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Say we have a standard thermal machine which operates in cycles. It absorbs energy in form of heat from a hot source, does some work and then transmits the remaining energy to the cold sink also in form of heat. The efficiency of this machine is given by:

$$\eta = 1-\frac{|Q_2|}{|Q_1|}$$

Therefore, an ideal efficiency of $\eta \longrightarrow 1$ could in theory be achieved if $|Q_2|\longrightarrow 0$ (I know this is deemed impossible according to the Kelvin-Planck statement, but this impossibility is precisely what I don't understand).

Since the machine only interacts with the source and the sink through the exchange of heat, and there can only be a flow of heat ocurring naturally between two bodies if they have different temperatures (as far as I'm concerend, at least), then in order to achieve $Q_2=0$ we would simply need a sink whose temperature matched that of the thermal machine. This is especially simple if the cycle the machine is operating on uses isothermal processes, since the temperature of the system would be constant (although different for each process).

I guess where I'm wrong is on the fact that some extra energy would be required for the cold sink to change its temperature over time?

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  • $\begingroup$ Try drawing a PV diagram for a complete ideal gas cycle where there is net work done but no heat rejected in the cycle $\endgroup$
    – Bob D
    Commented Oct 10, 2023 at 18:17

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Always think of thermal phenomenon as either entropy flowing from one place to another or entropy generated locally at some place. Entropy at some temperature has energy the same way as a piece of gravitational mass has gravitational energy in a gravitational potential at a certain height or electric charge has electrostatic energy at a certain electric potential. In this context the thermal potential is temperature, and when entropy is transported from one temperature to another it can do work the same way as when a certain mass falls from one height to a lower one can do work or electric charge is transported from one potential to a lower potential can do work. It also goes in reverse when moving entropy from a lower temperature to a higher one takes work, this is the job of a heat pump.

If you drop entropy from a higher temperature $T_1$ to a lower temperature $T_2$ then the difference between the two energies can be captured by an ideal Carnot engine as work $\Delta W=T_1 \Delta S -T_2 \Delta S$, where $\Delta S$ is the transported entropy that is absorbed at $T_1$ and rejected at $T_2$, while $\Delta W$ is the "mechanical" work done by the engine. Since the input thermal energy is $T_1 \Delta S$ the efficiency of the engine is $$\eta = \frac{\Delta W}{T_1 \Delta S}=\frac{T_1 \Delta S-T_2 \Delta S}{T_1 \Delta S}=1-\frac{T_2}{T_1}.$$ Since $T_2 < T_1$ we get $\eta <1$, and to achieve $\eta = 1$ you need to dump the entropy $\Delta S$ in a sink whose temperature is $T_2=0$. Unfortunately, there is also a 3rd Law that essentially states that teh temperature absolute zero, $T_2=0$, cannot be reached in a finite number of steps...

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  • $\begingroup$ Beautifully explained, thanks a lot! $\endgroup$ Commented Oct 10, 2023 at 19:06

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