# Final equilibrium temperature of a negative temperature body and positive temperature body

Suppose system A, with a positive temperature $T_A=|T_0|$, interacts with system B, with negative temperature $T_B=-|T_0|$.

What will be the final equilibrium temperature of the whole system(assuming everything is completely closed, etc...)?

From standard thermodynamics, we have that the heat lost by system B(since it is "hotter" than system A), $Q_B$, is equal to the heat gained by system A, $Q_A$. $$Q_B+Q_A=0 \rightarrow C_B(T_f-T_{B})+ C_A(T_f-T_{A})=0.$$ Assuming for simplicity $C_A=C_B$ we get that $$T_f = (T_A+T_B)/2 = 0.$$ So assuming this is correct, and nothing funny happens due to the negative temperature, I am not sure how to interpret this result.

Here is my attempt of making sense of this:

System B is losing heat, since it is hotter than system A. Thus it should become less hot. Its temperature therefore decreases, i.e becomes more negative (towards $-\infty$) after which it discontinuously jumps to $+\infty$ and continues to decreases until it reaches $T=0^+$. System A, on the other hand, gains heat and becomes hotter. Its temperature is therefore increasing, first passing through $T=\infty$, discontinuously changing to $T=-\infty$ and continuing to increase until it reaches $T=0^-$.

This is of course nonsense because we now have an absolutely hot object ($T=0^-$) with an absolutely cold object ($T=0^+$) which certainly cannot be called equilibrium.

• It seems that final temperature will be greater than 0 K. See this link: quantum-munich.de/research/negative-absolute-temperature – Deep Aug 19 '17 at 4:40
• I'm sorry, but why is system $B$ "hotter" than system $A$. You state "Suppose system A, with a positive temperature ... interacts with system B, with negative temperature." – Semoi Aug 19 '17 at 9:24