Thermodynamics: Violating Clausius => Violating Kelvin

[laser1]

Homework Statement

textbook paragraph

Relevant Equations

energy

I am not convinced that the heat entering $T_l$ from $T_h$ is equal to $Q_l$ in the Carnot engine. Why should it be? thanks

 

[haruspex]

Homework Statement: textbook paragraph
Relevant Equations: energy

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I am not convinced that the heat entering $T_l$ from $T_h$ is equal to $Q_l$ in the Carnot engine. Why should it be? thanks

That's how $Q_l$ is being defined here.

 

[TSny]

I am not convinced that the heat entering $T_l$ from $T_h$ is equal to $Q_l$ in the Carnot engine. Why should it be? thanks

You can assume that the Carnot engine is designed so that when it runs between the temperatures $T_h$ and $T_l$, it deposits exactly $Q_l$ of heat to the $T_l$ reservoir after running for some integer number of complete cycles. Here, $Q_l$ denotes the heat removed from the $T_l$ reservoir by the Clausius violator.

 

[Steve4Physics]

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I am not convinced that the heat entering $T_l$ from $T_h$ is equal to $Q_l$ in the Carnot engine. Why should it be? thanks

@laser1, if you are still unclear, maybe this will help.

First, imagine the setup without the 'Clausius violater'. Consider some example values:

$Q_h =100$ J. This is the heat transferred from the high temperature reservoir to the engine.
$W = 60$ J. This is the work done by the engine.
$Q_l~(= 100 - 60) = 40$ J. This is the 'waste' heat – the heat leaving the engine and entering the low temperature reservoir.

Now suppose you add the Clausius violater (CV) which (without any additional energy input) removes 40J from the low temperature reservoir and transfers the 40J to the high temperature reservoir. (This violates Clausius’ version of the 2nd law.)

Consider the overall system (engine+CV). Can you answer these questions:
1) What is the total heat transferred to/from the hot reservoir?
2) What is the total heat transferred to/from the cold reservoir?
3) What is the efficiency of the system (engine+CV)?

Click on the fuzzy spoiler below to check your answers:

1) the total heat removed from the hot reservoir is $Q_h -Q_l = (100-40) = 60$ J (because 100J was removed and 40J was added);

2) the total heat added to the cold reservoir is $Q_l +(-Q_l) = (40 - 40) = 0$ (because 40J was removed and 40J was added).

3) The overall effect is that we have converted 60J of heat from the hot reservoir to 60J of work. There is no waste heat transferred to the low temperature reservoir. Therefore the efficiency of the system is 100%, violating Kelvin’s version of the 2nd law.

 

[laser1]

Now suppose you add the Clausius violater (CV) which (without any additional energy input) removes 40J from the low temperature reservoir and transfers the 40J to the high temperature reservoir. (This violates Clausius’ version of the 2nd law.)

Here is my question: why does it have to remove 40 J from the low temperature reservoir? Why not add 50 J from the low temperature to the high. I know it will not make any difference when proving the Clausius-Kelvin equivalence, but then $Q_l$ won't be the same!

Edit: I think it would make a difference actually

 

[Steve4Physics]

Here is my question: why does it have to remove 40 J from the low temperature reservoir? Why not add 50 J from the low temperature to the high. I know it will not make any difference when proving the Clausius-Kelvin equivalence, but then $Q_l$ won't be the same!

Edit: I think it would make a difference actually

The Clausius violater (CV) is (an impossible) device that can transfer X joules of heat from a low temperarure reservoir to a high temperature reservoir with no additional energy required. We can choose X to be any convenient value.

We would like to show that use of the CV can lead to the (impossible) situation where we have 100% efficiency. 100% efficiency is achieved if there is no waste heat to be transferred to the low temperature reservoir; all of the heat taken from the hot reservoir is converted to work,

Using the Post #4 example, if we choose X= 40J then we can imagine directly feeding the CV with the waste heat from the engine – we wouldn’t even need a cold reservoir!

If we define % efficiency as:

$\frac {\text{work done}}{\text{net heat removed from hot reservoir}} \times 100$

you should be able to work out what the efficiency would be using X=50J. What do you get?

 

[laser1]

We can choose X to be any convenient value.

Fair enough, so "by definition" in disguise!

If we define % efficiency as:

work donenet heat removed from hot reservoir×100

you should be able to work out what the efficiency would be using X=50J. What do you get?

Yeah, >1 which is impossible. But also, I could've said X = 30 J and be fine! But yeah, "any convenient value" makes sense to me, thanks!

 

[laser1]

We would like to show that use of the CV can lead to the (impossible) situation where we have 100% efficiency

I still have some doubts. Would it still be correct to say (in the context of your example) that I can still transfer heat from a cold region to a hotter region, given that it is less than 40 J? Because that way, Clausius' statement is violated but Kelvin's is not! (In this case, $\eta \approx 86\%$)

 

[haruspex]

I still have some doubts. Would it still be correct to say (in the context of your example) that I can still transfer heat from a cold region to a hotter region, given that it is less than 40 J? Because that way, Clausius' statement is violated but Kelvin's is not! (In this case, $\eta \approx 86\%$)

So start from the other end. If Clausius is violated then there is some quantity of heat, Q, which can flow from the cold to the hot. Sending that back through the Carnot engine, extracting work W, returns Q-W to the cold side. Net result is to extract work W from the cold side.

 

[laser1]

So start from the other end. If Clausius is violated then there is some quantity of heat, Q, which can flow from the cold to the hot. Sending that back through the Carnot engine, extracting work W, returns Q-W to the cold side. Net result is to extract work W from the cold side.

Well, that doesn't really make sense to me after thinking about it. The overall process gives net of 0 J for the hot reservoir, and net of Q-W to the cold reservoir, with W work done. What does this prove?

 

[Lord Jestocost]

Maybe, Clausius’ statement becomes clearer if it is reworded as follows as done by Howard DeVoe in his book “Thermodynamics and Chemistry” (chapter 4.2 “Statements of the Second Law”):

“The Clausius statement of the second law: It is impossible to construct a device whose only effect, when it operates in a cycle, is heat transfer from a body to the device and the transfer by heat of an equal quantity of energy from the device to a warmer body.”