Entropy change of a reservoir after heating something up

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SUMMARY

The discussion focuses on calculating the change of entropy for a 1 kg silver sample heated from 273 K to 373 K using the equation ΔS = ∫dQ/T. The entropy change for the silver is calculated as 0.312C, while the change for the heat reservoir, which remains at constant temperature, is derived to be -0.268C. The overall entropy change for the universe is positive, confirming the second law of thermodynamics. The discrepancy in the reservoir's entropy calculation highlights the importance of considering heat transfer at constant temperature.

PREREQUISITES
  • Understanding of thermodynamic principles, particularly the second law of thermodynamics.
  • Familiarity with the concept of entropy and its mathematical representation.
  • Knowledge of heat transfer and specific heat capacity (C) calculations.
  • Proficiency in calculus, specifically integration techniques.
NEXT STEPS
  • Study the derivation of the entropy change formula ΔS = ∫dQ/T in detail.
  • Learn about the implications of irreversible processes in thermodynamics.
  • Explore the concept of heat reservoirs and their role in thermodynamic systems.
  • Investigate the relationship between temperature gradients and entropy changes in materials.
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Students and professionals in physics, particularly those studying thermodynamics, as well as engineers involved in heat transfer applications.

Robsta
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Homework Statement


1kg of silver is heated by a large heat reservoir at 373 K from 273K. Calculate the change of entropy in:
a) the silver
b) the reservoir
c) the universe.

Homework Equations



ΔS = ∫dQ/T

The Attempt at a Solution



calculating the change in the silver first

ΔS = ∫dQ/T
= C∫dT/T
= Cln(T2/T1)
= 0.312

This is fine, I understand it. Now I need to work out the entropy change in the large reservoir. Since the change happens spontaneously, it isn't reversible and the entropy of the universe increases. This means that I can't just take the negative of the change in the silver.

Since the temperature of the reservoir doesn't change, is there no effective change in the entropy? This obviously shouldn't be the case because it would violate the second law. I'd appreciate any insight that can be offered.
 
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The temperature of the reservoir doesn't change appreciably, but it does lose the same amount of heat that the silver absorbs. As far as the reservoir is concerned, because the heat transfer took place at constant temperature, it thinks it experienced a reversible change (actually, it really did experience a reversible change). On the other hand, the silver experienced an irreversible change (since there were substantial temperature gradients in the silver during the transient heating). The combined change in both the silver and the reservoir must be positive because of the irreversibility in the silver. So you have enough information now to determine the entropy change of the reservoir, and the overall entropy increase for the combination (i.e., the universe).

Chet
 
Great, thanks for your help. So doing the entropy integral for a fixed temperature, I get:

ΔS = ∫dQ/T = (1/T) * C(ΔTsilver) = (1/373)*C*100
= -61.66
This number is massive compared to the integral I did above for the silver with a varying temperature. Could you perhaps tell me which number is wrong and why?
 
You got the first part correct : 0.312C
But, it looks like you made an arithmetic error (or something) in the second part. I get -0.268C.

Chet
 
Last edited:

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