Entropy and Maximum work for two idential, finite sized bodies

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SUMMARY

The discussion focuses on the thermodynamic principles governing two identical, finite-sized bodies used in a heat engine. The change in entropy (ΔS) is calculated using the formula ΔS=nCv*ln(Th/Tc), where Cv represents the constant heat capacity. To determine the maximum work output, participants are advised to calculate the internal energy (U) for both bodies before and after the process, with the difference representing the maximum work that can be extracted.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically entropy and heat engines.
  • Familiarity with the formula for internal energy (U=CvT).
  • Knowledge of constant volume heat capacity (Cv) and its implications in thermodynamic processes.
  • Ability to apply logarithmic functions in thermodynamic equations.
NEXT STEPS
  • Study the derivation and applications of the entropy formula ΔS=nCv*ln(Th/Tc).
  • Learn how to calculate internal energy changes in thermodynamic systems.
  • Explore the concept of maximum work extraction in heat engines.
  • Investigate the implications of constant heat capacity in various thermodynamic cycles.
USEFUL FOR

This discussion is beneficial for students and professionals in thermodynamics, mechanical engineers, and anyone interested in the principles of heat engines and entropy calculations.

J.Welder12
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Two idential, finite sized bodies of constant volume and constant heat capacity are used to drive a heat engine- heat is taken from the hot (Th) body, work is done, and heat is ejected to the cold (Tc) body. Both bodies wind up at Tf

(a) What is the change in the entropy of the system?

(b) What is the maximum amount of work that can be done?


Relevant equations
U=CvT
Cv is not time dependent


(a) I got ΔS=nCv*ln(Th/Tc)
(b) No idea where to start, any help would be great!
 
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Welcome to PF, J.Welder12! :smile:

For (a) you need to consider the 2 bodies separately and apply your formula ΔS=nCv*ln(T2/T1) to them.
This gives 2 contributions to ΔS.

For (b) you should calculate U for both bodies before and after.
The difference is the maximum amount of work that can be extracted.
 

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