Is heat transfer $Q$ invariant under relativistic motion? This question was oddly enough inspired by a dream while studying for a thermodynamics exam, and I've been wracking my brain ever since. 
Consider a scientist at rest in a lab frame $S$. She observes the interaction between two gases separated by a barrier in an insulated container; that is, no work is done, and no change in chemical potential occurs, only thermal interactions. Suppose from the start of the interaction to the end at equilibrium, the scientist observes a net heat transfer $Q$ occurs between the gases. 
Now suppose the scientist in frame $S'$ moves at a high speed $v$, so that relativistic effects must be considered. My question is, will the scientist still observe the heat transfer being $Q' = Q$ in quantity? Thanks.
 A: Which is the heat for a moving system if $ Q_0 $ is the heat for a system at rest?
If you ask Planck, Einstein, von Laué, Pauli, or Tolman the heat $ Q $ for the moving system is given by
$$ Q = \frac{Q_0}{\gamma} $$
with $ \gamma $ the Lorentz factor, whereas Ott, Arzeliés, and Einstein (again) propose the alternative expression
$$ Q = \gamma \, Q_0$$
It is important to mention how Møller, in the first edition of his celebrated textbook on relativity, used the Planck expression, but replace it by the Ott expression in late editions. More recently Landsberg et al. introduced still another expression
$$ Q = Q_0$$
Thus, heat can decrease, increase, or be a Lorentz invariant depending on whom you ask.
A: Rotational energy is angular momentum times angular velocity. So in case where angular velocity is reduced by 1/gamma, because of time dilation, then rotational energy is reduced by 1/gamma. So the thermal energy in the rotational degree of freedom goes as 1/gamma.
Force on the 'wall' of an Einstein light clock is reduced by 1/gamma. So the amount of work the photon gas inside the clock can do goes as 1/gamma. So thermal energy of photon gas goes as 1/gamma.
A somewhat related thought:
A black body that moves very fast emits photons at rate that goes as 1/gamma, it emits at slow rate. But it still absorbs photons normally, so that the black body tends to absorb thermal energy, which is a definition of low temperature.
