# Contradiction/breach 2nd law? Simply with carnot engine + negative temperature

1. Sep 19, 2010

### nonequilibrium

Hello,

For a reversible heat engine between temperatures $$T_1$$ and $$T_2$$, the ratio of heat going in and out the engine is $$\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$$ (second law).

Say $$T_1 = 200 K$$ and $$T_2 = 273 K$$, then heat goes from 2 to 1 and as we can see, the ratio of heat that goes in the engine is indeed bigger than what comes out ( = logical, due to delivery of work).

Take two ideal paramagnets, each at a negative temperature (note: negative temperature > positive temperature; the scale goes $$0 \to \infty \to - \infty \to 0$$).

Now say $$T_1 = -200 K$$ and $$T_2 = -273 K$$, then heat will go from 1 to 2 (Because the entropy of paramagnetic reservoir 1 will raise more than the drop in that of paramagnetic reservoir 2). The ratio of heat going in and out hasn't changed due to the minus cancelling in top and bottom. This says that the heat going in from 1 into the engine is less than the heat deposited from the engine into 2.

What's the explanation?

2. Sep 19, 2010

### Drakkith

Staff Emeritus
You cannot have a negative temperature on the Kelvin scale. 0 is absolute zero, you cannot get any colder than that.

3. Sep 19, 2010

### nonequilibrium

:( You missed my point completely, did you even read my post? I said negative temperature is even hotter than any positive temperature!

4. Sep 19, 2010

### Drakkith

Staff Emeritus
What does that even mean? How can a negative temperature be hotter than a positive temp?

5. Sep 19, 2010

### Pythagorean

Abstract math is not physics

6. Sep 19, 2010

### nonequilibrium

That if it's placed in contact with any system with a postive temperature, heat will flow from the negative to the positive system.

EDIT: (to pythagorean) What do you mean? With what part do you not agree? Please don't interpret me as somebody who is saying "the 2nd law is false"; of course not. I'm just trying to interpret the above. The two two things that seem contradictory to me: 1) the ratio of heat directly derived from the 2nd law; 2) heat flow for idealized paramagnets.

7. Sep 19, 2010

### Pythagorean

Kelvin is bound between 0 and a finite number because it's based on energy. You can't have negative energy and you can't have infinite energy.

8. Sep 19, 2010

### Pythagorean

This is the faulty application of abstract math that I'm referring to:

.

9. Sep 19, 2010

### Drakkith

Staff Emeritus
Why would heat flow from a lower temperature to a higher temperature? Or does this go back to saying that the negative temperature has more heat than the positive temperature?

10. Sep 20, 2010

### Gerenuk

Please excuse the ignorance of the previous posters. They really should read up some concepts beyond the very basics before they complain.

I haven't thought too much, but my first guess is that you are completely right, but there is no problem. In one case work goes in for a reversible operation and in the other case work goes out. Is that a problem?

11. Sep 20, 2010

### Staff: Mentor

Negative temperatures are only defined for systems with a limited number of energy states. I am not sure that the derivation for the equation of the heat engine is valid for such a system. I would think you would need to work this with statistical mechanics.

12. Sep 20, 2010

### Gerenuk

I think the equation still works for reversible processes in any system. What about my proposal that indeed there is a difference, but it's not a problem?

13. Sep 20, 2010

### Staff: Mentor

I suspect (just a guess really) that is correct, but I would have to actually work through the derivation of the heat engine for a negative temperature system before I would feel comfortable making that assertion.

14. Sep 20, 2010

### nonequilibrium

Hello, thank you for the replies.

Gerenuk, I'm not sure how it would not be a problem? If $$T_1 = -200K$$ and $$T_2 = -273K$$, then heat will flow (if allowed...) spontaneously from system 1 to system 2, we all agree on that. This means I can put in an engine with system 1 acting as the hot reservoir and 2 as the cold reservoir, where $$W > 0$$. Obviously the heat coming out of 1 must thus be greater than the heat going into 2, if work is delivered, yet the reversible equation $$\Delta S_{universe} = - \frac{Q_1}{T_1} + \frac{Q_2}{T_2} = 0$$ implies this is not the case (if you're wondering about the signs: the first term in its totality presents the entropy GAIN in the first system, so since the temperature is negative, I had to add a minus [N.B.: all $$Q$$ symbols are taken as absolute values]; reverse the situation for system 2).

DaleSpam, fair point! So you can't actually make them reservoirs without destroying their weird behavior, but okay, let's imagine the two pseudo-reservoirs really big and only a small engine in between. The equation $$\Delta S_{universe} = - \frac{Q_1}{T_1} + \frac{Q_2}{T_2} = 0$$ won't be exact anymore, but we can approach zero as close as we want, and then even for finite reservoirs we'll get the weird situation where the heat deposited is greater than the heat absorped, although work was delivered...

Really weird situation, cause it's such a simple set-up, but I can't pinpoint where it goes wrong...

15. Sep 20, 2010

### Gerenuk

Why should work be delivered? The only constraint is that it is reversible. But maybe this transfer actually requires heat?
Just a random thought.... not sure.

16. Sep 20, 2010

### nonequilibrium

Okay, let's do the following thought experiment:

Q_h is the heat going from 1 into the engine, Q_c is the heat going from the engine into 2. Now let's presume the engine USES work W as to fix our anomaly. Then Q_h + W = Q_c. Now stop the engine. Call the state of the universe we have now A. Now reverse the engine: now we get out W and Q_c leaves 2 and Q_h goes into 1, okay. Now as we know: heat can spontaneously flow from 1 to 2, so let the Q_h flow back spontaneously into 2. Stop. Call this state B. Now if you've followed closely, the net result of going from A to B: heat has been taken out of 2 and work has been delivered without any other effect in the universe.

The Kelvin-Planck statement says "It is impossible to convert heat completely into work in a cyclic process." We have just breached this in our thought experiment, thus W must be going out instead of in.

Or the summary of this post for the lazy ones: if it costs work rather than gives work, we'd directly be breaching the kelvin-planck version of the 2nd law.

17. Sep 20, 2010

### JaredJames

How do you achieve a negative temperature (-K)?

The results from putting a negative Kelvin figure into this equation are no different to a person accidentally putting the Celsius value in (when negative). Putting the Celsius value into this equation doesn't work.

Just because you can put other values in (particularly Celsius as they can be equal to the Kelvin Value - 0C = 273K etc and this is a common mistake) doesn't make the result that comes out correct.

You can prove any law of physics wrong if you invent values to go into it, but unless you can prove the existence of negative temperatures (below 0K - or negative energy), I think you're breaching PF guidelines.

Or am I just totally missing the point here?

Last edited: Sep 20, 2010
18. Sep 20, 2010

### Staff: Mentor

19. Sep 20, 2010

### nonequilibrium

jarednjames: Indeed you're missing the point, as you say yourself, but I don't mean that impolitely: I'd also be very skeptical if I hadn't heard of statistical mechanics before. The main difference between the temperature you're used to is that this temperature's no longer a degree for kinetic energy, but something more general (still dependent on the concept of energy). The important thing is that the equation I use is not dependent on the kinetic energy temperature and should also be applicable to the general situation, or at least I thought so. The conclusion seems to be that $$\Delta S = \int \frac{dQ_{rev}}{T}$$ does not work in general?

20. Sep 20, 2010

### JaredJames

So does this have a basis in real life? Is it useful or just some piece of maths that shows A but we can only achieve B due to the laws of physics?

"By the definition of temperature such a system has a negative temperature."

Does the temperature actually drop below 0K or is it purely showing it as a negative temperature (in reality, the temperature is +K (any positive temperature) but due to the maths it comes out as negative)?

21. Sep 20, 2010

### Staff: Mentor

I don't know what you mean by this. What is an "actual" temperature as opposed to the temperature you get "due to the maths"? Physical quantities have some definition, I don't know what an "actual" physical quantity would be in the absence or contradiction of its definition, especially not one so abstract as temperature.

22. Sep 20, 2010

### nonequilibrium

Well I'm talking about an ideal paramagnet, which doesn't exist, but that's like an ideal gas doesn't exist: the fact it doesn't exist doesn't make it useless, it's just an easy way to understand the basics. It does have a physical significance and should thus be taken seriously, IMO. Anyway,

(ignore the names on it, just took the right shape from google) Say the energy is the x-axis and the entropy is the y-axis. Then this is the graph for a paramagnet. As you can see, if you keep adding energy, eventually the entropy will go down (this can be proven, this is just the result). If $$\beta$$ is the derivative of the graph at a certain point, then according to the established relation $$\beta = 1/kT$$ we get that the right portion of the graph represents a negative temperature. If you think about it, using this graph, you'll understand why a negative temperature is actually hotter than any positive temperature. If you want more exact info, I direct you to a text book on statistical mechanics :) very interesting

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Anybody any ideas ? :/

EDIT: I PM'd jared to reply to his post below; didn't answer here cause I don't want to get too much off topic

Last edited: Sep 20, 2010
23. Sep 20, 2010

### JaredJames

I mean, does the temperature drop below 0K? Bare in mind no one can get to 0K, let alone below.

When I say the 'maths showing it as negative', I mean does the temperature show as negative in the equation but in real life there is always a positive temperature. If I have understood the graph above, the right side of the graph shows a negative temperature. But the actual temperature of the system is always positive. I can't really put it into words what I'm thinking.

I'll chalk this up to me not knowing what statistical mechanics is.

24. Sep 20, 2010

### D H

Staff Emeritus
As mr. vodka already mentioned, the temperature of systems with a negative temperature is higher than that of any system with a positive temperature. The discovery of such systems did lead to a slight modification of the second law of thermodynamics. Google "Kelvin-Planck-Ramsey statement".

25. Sep 20, 2010

### D H

Staff Emeritus
Oooooh! I got a googlewhack! Searching for the exact phrase "Kelvin-Planck-Ramsey statement" yields exactly one hit (it's gonna be two in short order when Google finds this thread). Getting rid of the quotes yields a lot more hits.