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What is the minimum energy required to move a unit of heat from the cold side of a system to the hot side?

One obvious lower bound is that it must be at least as much as the amount of work that can be done by a maximum efficiency heat engine when that heat moves back. So for example, for a system with the cold end at 0°C and the hot end at 50°C (323K), maximum heat engine efficiency is 50/323 = ~15%. By this logic, since 1J of heat moving from hot to cold side can perform at most 0.15J of work, therefore it must take at least 0.15J of energy to move 1J of heat back from the cold end to the hot end at these given temperatures. Assuming I made no mistakes here.

Is this the actual theoretical lower bound on the energy required, or is there a higher theoretical lower bound?

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By this logic, since 1J of heat moving from hot to cold side can perform at most 0.15J of work, therefore it must take at least 0.15J of energy to move 1J of heat back from the cold end to the hot end at these given temperatures. Assuming I made no mistakes here.

That is correct, because then you are talking about a heat pump, i.e., a heat engine operated in reverse.

The minimum amount of energy (work) needed to move a unit of heat from a cold environment to a warm environment is determined by the maximum possible coefficient of performance (COP) of the heat pump, which is the Carnot COP.

For any heat pump,

$$COP=\frac{Q_H}{W}$$

The maximum possible COP for a heat pump is the Carnot COP

$$COP=\frac{T_H}{(T_{H}-T_L)}$$

So if $T_{L}$=273 K (0 C), and $T_{H}$ = 323 K (50 C), $COP_{max}$ = 6.46

Substituting into the first equation,

$$W=\frac{Q_H}{6.46}$$

So if $Q_H$ = 1J, the minimum work required is $W=0.155 J$, which is what you calculated.

Hope this helps.

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    $\begingroup$ To expand on this answer, the lower bound arises from the impossibility of destroying entropy, which is forbidden by the Second Law. If you wish to move $Q/T=1\,\mathrm{J}/273\,\mathrm{K}$ of entropy into the hot end, you can remove no more than $0.845(=273/323)\,\mathrm{J}$ from the cold end, or you'll end up with excess entropy. The energy remainder of $1-0.845=0.155\,\mathrm{J}$ must then come from reversible work you do on the system. $\endgroup$
    – Chemomechanics
    Commented Aug 11, 2020 at 21:34
  • $\begingroup$ @Chemomechanics Thanks for providing your additional insight. OP take note. $\endgroup$
    – Bob D
    Commented Aug 11, 2020 at 21:40
  • $\begingroup$ That is indeed a very interesting additional insight. Thanks both! $\endgroup$
    – Roman Starkov
    Commented Aug 11, 2020 at 22:26

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