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MarcusAgrippa

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Clue: what is the definition of entropy?

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MarcusAgrippa

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ds=dw/t

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MarcusAgrippa

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dS = dQ/T

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Chestermiller

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MarcusAgrippa

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True. But most elementary exercises assume that dS is calculated in this way. Anyhow, I was correcting Piston's formula, not responding to your comments - which are valid. The question is not optimally worded. But this is a common fault in elementary courses. Perhaps Piston should check that he has transcribed the question correctly.

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Chestermiller

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Says who? And, do you really think that cooling a pile of bricks from 500 C to 20 C in air is a reversible process?True. But most elementary exercises assume that dS is calculated in this way.

Chet

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MarcusAgrippa

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i assume dw is the lost work which is an irrevesible process

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Chestermiller

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Chet

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Chet

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MarcusAgrippa

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Smith and Van Ness (Introduction to Chemical Engineering Thermodynamics, Chapter 16) have presented a precise and detailed analysis of how to determine the Lost Work, and have derived the following equation for the lost work:

$$LW=T_0ΔS-Q$$

where ΔS is the change in entropy of the system (in this case the bricks) between the initial and final equilibrium states, T

In the case of our process,

##ΔS=100C\ln (\frac{293}{773})=-97.01C## (kJ/K), where C is the heat capacity of the bricks.

and

##Q=100C(293-773)=-48000C## (kJ)

So, ##LW = -28424C+48000C=19576C## kJ##=19.6C## MJ

So, ##19.6C = 18.8##

So, ##C = 0.96 kJ/kgK##

and, ##ΔS = -93.2 (kJ/K)##

Hope this makes sense.

Chet

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Chestermiller

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Chet

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MarcusAgrippa

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If one accepts that "process irreversibility" = lost work, and that the formula LW=T_0 ΔS − Q is valid, then your solution seems inevitable. So my first question is how do you get "process irreversibility" = lost work? Is this standard engineering jargon? I have not come across this term before.

My second question is on the validity of the formula quoted. I would have calculated the lost work by first calculating how much heat is conducted from the bricks to the surroundings. I would then ask, how much work might have been gotten out of this heat had a Carnot engine been driven by the heat from the bricks being exhausted to the surroundings. Of course, this required the hotter reservoir (the bricks) to have a continually changing temperature, so the calculation is not straightforward.

I am trying to locate Smith and van Ness to see how they prove their formula. None of the numerous physics books that I have consulted address this problem, except perhaps for Zemansky who gives a formula that I am not able to reconcile with the one you have quoted.

This topic is more interesting than I first imagined. This issue of lost work may address a gap in my knowledge that has bothered me for some time: what is the difference in content of 1st and 2nd laws in the case of irreversible processes. The mismatch between T dS and dQ appears to be nicely illustrated by the formula you quote. It also emphasises the fact that the "T" that appears in the Clausius theorem is not the system temperature, but the temperature of the reservoir that is used to dump the rejected heat.

My second question is on the validity of the formula quoted. I would have calculated the lost work by first calculating how much heat is conducted from the bricks to the surroundings. I would then ask, how much work might have been gotten out of this heat had a Carnot engine been driven by the heat from the bricks being exhausted to the surroundings. Of course, this required the hotter reservoir (the bricks) to have a continually changing temperature, so the calculation is not straightforward.

I am trying to locate Smith and van Ness to see how they prove their formula. None of the numerous physics books that I have consulted address this problem, except perhaps for Zemansky who gives a formula that I am not able to reconcile with the one you have quoted.

This topic is more interesting than I first imagined. This issue of lost work may address a gap in my knowledge that has bothered me for some time: what is the difference in content of 1st and 2nd laws in the case of irreversible processes. The mismatch between T dS and dQ appears to be nicely illustrated by the formula you quote. It also emphasises the fact that the "T" that appears in the Clausius theorem is not the system temperature, but the temperature of the reservoir that is used to dump the rejected heat.

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Chestermiller

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Me neither.If one accepts that "process irreversibility" = lost work, and that the formula LW=T_0 ΔS − Q is valid, then your solution seems inevitable. So my first question is how do you get "process irreversibility" = lost work? Is this standard engineering jargon? I have not come across this term before.

Yes. This is exactly what they addressed in Smith and van Ness. They also provided an example to give a better idea how the continuously varying temperature is taken into account. It was very interesting.My second question is on the validity of the formula quoted. I would have calculated the lost work by first calculating how much heat is conducted from the bricks to the surroundings. I would then ask, how much work might have been gotten out of this heat had a Carnot engine been driven by the heat from the bricks being exhausted to the surroundings. Of course, this required the hotter reservoir (the bricks) to have a continually changing temperature, so the calculation is not straightforward.

Yes. Same here. To work this problem, I had to read up on the subject in Smith and van Ness, and I found the development very intriguing. The best relationship that I have found up to now for expressing the 2nd law mathematically is the Clausius inequality (which still is a very powerful relationship). You can see the close relationship between the lost work equation and the Clausius inequality. But this turns it into an actual equality. The development is, for some reason, presented in one of the latter chapters of Smith and Van Ness.This topic is more interesting than I first imagined. This issue of lost work may address a gap in my knowledge that has bothered me for some time: what is the difference in content of 1st and 2nd laws in the case of irreversible processes.

This is something that I have been trying to hammer home for a long time in Physics Forums, but the textbooks out there fail to emphasize this point. As a result, student after student is profoundly confused by the Clausius inequality. In the descriptions that I write, I use TThe mismatch between T dS and dQ appears to be nicely illustrated by the formula you quote. It also emphasizes the fact that the "T" that appears in the Clausius theorem is not the system temperature, but the temperature of the reservoir that is used to dump the rejected heat.

Chet

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MarcusAgrippa

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I would also appreciate the page number in Smith and van Ness where they develop that formula. I now have the book and am reading it with interest. Converting the inequality into an useful equality is something that I have searched for in vain in standard texts.

The idea of the temperature of an interface seems a bit murky.

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Chestermiller

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The interesting part is sections 16.2 and 16.3, and particularly, the part of example 16.1 discussed on page 552. As far as my write up is concerned, I will send it to you in a Conversation.Very interested. Do you need an email address, or can you post it here? I would also appreciate the page number in Smith and van Ness where they develop that formula. I now have the book and am reading it with interest. Converting the inequality into an useful equality is something that I have searched for in vain in standard texts.

The idea of the temperature of an interface seems a bit murky.

Chet

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