The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

[Fool]

Hot gas doesn't contain heat any more than it contains work when compressed.

[Fool]

If that were the case, I think the universe would have reached equilibrium eons ago.

Ha ha stay on subject, Stirling Engines. The universe is way larger than our little puny, understatement, Stirling Engines, and probably takes way more time to cool. In fact the big explosion has cooled to 3 Kelvin long ago. Space is very cold.

The rest is heated by remote nuclear furnaces with calculable lifetimes, stars.

Our Stirling Engines cool off in a way way way, understatement, shorter amount of time.

[Fool]

(Fool, no help please)

Apparently Fool is no help. LOL.

I keep thinking about the old saying that has a horse and water. We need more mathematicians.

I get it, but doubt anyone else here does...

I keep thinking about Feynman's method of making people think by leaving errors in textbooks. And how much we should care about the "others". Many will also get it, but probably not the vocal ones.

Hey Fool, read this expression out loud and get back to us...

Ha ha. Good one.

Q=M•Cv•T=U

Que equals You with Mee See Vee Tea in the middle. LOL

Equating a Q with internal energy in an attempt to explain, as you pointed out "usually" means something else, that variables are just that, variable so I can make mistakes. He he he. By the way, thanks for the compliments and constructive corrections.

Q = input, output, etc
U = internal energy
T = temperature

Very good points. It still goofs me thinking about how ∆Q, ∆U etc differ from Q and U. I guess that is why variable definitions are given for each problem. Assuming standards gets into my dislecksia quickly.

[Fool]

Just to be clear, cooling due to transfer of kinetic energy is over and above cooling that may be due to "expansion", simply spreading out the heat over a larger volume.

IMO there is no such thing as heat attempting to reach equilibrium.

The idea that heat wants to combine with cold, seeks out or finds itself compelled to flow to cold is just Caloric/Carnot theory mythology.

This is probably the crux of your misunderstanding. You seem to think that heat gets conducted into an engine. But it won't conduct to anything cold.

All I see in your post is contradictions.

Throw a red hot bolt into a cup of water. The water gets hotter. The bolt gets colder. The two reach a common final temperature between the two.

Energy dissipates from concentrated, to less concentrated. Hotter to colder. The rate it does so has been observed to depend on temperature difference, conductivity (type of material) and area of path.

Gas solid or liquid.

[Fool]

But I think we've had this conversation before. If you keep adding wood to a fire it doesn't mean wood is not being added or burned if the fire looks the same after three days.

What the fire looks like is internal energy. The wood added is heat in. The wood burnt is heat out.

[Tom Booth]

...

Throw a red hot bolt into a cup of water. The water gets hotter. The bolt gets colder. The two reach a common final temperature between the two.

...

My point is, you have to throw the hot bolt into the water.

The heat doesn't just jump off the bolt and head for the water on its own.

[Tom Booth]

But I think we've had this conversation before. If you keep adding wood to a fire it doesn't mean wood is not being added or burned if the fire looks the same after three days.

What the fire looks like is internal energy. The wood added is heat in. The wood burnt is heat out.

As usual you're missing the point entirely.

[matt brown]

Hey Fool, read this expression out loud and get back to us...

Ha ha. Good one.

Q=M•Cv•T=U

Que equals You with Mee See Vee Tea in the middle. LOL

Equating a Q with internal energy in an attempt to explain, as you pointed out "usually" means something else, that variables are just that, variable so I can make mistakes. He he he. By the way, thanks for the compliments and constructive corrections.

Overall, they're all variables, but expressions are meant to limit any 'solution set' according certain rules. You're using what I call new age thermo (akin new age math) which fudges many long held rules. Yes, I get it, but take another look at your Q=.................=U and try to justify how any path function can (ever) equal a state function ??? Ultimately, this is Tom's biggest problem, simply mixing state functions PVTUm (etc) with path functions QW (ok, qw properly) to arrive at some wacky conclusions. And yes, boys and girls, this is the hallmark of thermo amateur hour.

Very good points. It still goofs me thinking about how ∆Q, ∆U etc differ from Q and U. I guess that is why variable definitions are given for each problem. Assuming standards gets into my dislecksia quickly.

I've seen a lot of bogus expressions creeping online lately across the web, and most include creative use of delta mod. I tend towards RA (ratio analysis) which ovoids these pitfalls and back it up with GA (graphical analysis) to catch anything that falls thru the cracks (RA has some shortcomings).

[Tom Booth]

...
this is Tom's biggest problem, simply mixing state functions PVTUm (etc) with path functions QW (ok, qw properly) to arrive at some wacky conclusions.
...

You mean my biggest gripe?

Because that is what the "Carnot Limit" calculations do. With the definition of "heat".

[Fool]

Yes, I get it, but take another look at your Q=.................=U and try to justify how any path function can (ever) equal a state function ???

I said I was wrong. Good catch. I don't think Tom will ever understand how a state variable differs from a path variable.

The heat doesn't just jump off the bolt and head for the water on its own.

It sounds like you are demanding proof that a red hot bolt in a comfortable, cool, room to suddenly radiate heat faster with a block of ice there too, just because the ice is colder, to verify the law of heat traveling from hotter to colder and faster with a grater difference.

[Tom Booth]

Yes, I get it, but take another look at your Q=.................=U and try to justify how any path function can (ever) equal a state function ???

I said I was wrong. Good catch. I don't think Tom will ever understand how a state variable differs from a path variable.

...

LOL...

Try looking at the thread title:


The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

[Fool]

I also don't think that Tom will ever understand how the differences between a state variable and path variable cancel when put into a ratio, and how they are depicted on a PV diagram.

Still he tries to equate, heat a path variable with, internal energy a state variable. He just loses it every time.

[Fool]

The heat doesn't just jump off the bolt and head for the water on its own.

Heat transfer in the air is by conduction, convection, and radiation. Which one of these are not the heat "jumping off and heading to the water"? None.

For your statement to pass scientific testing, all three must not be true, yet they all three are happening, been observed, measured, felt, and are true. Your statement is false.

When you surround, block, a hot mass and cold mass with an insulator you must expect the process to slow, unless your are below the critical insulation thickness.

[Tom Booth]

I also don't think that Tom will ever understand how the differences between a state variable and path variable cancel when put into a ratio, and how they are depicted on a PV diagram.

Still he tries to equate, heat a path variable with, internal energy a state variable. He just loses it every time.

So you claim. and have wasted 45 pages of forum space with your number juggling without being able to prove it.

"State variables" and "path variables" ARE constituent parts of a PV diagram. Hardly a change of venue.

It is you Carnot limit advocates who "equate, heat" (Q) "a path variable with, internal energy" (T) "a state variable".

No basis or justification for that whatsoever. A completely false equivalency.

[Fool]

Only the ratios are equivalent. Because the differences cancel.

That was shown many times over. The 45 pages of waste is from you not seeing the obvious.