# Carnot cycle with efficiency > 1

1. Jul 14, 2010

### paweld

Is it possible (at least theoretically) to construct engine which uses Carnot cycle
and two reservoirs with temperatures $$T_1<0<T_2$$ (one of them has
negative temperature which is possible e.g. in case of two-state paramagnet).
Such cycle would have efficiency greater then 1.

(of course the reservoir with negative temperature is "hotter")

2. Jul 14, 2010

### Andy Resnick

3. Jul 14, 2010

### Staff: Mentor

I'm not following - you want the cold reservoir to be hot and the hot reservoir to be cold? If you plug that into the efficiency equation, you get an efficiency below zero! In other words, nothing.

And a temperature below zero....you mean below absolute zero?

http://en.wikipedia.org/wiki/Carnot_cycle

4. Jul 14, 2010

### Pythagorean

Temperature conventions that use negative values do not do so for any absolute physical reasons. They are more designed around human convenience (water's state changes for Celsius and I think human comfort for Farenheit).

That is why we have the Kelvin scale. You can't go below 0 Kelvin because the Kelvin temperature relates directly to energy, which can't be negative. Kelvin best represents the actual thermodynamics, which doesn't allow for a negative number.

5. Jul 15, 2010

### xxChrisxx

Sounds like meaningless twaddle to me.

6. Jul 18, 2010

### jack action

No. The reason is that negative temperatures, although numerically lower than zero, are higher than infinite temperature. From http://en.wikipedia.org/wiki/Negative_temperature#Heat_and_molecular_energy_distribution":

On the "other side", everything works in reverse: when you add energy, the temperature decreases; when you remove energy, the temperature increases. So let's assume you have a Thot that is negative and a Tcold that is positive. Going from Thot to −∞ K, you will need to add energy. Then going from +∞ K to some temperature Tx you will recover the amount of energy you spent "cooling" your negative temperature, making your net energy balance = 0. Finally going from Tx to Tcold will represent the actual energy you will recover. So the actual Carnot cycle have to be calculated with Tx as the hotter temperature.

Not a pro on the subject, that is just my understanding.

Last edited by a moderator: Apr 25, 2017