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

[Fool]

Stroller,

VincentG's finger will absorb negligible heat. You described a isentropic process. Only one mass, the internal gas.

[Stroller]

Stroller,

VincentG's finger will absorb negligible heat. You described a isentropic process. Only one mass, the internal gas.

True. It'll absorb enough for him to feel the increase in temperature of the gas though.

[Stroller]

Yes. This is what Charles law tells us:

"When the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be in direct proportion."

Hence:

V1/T1 = V2/T2

The equation shows that, as absolute (Kelvin) temperature increases, the volume of the gas also increases in proportion.

Ok, it may just be me, but I don't have this completely straight.

The joules needed to raise 100cc from 300k to 600k at constant volume = ?

That's a different and more complex case. Since the pressure will be raised as the gas is heated in a constant volume, the rise in pressure will add to the rise in temperature. So less Joules of input heat energy will be required than we calculated with Charles law.

Then, the temperature of the gas after free adiabatic expansion(with no additional heat input) from 100cc at 600k to 200cc is = ?

Lower than 600K

The joules needed to heat 100cc at 300k through a free expansion to 200cc at 600k = ?

We did this one already. It's 3.65J

Now of course, "free expansion" is still fighting the 14.7psi of the atmosphere. But that is the baseline for this correct?

Correct.

Thanks for your patience. These are things I think need to be established before discussing any further dynamic scenarios.

No problem. It's easy to get off track and end up at cross purposes (and cross) when discussing thermodynamics. Best taken in small steps

[Stroller]

Ok, it may just be me, but I don't have this completely straight.

The joules needed to raise 100cc from 300k to 600k at constant volume = ?

That's a different and more complex case. Since the pressure will be raised as the gas is heated in a constant volume, the rise in pressure will add to the rise in temperature. So less Joules of input heat energy will be required than we calculated with Charles law.

Then, the temperature of the gas after free adiabatic expansion(with no additional heat input) from 100cc at 600k to 200cc is = ?

Lower than 600K

The joules needed to heat 100cc at 300k through a free expansion to 200cc at 600k = ?

We did this one already. It's 3.65J

It looks like I goofed my earlier calc by a factor of 10, so it's 36.5J, not 3.65.

In the more complex case of joules needed at constant volume, a reasonable engineering approximation would be to multiply the Cv value of air by the mass of air and the change in Temperature.

0.718 x 0.12g x 300 = 25.85J

This is 9.65J less than the free expansion case. So to answer your other question of what T the air will be once we allow it to expand freely after it reached 600K at constant volume, we just need to work out how much 100cc of freely expanding air will heat up if we put 25.85J of heat energy into it.

Q 25.85J / (Cp value of air is 1.01 x mass of air 0.12g) = Delta T = 213.28K
We add this result to the initial temperature of 300K and we get 513.28K

[VincentG]

It looks like I goofed my earlier calc by a factor of 10, so it's 36.5J, not 3.65.

In the more complex case of joules needed at constant volume, a reasonable engineering approximation would be to multiply the Cv value of air by the mass of air and the change in Temperature.

0.718 x 0.12g x 300 = 25.85J

This is 9.65J less than the free expansion case. So to answer your other question of what T the air will be once we allow it to expand freely after it reached 600K at constant volume, we just need to work out how much 100cc of freely expanding air will heat up if we put 25.85J of heat energy into it.

Q 25.85J / (Cp value of air is 1.01 x mass of air 0.12g) = Delta T = 213.28K
We add this result to the initial temperature of 300K and we get 513.28K
Top

Thanks for rechecking. The thing about the rise in pressure causing a rise in temperature during Cv heating, and subsequent reduction in required heat input energy is something I'd like to focus on in the future as I think it's relevant to internal cooling.

513k at the end of free expansion is actually higher than I would have thought. I don't imagine it would still expand to 200cc at only 513k would it? That would seem to violate pv=nrt.

Using my gas calculator it would expand to about 170cc.

[VincentG]

36.5j for free expansion to 200cc at 600k.

Now how would we calculate additional heat input for the 102g piston in your first example?

[Stroller]

36.5j for free expansion to 200cc at 600k.

Now how would we calculate additional heat input for the 102g piston in your first example?

That will still be 1J. The amount of energy required to lift 1Newton (102g at Earth's surface) through 1meter.

[VincentG]

That will still be 1J. The amount of energy required to lift 1Newton (102g at Earth's surface) through 1meter.

I understand, but how do we show that extra energy being applied to the gas, mathematically?

[Stroller]

Energy required to expand the gas to 200cc AND lift a 1N weight by 1m = 36.5J + 1J = 37.5J

[VincentG]

Energy required to expand the gas to 200cc AND lift a 1N weight by 1m = 36.5J + 1J = 37.5J

How does the 1J interact with the gas to lift 1N? It seems too simple to just add 1.

Where does the extra Joule go?

[Stroller]

The extra 1J gives the gas the pressure it needs to lift the weight of the piston through 1m.
The extra 1J is stored in the piston at it's new altitude as gravitational potential energy.

[VincentG]

Often times science is math that cannot be put into words.

In this case, it's words that cannot be put into math?

If I were a teacher I would ask to show your work. And I'm not trying to single you out. I would just like to know.

[Stroller]

We already did the math in previous posts.

The joule symbol: J, is the unit of energy in the International System of Units (SI). It is equal to the amount of work done when a force of one newton displaces a mass through a distance of one metre in the direction of that force.
https://en.wikipedia.org/wiki/Joule

So what is the weight of that mass? I chose the piston to be 102g because in the earth's gravitational field at the surface, that is the weight of the mass that 1J of energy will lift by 1 meter. 1J is equivalent to 1 Newton-meter (and many other things - see the list on wikipedia).

It makes no difference whether the weight is lifted by a spring, the gas pressure, or any other means. The energy required is still 1J.

If we take the 102 piston and drop it on the ground from a height of 1meter, 1Joule of energy will be dissipated in the forms of sound, and heat (it'll warm the ground it hits a little bit).

But while it's held up by the gas pressure, that 1J is stored in the piston's 102g mass as gravitational potential energy.

[VincentG]

If the starting pressure at 300k and 100cc was 1atm + the added pressure of the 102g piston, it would be a basic pv=nrt calc.

But if the temperature of 100cc at 300k and just 14.7psi is doubled to 600k, any additional weight will no longer allow expansion to 200cc. The only way to continue expansion to 200cc with the load of the 102g piston would be to increase gas temperature beyond 600k.

That is no longer a case of just adding "Joules". Yes, Joules must be added, but Tmax needs to rise above 600k, rendering the exercise invalid from the start.

Which leads back to my original question of how to prove that heat is destroyed when work is done.

[Stroller]

I'm an engineer, not a forensic nitpicker. I do practical calcs good enough to get the job done, and don't bother about negligible additional forces. That's because nothing ever lives up to its theoretical promise anyway.

If you want to know how many millimetres or extra pascals the gas compressed when the piston was placed on it, ask someone else. ;)

Which leads back to my original question of how to prove that heat is destroyed when work is done.

Heat isn't a substance, it's manifested as a flow of energy from a hotter body to a colder body.
Energy is neither created nor destroyed, but it can get translated from one thing to another. In our example, the extra Joule started out as additional energy passed from the external energy source into the gas, and ended up as one Joule's worth of gravitational potential energy stored in the piston. So the extra Joule didn't make the gas any hotter, because bits of it were getting stored in the piston as GPE as it rose.