Relevance of S q / T

[MedPR]

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Found it on mcat-review as part of a discussion on entropy and the fact that universal entropy will always increase.

What is the significance?

 

[aSagacious]

I'm sorry that I won't answer your question directly, but your question reminded me of a great YouTube video that I saw a while ago. Just thought I'd share.

[YOUTUBE]http://www.youtube.com/watch?v=7ImvlS8PLIo[/YOUTUBE]

 

[MrNeuro]

Found it on mcat-review as part of a discussion on entropy and the fact that universal entropy will always increase.

What is the significance?

significance of it is you can never get 100% efficiency w/ an engine or heat pump

its the 2nd law of thermodynamics

 

[MedPR]

significance of it is you can never get 100% efficiency w/ an engine or heat pump

its the 2nd law of thermodynamics

What does efficiency have to do with entropy?

 

[chiddler]

what does q/T mean?

how do you translate S>q/t into second law of thermodynamics?

 

[MrNeuro]

What does efficiency have to do with entropy?

if the entropy of the universe always has to increase or remain the same then if you have an engine which takes in heat (lowering the heat of the universe) and turns it all into work then what we've done is decreased the entropy of the universe.

ΔS = q/T

So we know that the efficiency of an engine is 1- Qc/Qh or 1 - Tc/Th

so in order to have 100% efficiency we'd need the temperature of the cold sink to be 0K (which i believe is near impossible) ...

Why is efficiency of an engine dependent upon temperature and what does that have to do w/ the second law of thermodynamics?

We know the 2nd law is what you originally posted ΔS ≥ q / T so if the cold sink is colder then T is lower then a smaller heat transfer to the cold sink would be equivalent to a larger heat transfer to a hotter cold sink.

For example if the Temperature of the cold sink is 2 K then in order to increase the entropy of the universe by 50 percent we'd need to transfer 1 J of heat whereas if the temperature of the cold sink is 10 K in order to increase the entropy of the universe we'd need to transfer 5 J of heat.

So this all ties back into thermal efficiency because every joule of heat can be potentially used as work however due to the 2nd law of thermodynamics work is limited by the entropy of the universe as it must always remain the same or increase.

Make sense?

 

[chiddler]

Not making too much sense.

"We know the 2nd law is what you originally posted ΔS ≥ q / T so if the cold sink is colder then T is lower then a smaller heat transfer to the cold sink would be equivalent to a larger heat transfer to a hotter cold sink. "

reread this sentence! @_@

can you take it a little slower, please.

 

[MrNeuro]

Not making too much sense.

"We know the 2nd law is what you originally posted ΔS ≥ q / T so if the cold sink is colder then T is lower then a smaller heat transfer to the cold sink would be equivalent to a larger heat transfer to a hotter cold sink. "

reread this sentence! @_@

can you take it a little slower, please.

yeah sure

2nd law is ΔS ≥ q / T so the efficiency of the engine impinges on the entropy of the universe reamining constant or increasing

the cold sink of our engine (aka where our heat goes after work is performed) is part of the universe

So why does a colder sink increase the efficiency of our engine?

Well b/c if the engine is colder (t is smaller) that means that our ΔS is more sensitive to heat absorption.

Example ΔS ≥ q / T if we take -5J from a 10 K Hot sink were changing the ΔS of the universe by -0.5 J/K we know that we must return at MINIMUM 0.5 J/K so lets imagine three cases
Case 1 our cold sink is 10 K
Case 2 our cold sink is 5 K
Case 3 our cold sink is 1 K


Case 1: if the cold sink is 10 K then in order to return at least a 0.5 J/K to the universe we must input 10 J of heat into our cold sink which means that there is 0 J left to do work

Case 2: if the cold sink is 5 K then in order to return at least 0.5 J/K to the universe we must input 2.5 J of heat into our cold sink which means that there is 7.5 J left to do work

Case 3 If the cold sink is 1 K then in order to return at least a 0.5 J/K to the universe we must input 0.5 J of heat int our cold sink which means there is 9.5 J left to do work

Now the J left over is the maximum amount of J we could possibly have left over as the entropy of the universe is not increasing its remaining constant.

So do you see if the output sink which is part of the universe is colder that means that in order to obey the 2nd law of thermodynamics we must give some heat back to the universe and that this heat flow back to the universe can be minimized by making the cold sink VERY VERY COLD. Hence its impossible to get 100 % heat engine efficiency because a 0 K cold sink doesn't exist.

 

[chiddler]

yes thank you very much.