2 body engine, final equilibrium temperate and work produced

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Discussion Overview

The discussion revolves around a heat engine operating between two bodies with different heat capacities and initial temperatures. Participants are exploring the final equilibrium temperature of the bodies and the work produced by the engine, focusing on the principles of a reversible Carnot cycle.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant presents a problem involving two bodies with constant heat capacities and asks for the final equilibrium temperature and work done by the heat engine.
  • Another participant suggests calculating the change in entropy for each reservoir in terms of the final temperature, T*, and notes that the change in entropy of the working fluid is zero in a Carnot cycle.
  • There is a repeated inquiry about whether the total change in entropy of the two reservoirs is zero, with one participant affirming that it is indeed zero.
  • A participant questions whether a solution has been reached yet, indicating a lack of progress in solving the problem.

Areas of Agreement / Disagreement

Participants generally agree on the concept that the total change in entropy of the two reservoirs is zero, but the overall problem of finding the final temperature and work done remains unresolved.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the heat capacities and the specific conditions of the Carnot cycle, which may affect the calculations of equilibrium temperature and work done.

Who May Find This Useful

Students and enthusiasts interested in thermodynamics, particularly those studying heat engines and Carnot cycles, may find this discussion relevant.

mrmerchant786

Homework Statement


2 bodies with contant heat capacity C and 2C, are initially at temp T and 2T. if a heat engine executing a reversible carnot cycle operates between the two bodies until their temperatures are equal, what is the final temp of the bodies and how much work is preformed by the heat engine

Homework Equations


W=Qη dQ=CdT

The Attempt at a Solution


q2 = C2 * (Tf-T2)
q1 = C1 * (Tf-T1)
So far this is where I'm at, I don't know how to proceed further to find the equilibrium temperature and then to fine the work produced
 
Physics news on Phys.org
Suppose T* is the final temperature. In terms of T*, what is the change in entropy of each of the two reservoirs. Since you are operating using a Carnot cycle, the change in entropy of the working fluid is equal to zero. What does that tell you about the total of the entropy changes of the two reservoirs?
 
Chestermiller said:
Suppose T* is the final temperature. In terms of T*, what is the change in entropy of each of the two reservoirs. Since you are operating using a Carnot cycle, the change in entropy of the working fluid is equal to zero. What does that tell you about the total of the entropy changes of the two reservoirs?

will it be zero?
so s1 +s2 = 0
 
mrmerchant786 said:
will it be zero?
so s1 +s2 = 0
Yes. ##\Delta S_1+\Delta S_2=0##
 
So, have you solved it yet?
 

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