Impossible Mono-thermal Source: Applying Laws of Thermodynamics

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Homework Help Overview

The discussion revolves around the application of the first and second laws of thermodynamics to demonstrate the impossibility of constructing a heat engine that operates with only one heat reservoir at a constant temperature.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the first law of thermodynamics regarding internal energy changes and the relationship between heat and work. Questions are raised about the meaning of a constant temperature in this context and its effect on work and efficiency.

Discussion Status

Some participants have offered insights into the implications of constant temperature on work output and efficiency, while others are questioning the foundational assumptions of the problem. Multiple interpretations of the thermodynamic principles are being explored without reaching a consensus.

Contextual Notes

There is an ongoing discussion about the meaning of "T ext." and its relevance to the problem setup, as well as the constraints imposed by the laws of thermodynamics on the operation of a heat engine.

Nusc
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Applying the 1st and 2nd law of thermodynamics to a cycle, show that it is impossible to build a heat engine with only one heat resevoir (T ext.)

First law : dU=dQ - dW

2nd law : It is impossible for any system to undergo a process in which it absorbs heat from a reservoir at a single temperature and converts the heat completely to mechanical work, with the system ending in the same state in which it began.

Where do I begin and is T ext.?
 
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Nusc said:
Applying the 1st and 2nd law of thermodynamics to a cycle, show that it is impossible to build a heat engine with only one heat resevoir (T ext.)

First law : dU=dQ - dW

2nd law : It is impossible for any system to undergo a process in which it absorbs heat from a reservoir at a single temperature and converts the heat completely to mechanical work, with the system ending in the same state in which it began.

Where do I begin and is T ext.?
First Law: Consider the change in internal energy if the heat is flowing from one reservoir to another at the same temperature. What does that say about the relationship between dQ and dW?

Second Law: All the heat cannot be converted to work. So what does that tell you about the value of dQ (and dW)?

AM
 
Okay so dU must be 0 and if there were true then by first law

W = -Q but if temp is constant than

Q = mCdelta T => Q =0

So W = 0

So the efficiency is 0

Is that why?
 
Nusc said:
Okay so dU must be 0 and if there were true then by first law

W = -Q but if temp is constant than

Q = mCdelta T => Q =0

So W = 0

So the efficiency is 0

Is that why?
There is no work done. So it is not an engine. (The efficiency is undefined).

AM
 

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