Are you predicting that both electrodes will heat up equally?
@freemo Ok, I forgot to explicitly state that I assume everything is perfectly conductive~~. But what's wrong with the reasoning that the electrons that escape are the ones that have the highest kinetic energy, and thus their escape will lower the average kinetic energy of electrons in the electrode and thus the temperature of the electrons in the electrode, so also the temperature of the electrode?
@freemo Why should the positive electrode heat up at all?
Let's say that you have two electrodes in vacuum and apply a potential difference to them. This will cause electrons to sometimes be "pulled out" of one of them. How will that impact the temperature of both electrodes?
I would expect the negative electrode to be heated by incoming electrons: after all they've had at least ~0 kinetic energy after leaving the positive electrode, so they bring in extra kinetic energy due to the potential difference.
I'm not sure what should happen to the positive electrode. On one hand, I would intuitively expect ones that have higher kinetic energy of thermal motion to be more likely pulled out, which would decrease the temperature (because we're skimming the top of the kinetic energy distribution, which should decrease the expected value). On the other hand, that would mean that this mechanism is a heat pump that should work regardless of the current temperatures of the electrodes, which is obvious violation of 2nd law of thermodynamics.
Where am I being foolish?
I started thinking about this after reading that one should not apply DC to fluorescent tubes, "because otherwise one filament cools off while the other overheats, evaporates and darkens one end of the tube" (http://ludens.cl/Electron/Fluolamp/fluolamp.html).
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