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

[Tom Booth]

Adding heat increases internal energy, work decreases internal energy.

In reality for transfers to or from a gas, work or heat are just different words to describe a transfer of kinetic energy in or out of the gas. Actually identical on a molecular or atomic level.

Fool likes to imagine some significant distinction between heat and internal energy. For a gas, there really isn't any.

The "vibrations" of the hot plate transfer energy to the gas which increases the ",vibration" of the gas. The gas transfers the same to the piston.

A simple transfer of kinetic energy

Hot plate molecule to gas molecule, gas to piston.

Xlnt description except for the caveat that adding heat TENDS to increase internal energy, and work TENDS to decrease internal energy.

Not sure what you mean.

Either energy is transfered into (or out of) the gas or it isn't.

[matt brown]

Temperature equates to internal energy through the following equation:

Q=M•Cv•T=U

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

Q notation requires extreme care and is usually restricted to a finite value in a relatively 'narrow' expression.

I have no problem with common heat usage, but always (ok usually) distinguish between

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

My problem is when notation strays far afield from something like dT=dU and stuff gets fuzzy, fast.

Heat is:
∆Q=M•Cv•∆T not equal to U

The equation:
n = (Qh-Qc)/Qh

Could be rewritten:
n = (Uh-Uc)/Uh

Which has nothing to do with heat, only the currently contained internal energy when at the hottest and coldest. For a full cycle:

But Tom is assuming it is written:
n = (∆Qh-∆Qc)/∆Qh

I hope this points out how they are different, but can be interchanged in a ratio equation, a fraction.

n=(Uh-Uc)/Uh

Uh = M•Cv•Th
Uc = M•Cv•Tc

Mathematically showing some thermal energy can be converted to work, but typically not all.

This equation says it all: W=Qh-Qc

Tom likes to equate an energy river to an energy lake. Not the same but can be interchanged in some equations.

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

[matt brown]

Not sure what you mean.

Either energy is transfered into (or out of) the gas or it isn't.

Any isothermal process has no net gain or loss of internal energy, only an energy transfer 'thru' gas.

So, your "Adding heat increases internal energy, work decreases internal energy." is not always valid.

[Tom Booth]

Not sure what you mean.

Either energy is transfered into (or out of) the gas or it isn't.

Any isothermal process has no net gain or loss of internal energy, only an energy transfer 'thru' gas.

So, your "Adding heat increases internal energy, work decreases internal energy." is not always valid.

"Net" does not mean energy is not transfered into and out of the gas.

The heat input is just balanced by the work output.

Or heat and work input is balanced by the heat and work output. Whatever the case may be.

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.

Why waste time on these constant idiotic nonsensical debates?

[Tom Booth]

....
The equation:
n = (Qh-Qc)/Qh

Could be rewritten:
n = (Uh-Uc)/Uh

Which has nothing to do with heat, only the currently contained internal energy when at the hottest and coldest. For a full cycle:

But Tom is assuming it is written:
n = (∆Qh-∆Qc)/∆Qh
....
Tom likes to equate an energy river to an energy lake. Not the same but can be interchanged in some equations.

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

Please don't put words in my mouth declaring what I think or assume or like, because you don't know what your talking about.

[VincentG]

Why waste time on these constant idiotic nonsensical debates?

If it can't be settled whether or not work lowers internal energy, it's not idiotic or a waste of time as long as it stays on track and civil.

[Tom Booth]

Why waste time on these constant idiotic nonsensical debates?

If it can't be settled whether or not work lowers internal energy, it's not idiotic or a waste of time as long as it stays on track and civil.

I was referring to Matt's comment:

... adding heat TENDS to increase internal energy, and work TENDS to decrease internal energy.

And then:

Any isothermal process has no net gain or loss of internal energy, only an energy transfer 'thru' gas.

So, your "Adding heat increases internal energy, work decreases internal energy." is not always valid.

These constant hair splitting, make wrong, frankly, stupid, inconsequential non-issues that Matt and "fool" will both latch onto and go on and on and on about.

It's like a bank balance spending and earning.

Adding money to your account raises the balance, spending money lowers the balance.

A simple statement of fact.

But let's get technical and say if you make a deposit of $100 at the same instant you spend $100 then that statement is not really true!

So... I didn't earn any money this week?

I couldn't buy anything with the money I earned?

Because my deposit and debit canceled each other?

A completely vacant argument, but the two of them have gone on and on making this argument over and over, wasting time and forum space

That, and other semantic quibbling over the meaning of words, like "heat", "vacuum", "pressure" etc. etc. seem to never end.

As far as I'm concerned, just making an argument for arguments sake.

These nit picking made up non-arguments, over and over, along with fools walls of meaningless equations and Matt's walls of meaningless charts and graphs injected into and disrupting nearly every conversation all the time, over and over.

[Fool]

Just because you don't understand it, doesn't make it meaningless. Seems like something you've said in the past, about refrigerators. If you want to eliminate squabbling, you probably don't want to enter any discussions.

Do you consider a bank balance the same as dollars in?

Do you consider a bank balance dollars out?

Do you consider a bank balance dollars saved?

Do you consider a bank balance dollars?

Dollars in minus dollars out equals the change in dollars, but it doesn't equal the current total balance.

Balance is not the same as dollars in.

Heat in minus heat out equals the change in internal energy, but it doesn't equal current internal energy stored.

Water into a lake (a river), minus water out of a lake (another river) equals the change in size of the lake, but it doesn't equal the total size of the lake.

You are the one wrapped up in semantics, Matt's point was crystal clear.

[Tom Booth]

Just because you don't understand it, doesn't make it meaningless. Seems like something you've said in the past, about refrigerators. If you want to eliminate squabbling, you probably don't want to enter any discussions.

Do you consider a bank balance the same as dollars in?

Do you consider a bank balance dollars out?

Do you consider a bank balance dollars saved?

Do you consider a bank balance dollars?

Dollars in minus dollars out equals the change in dollars, but it doesn't equal the current total balance.

Balance is not the same as dollars in.

Heat in minus heat out equals the change in internal energy, but it doesn't equal current internal energy stored.

Water into a lake (a river), minus water out of a lake (another river) equals the change in size of the lake, but it doesn't equal the total size of the lake.

You are the one wrapped up in semantics, Matt's point was crystal clear.

See what I mean?

VincentG, if you want to continue the discussion, I think you might want to make a fresh start of it with a new topic.

[Fool]

....
The equation:
n = (Qh-Qc)/Qh

Could be rewritten:
n = (Uh-Uc)/Uh

Which has nothing to do with heat, only the currently contained internal energy when at the hottest and coldest. For a full cycle:

But Tom is assuming it is written:
n = (∆Qh-∆Qc)/∆Qh
....
Tom likes to equate an energy river to an energy lake. Not the same but can be interchanged in some equations.

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

Please don't put words in my mouth declaring what I think or assume or like, because you don't know what your talking about.

On page 42, "The answer to life the universe and everything.".

I added the bolding:

In reality for transfers to or from a gas, work or heat are just different words to describe a transfer of kinetic energy in or out of the gas. Actually identical on a molecular or atomic level.

Fool likes to imagine some significant distinction between heat and internal energy. For a gas, there really isn't any.

The "vibrations" of the hot plate transfer energy to the gas which increases the ",vibration" of the gas. The gas transfers the same to the piston.

A simple transfer of kinetic energy

Summarizing what we think each other says is a valid form of trying to see another person's viewpoint, and maybe them seeing it too. If you disagree with all or part of the summery maybe you could discuss the misconception you perceive, rather than demanding things of others you are, apparently, not willing to do yourself.

I'm trying to explain and separate the difference between internal energy's total current level, and, energy input/output in the form of work, chemical, mass flow, and heat, E.C.T... All are different in some very important ways, and similar in that they are all energy measured at time in Joules.

As they say, you aren't going to understand quantum mechanics or Einstein Relativity unless you understand the mathematics. You aren't going to understand the difference between heat, temperature, and internal energy, unless you understand the mathematics.

[Fool]

"See what I mean?"

At least I was adding some science too, a point you didn't address choosing instead to just squabble. What kind of a comment would you consider yours? Sarcasm, taunting, gloating? Squabbling...

Rhetorical questions are questions where the obvious answer is meant to demonstrate a viewpoint. You haven't yet acknowledged that viewpoint. Or that clarification would help, if you don't understand.

[Tom Booth]

...
I'm trying to explain and separate the difference between internal energy's total current level, and, energy input/output in the form of work, chemical, mass flow, and heat, E.C.T... All are different in some very important ways, and similar in that they are all energy measured at time in Joules.
...

You are unduly complicating the issue.

There are no chemical reactions in a Stirling engine and there is no mass flow through the engine.

That leaves heat and work

The problem can be further simplified by realizing, as I pointed out before, gases are in their simplest form, so have no "hidden" or latent "internal energy".

Joule's Law:

The internal energy of a given quantity of a gas depends only on the temperature

[Fool]

gases have a temperature. Temperature is manifest by kinetic energy. Contained Kinetic energy is stored energy. It is not heat. It is not work. It is called internal energy.

A spinning flywheel doesn't have work in it, it has stored kinetic energy, internal energy. Work is just how it got there, or more precisely force, pushing. The flywheel doesn't contain "pushing".

A hot bolt isn't full of heat. It only radiates or conducts heat.

[Tom Booth]

gases have a temperature. Temperature is manifest by kinetic energy. Contained Kinetic energy is stored energy. It is not heat. It is not work. It is called internal energy.

A spinning flywheel doesn't have work in it, it has stored kinetic energy, internal energy. Work is just how it got there, or more precisely force, pushing. The flywheel doesn't contain "pushing".

A hot bolt isn't full of heat. It only radiates or conducts heat.

A flywheel or hot bolt isn't gas.

Try sticking to the subject and stop trying to complicate the issue.

[Fool]

And Tom, I don't see any evidence that heat will not flow as fast as it can to reach equilibrium, depending on the resistance of the medium it has to work with.

Heat a nut or bolt up to red hot. Feel the heat rushing out of the metal to your colder hand.

See how long you can hold a nut that is a water boiling temperature 100° C. And see how much faster that same nut warms up your hand if it is red hot. Higher temperature differences transmit energy faster.

The confusion is in conductivity verses R-value. All materials have both. But the area of the energy transfer path is very important.

A long skinny iron wire poked into a fire will get red hot on one side, because energy went from hot flame to cold metal. Leaving it in the fire for a long time the hot end stops getting hotter because it is now a flame temperature. The end you are holding doesn't, because energy being conducted up the wire gets lost to the cold surroundings.

Try a copper wire. It needs to be longer, or the iron wire can be shorter.

Because of the R-value.and path area the, "as fast as it can", might seem really slow or not at all.