Maximizing Heat Pump Efficiency with Constant Entropy

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Homework Help Overview

The discussion revolves around maximizing the efficiency of a heat pump while considering the principles of thermodynamics, particularly the second law. Participants are exploring the implications of heat exchange between objects at different temperatures within an isolated system.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants are attempting to solve for the final temperature of objects based on their specific heat capacities and initial temperatures. There are questions about the nature of heat exchange (radiation vs. conduction) and whether the objects can expand or contract. Some participants are also questioning the assumptions regarding the second law of thermodynamics and its implications for heat flow.

Discussion Status

The discussion is ongoing, with participants providing hints and references to the second law of thermodynamics. There is an exploration of the concept of entropy and its role in the problem, but no consensus has been reached on the correct approach or solution. Some participants express uncertainty about the equations needed to proceed.

Contextual Notes

Participants note that the problem does not specify how heat is exchanged between the objects and that it involves an isolated system, which raises questions about the spontaneity of heat flow and the potential for achieving a higher temperature than initially calculated.

  • #31
haruspex said:
No, I mean the point at which there is no gradient. When no more heat can be pumped into B, the temperatures of A and C will be equal, right? That gives you your third equation.
From these you can extract a cubic in one unknown. Now, solving a cubic is a lot easier if you happen to know one root. Here's where the problem setter has been kind to you - you do already have one solution to it. (Hint)

I got a cubic in one unknown,
2T_{A1}^3-1000T_{A1}^2+32\times 10^6=0
Honestly, I can not figure out that one root and had to use a calculator. The calculator gave me 2 positive roots, 256.16 and 400. The second root indicates that no heat is extracted from A, I don't see how this could have been obvious to me. :(

Corresponding to these two values of ##T_{A1}##, I get ##T_{B1}## equal to 200 and 487.67. Is 487.67 the correct answer?

haruspex said:
...the max temp achievable will correspond to no change in entropy..
I haven't been able to understand why the total change in entropy is constant? How did you even bring this relation? Have I missed out anything in my notes? If so, please give me a link which explains this.
 
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  • #32
Pranav-Arora said:
I got a cubic in one unknown,
2T_{A1}^3-1000T_{A1}^2+32\times 10^6=0
Honestly, I can not figure out that one root and had to use a calculator. The calculator gave me 2 positive roots, 256.16 and 400. The second root indicates that no heat is extracted from A, I don't see how this could have been obvious to me. :(
You had three equations for the three temperatures: known product, known sum, and two of them the same. The initial conditions met all these criteria, so must be a solution.
Corresponding to these two values of ##T_{A1}##, I get ##T_{B1}## equal to 200 and 487.67. Is 487.67 the correct answer?
That's what I got, as I recall. (Did it in my head while cycling.)
I haven't been able to understand why the total change in entropy is constant?
You were asked for the max temp that could be achieved (in principle). If entropy increases that will reduce your ability to pump heat up the gradient, so to get the max take entropy as constant.

This was a pretty problem. I hope you enjoyed the journey as much as I did.
 
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  • #33
Thank you haruspex for your input and patience. It was fun discussing the problem here, got to learn something new. :smile:

haruspex said:
If entropy increases that will reduce your ability to pump heat up the gradient, so to get the max take entropy as constant.

This cleared all my doubts, thank you again.
 

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