When the low $T$ reservoir with negative temperture (Kelvin), e.g. in Ising lattice, will the efficiency of ideal heat engine be larger than 1?

Question

The ideal Carnot engine works between two heat reservoir with two temperatures $t_h>0$(high temperature) and $t_l>0$(low temperature). Its efficiency is then

$$\eta=1-\frac{t_l}{t_h}<1$$

For example, given the low temperature reservoir which is an infinite large Ising lattice with Hamitonian( $B>0$) $$H= -\sum_{i} B s_i$$ with binary variable $s_i\in\{-1,1\}$. There exist configuration such that the absolute temperature of this system is negative in Kelvin.

If somehow you can couple the heat engine to the low temperature reservoir with above Ising lattice with temperature $-|t_l| < 0$), what's the efficiency of ideal Carnot engine with high temperature heat reservior $t_h>0$.

Answer 1

Yes, the efficiency is then higher than 1 but that is fine because, although it can be described by a negative temperature, it does not per se correspond to a heat reservoir but rather to a very peculiar (statistically) excited state of a macroscopic system that is just waiting to release this energy.

Now, this is not conceptually so problematic overall, because you can also worry about how you prepare your spin system in such a high energy state in the first place. And basically, you will find that most of the technologies we can think of will be ultimately bound by Carnot efficiency to get the electricity needed to generate the electromagnetic-wave that will generate the population inversion of your spin system.

That being said, there is an ongoing debate on the subject with two different schools of thought literally clashing against one another.

The first one claims that there is no such thing as negative (absolute) temperature and we need to change our definition of entropy in the microcanonical ensemble while the second one claims that negative (absolute) temperatures are perfectly fine and that we shouldn't try to extend thermodynamics as is for small systems.

I have personally a problem with the first school because it gives too much importance to the ground state of the system for the definition of the entropy (and it doesn't make sense to me...there are other technical reasons but that's not the topic at hand).

Answer 2

Yes, the efficiency can be greater than one. (At least, it can in principle - I'm not aware of any experimental verification of this at the present time.)

One way to define temperature is as follows: $$ \frac{1}{T} = \frac{\partial S}{\partial U}. $$ This means that if $T$ is negative then the entropy increases when you take energy out of the system. Therefore, in principle, you can take heat out of a negative-temperature system and convert it directly into work without violating the second law. But this is an irreversible process (it increases the entropy), so in theory you can even do even better - you can take energy out of some other reservoir (at a positive temperature) in order to offset the entropy increase and end up with a reversible process.

If you could solve the (probably formidable) technical issues involved in doing this, you'd end up with a heat engine that can take $Q$ units of heat out of the negative-temperature reservoir at $T_-$, and take $-Q\frac{T+}{T_-}>0$ units of heat out of the positive-temperature reservoir, and turn it all into $W\left(1-\frac{T_+}{T_-}\right)Q$ units of work. Since $W>Q$ in this case we can say that this engine has an efficiency greater than 1, without breaking the second law.

By the way, note that $T_-$ plays the role of the hot reservoir in this engine, not the cold reservoir. Negative temperatures are hotter than positive temperatures, which is really just a consequence of the definition above.

In his answer, gatsu quite rightly mentions the debate about whether negative temperatures exist at all. (I'm very much in the camp that says they do.) If there were an experimental demonstration of this greater-than-unity efficiency effect then that would just about wrap this debate up - but as I said, I think the technical challenges involved in doing so are probably quite extreme.

Answer 3

The solution to the novice's version of this problem is to switch to using SI units of temperature $K$ (kelvin). Temperatures can't go below absolute temperature, i.e. 0$K$.

In your case, when we take $T$ as negative it's not really negative but its much hotter than the hottest thing you can get. The relation $$T=\frac{dQ}{dS}$$ gives the temperature negative after the case when entropy is saturated and start to decrease when energy is added to the system.

So,the energy will start flowing in other direction and the source you are considering will become sink, and the -ve temperature sink would actually be the source.

Now if you'll consider the derivation of Carnot's Efficiency in terms of net entropy change in cycle is Zero, considering negative entropy of the source(-ve temprature), you'll get $$\epsilon= 1-\frac{T_+}{T_-}$$