change should account for all these stages to reach the accurate total of 865 J/K.
elements
Messages
29
Reaction score
0

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 don't 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 ?
 
Physics news on Phys.org
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.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top