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

[Drakkith]

Because temperature is defined by T=dE/dS. And all equations are based on this identity. You cannot arbitrarily introduce a minus sign there. Or at least it wouldn't make sense, because you had to rewrite all the other thermodynamics equation and thus going back to negative temperature.

You misunderstand what I'm saying. The temperature is still a negative temperature. However, you can calculate the energy that the system would deliver to another system and from that you could convert that negative temperature into a positive temperature for use in the math.

For example, if a 500k system delivered 50 joules of energy to a 200k system, and -100k system ALSO delivered 50 joules of energy to a 200k system, then for the purposes of math you could simply insert 500k into all the spots where you would have put -100k.

200k(T1)/500k(T2) = Q1/Q2.
200k(T1)/-100k(T2) = Q1/Q2.
You could also rewrite this as 200k/500k=200k/-100k. In terms of energy content, then both 200k/500k and 200k/-100k would be the same in this circumstance.

If QH is the energy transferred from the hot resevoir, and QC is the energy being transferred to the cold reservoir, then:
QH in the 200k/500k system would equal QH in the 200k/-100k system. Its the same for QC.

Note that in relation to value of 200k for T1, if i used 300k instead then -100k would still impart 50 joules into T1, so you would have to change the value of T2 to a higher value that would transfer 50 joules into 300k. If we say that 700k would do that, then:
300k/700k = 300k/-100k.

The total amount of thermal energy transferred between the hot reservoir and the system will be: QH = TH(SB - SA) If TH is a negative value, then QH, the amount of energy transferred to the system would be NEGATIVE according to this equation. Obviously that doesn't happen because the value for TH is NEVER a negative value, as you cannot have less energy than absolute zero at 0K.

I am NOT saying that negative temp doesn't exist. The values used in the math represent the energy content of systems at those temperatures. THAT is what is key here. THAT is why you cannot use a negative value to calculate the carnot cycle, because you cannot have absolute NEGATIVE energy in any system. Negative temperature does not imply negative energy, but if you use a negative value for T in your math, then that is implying that you have negative energy.Now, before anyone takes 1 little snippet out of my entire post and try to use that to tell me why I'm wrong like what's been happening, I request that you do the math first. Goodnight all.

 

[Dale]

But to use these equations for a heat engine you can't have it negative, it must be positive. Otherwise weird things happen, like heat flowing from the cold to the hot. Or the engine doing work using less energy than was put in. Wasn't that the problem on the 1st page of this thread? The math wasn't working out correctly?

I agree, but the solution isn't to mess with the definition of temperature but rather to derive an equation for a generalized Carnot cycle for a negative temperature system. That was my point way back in post 11.

Temperature and entropy are much more fundamental concepts, and the Carnot cycle equations are derived from them under certain assumptions. We are violating one of those assumptions (ideal gas -> positive temperatures) in this system, so we have to re-do the derivation, not redefine the fundamentals.

 

[Andy Resnick]

Temperature and entropy are much more fundamental concepts, and the Carnot cycle equations are derived from them under certain assumptions. We are violating one of those assumptions (ideal gas -> positive temperatures) in this system, so we have to re-do the derivation, not redefine the fundamentals.

The Carnot cycle (and thermodynamics generally) does not make any assumptions about ideal gases; it does not require the existence of an ideal gas (or atoms). The utility of Kelvin's (second) absolute temperature scale is that gas thermometers (using arbitrary gases) behave as ideal gas thermometers (i.e. identically) at low temperature. Thus, we can compare different thermometers by use of a material independent scale.

Assigning a unique temperature to a substance requires the substance to be at thermal equilibrium. Thus, applying temperature to a variety of systems is problematic- any system far from equilibrium, for example. Negative temperatures in physical systems arise when considering metastable systems- systems that can be held at steady-state conditions far from equilibrium, such as inverted atomic populations or spin states.

The situation with the proposed negative temperature heat engine is formally identical to that of a machine made of antimatter- as long as the machine does not interact with 'normal' matter, all is well.

 

[Dale]

The situation with the proposed negative temperature heat engine is formally identical to that of a machine made of antimatter- as long as the machine does not interact with 'normal' matter, all is well.

Yes, I like that analogy.

 

[Gerenuk]

The difference is that it is very easy to create negative temperature. As I said just take any system with finite energy capacity and almost fill it. You don't need metastability or any other anomaly. It can be an ordinary system as long as the maximum allow total energy is finite.

Of course it's a trivial matter to argue that if you put such a system in contact with gases, it will not keep it's temperature. But I wouldn't exclude negative temperature from consideration just because it cannot be in equilibrium with air.

 

[Drakkith]

I agree, but the solution isn't to mess with the definition of temperature but rather to derive an equation for a generalized Carnot cycle for a negative temperature system. That was my point way back in post 11.

Temperature and entropy are much more fundamental concepts, and the Carnot cycle equations are derived from them under certain assumptions. We are violating one of those assumptions (ideal gas -> positive temperatures) in this system, so we have to re-do the derivation, not redefine the fundamentals.

What your proposing is pretty much what I'm doing. Instead of coming up with a new equation, which would use -Tk, i am taking that -Tk and simply converting it to a positive number before i put it into the equation. Both ways would accomplish the same thing, calculating the energy transfer between 2 sources and the amount of work one would get out of the engine.

 

[Dale]

Both ways would accomplish the same thing, calculating the energy transfer between 2 sources and the amount of work one would get out of the engine.

Are you certain of that? I am not. To prove that would require a derivation of the Carnot cycle for a negative temperature system anyway.

 

[Drakkith]

Are you certain of that? I am not. To prove that would require a derivation of the Carnot cycle for a negative temperature system anyway.

I am certain enough to explain why I think so, whether it is wrong or right. I am also certain that I have addressed the underlying problem shown in the original post of this thread and how to avoid it. If not, please tell me why.

I think everyone can safely agree that the way temperature is defined for a negative temperature, and the use of that temperature in thermodynamics math needs to be clarified or expanded on. This whole thread is a case in point.

 

[DrDu]

Mr.Vodka, I still don't see any contradiction in your thought experiment.
The point is that a system with negative temperature allways has a tendency to emit energy spontaneously (that is irreversibly) and become colder and colder until its temperature becomes - infinity and then + infinity and finally positive, as you said in your first post. Think of a system of spins with population inversion. You will agree that the whole process is driven by the increase of entropy of the system. On the other hand, you can only keep up the negative temperature of the system by
isolating it to stop it loosing energy or, as this is not possible completely, you have to continuously pump the system to maintain inversion. Think e.g. of a laser.

In the course of the Carnot cycle you consider, the total entropy of the two systems remains constant and they finally reach the same temperature.
You could also consider mixing the two systems directly, but in that case entropy will increase, as it is a spontaneous irreversible process.
For a system with negative temperature, a higher entropy means a lower temperature. You may check this in a spin system. Formally, it follows from the positivity of heat capacity C, which is required for the stability of the system. C=T dS/dT.
That is what we expected from the considerations in the first paragraph. The system uses every chance to increase its entropy and reduce its temperature. To keep entropy constant in the Carnot process, we have to invest work.

 

[nonequilibrium]

Hello Dru, thanks for the post.

I'm not sure if I'm getting your point. Is it the following?
"The reason that when you have a reversible cycle using 1 as the hot reservoir (at -200K) and 2 as the cold reservoir (at -273K), the heat going into the cycle is LESS than the heat going out (into the cold reservoir), is because you have to supply work into the cycle to make it reversible"?

 

[Drakkith]

Hello Dru, thanks for the post.

I'm not sure if I'm getting your point. Is it the following?
"The reason that when you have a reversible cycle using 1 as the hot reservoir (at -200K) and 2 as the cold reservoir (at -273K), the heat going into the cycle is LESS than the heat going out (into the cold reservoir), is because you have to supply work into the cycle to make it reversible"?

Mr V, you were you saying originally that the heat coming out of T1, is LESS than the heat going into T2 right? If so, could you elaborate why you think that? I've reread your posts a dozen times now, but i just want to make sure I'm understanding you correctly.

 

[DrDu]

Hello Dru, thanks for the post.

I'm not sure if I'm getting your point. Is it the following?
"The reason that when you have a reversible cycle using 1 as the hot reservoir (at -200K) and 2 as the cold reservoir (at -273K), the heat going into the cycle is LESS than the heat going out (into the cold reservoir), is because you have to supply work into the cycle to make it reversible"?

Exactly!

 

[nonequilibrium]

Exactly!

But if it is reversible, then turn it around and now you get work OUT of it, and you end up transferring heat from a cold to a hot source. This is a clear breach of the 2nd law.

 

[Drakkith]

But if it is reversible, then turn it around and now you get work OUT of it, and you end up transferring heat from a cold to a hot source. This is a clear breach of the 2nd law.

No, you dont. The heat would only flow from the hot to the cold unless you applied work. The problem is in your math. Like I've said. Multiple times.

 

[DrDu]

Well, at least of a formulation of the second law which was stated when people couldn't think of states with negative temperature.
Considering the change of entropy, everything is ok.

Note that also other older statements of the second law that are violated, e.g. the statement of Kelvin and Planck:

"There is no change of state whose only result is a body becoming colder and a weight being risen."
However, for a system with negative temperature this is possible as I already showed.
Again this at once becomes obvious when considering a collections of spins in the field of a magnet. The energetic splitting of the two orientations of the spin depends on the strength of the magnetic field B. In a system with negative temperature, the upper levels are occupied more than the lower levels.
The temperature of a collection of spins is T= \frac{\epsilon}{k} /\ln \frac{N-n}{n} where2\epsilon=\mu B is the energetic splitting of the two levels of the spin with magnetic moment \mu in the field B. and N is the total number of spins, n the number of upper levels occupied and N-n the number of lower levels.
Apparently, if n>N/2, the temperature is negative. If the field strength B is reduced at fixed n, the temperature decreases (given that T is negative). But as the energy of the whole system U=n\epsilon/2 -(N-n)\epsilon/2=(n-N/2)\epsilon decreases, too, with decreasing B, a force is excerted on the magnet or work is done.

There is a third statement of the second law due to Clausius, according to which it is impossible to do nothing else than transfer heat from a colder to a hotter body.
This statement holds true for systems with negative temperature.
From the picture above you can convince yourself that then the other two statements have to be wrong for states with negative energy, simultaneously.
http://de.wikipedia.org/w/index.php...bile_2._Art_.jpg&filetimestamp=20070522114239

 

[DrDu]

Btw, I remember having read a popular article about negative temperatures in Scientific American (or in its German edition, Spektrum der Wissenschaft) some eons ago, probably in the late 80ies. It is well possible that they discussed also the modifications made necessary on the various formulations of the second law.
This forum is powered by Scientific American, so maybe someone can dig it out?

 

[DrDu]

Scientific American
Volume 239, Number 2, August, 1978
W. G. Proctor Negative absolute temperatures . . . . . 78--85