Entropy Change in a Reversible Process multiple phase change

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SUMMARY

The discussion focuses on calculating the total entropy change for a reversible process involving a block of frozen ammonia transitioning from -100°C to gas at 80°C. The heat required for this transformation is determined to be 2.36 x 105 J. The correct approach to calculate the total entropy change involves integrating the heat capacities at different phases: solid, liquid, and gas, as well as accounting for phase changes at the melting and vaporization points. The final entropy change is calculated to be 865 J/K, necessitating the evaluation of multiple integrals across different temperature ranges.

PREREQUISITES
  • Understanding of thermodynamics, specifically entropy and reversible processes.
  • Familiarity with phase changes and associated heat transfer concepts.
  • Knowledge of integration techniques in calculus for thermodynamic equations.
  • Proficiency in using specific heat capacities and latent heats for different phases of matter.
NEXT STEPS
  • Study the derivation and application of the entropy formula ΔS=∫dQ/T in thermodynamics.
  • Learn about phase change calculations, including heat of fusion and heat of vaporization.
  • Explore the integration of piecewise functions for calculating thermodynamic properties across different phases.
  • Investigate the specific heat capacities of various substances, focusing on ammonia in different states.
USEFUL FOR

This discussion is beneficial for students and professionals in thermodynamics, particularly those studying heat transfer, entropy calculations, and phase change processes in substances like ammonia.

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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 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.
 

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