Carnot efficiency on the thermodynamic scale: $\eta = 1 - T_\mathrm{C}/T_\mathrm{H}$

Throughout, $\Theta$ is a type of empirical temperatures carrying a linear order ($s<t$ means the reservoir at $s$ is colder than the one at $t$), and a cycle is a finitely supported function $c:\Theta\to\mathbb R$: for each reservoir temperature $t$, the number $c(t)$ is the net heat absorbed by the device from the reservoir at $t$ in one complete cycle (negative when heat is rejected to it), and only finitely many reservoirs are touched. The work delivered in the cycle is

which is the first law for a device returning to its initial state. A set $\mathcal C$ of cycles stands for the physically realizable ones; it is assumed to be a convex cone: $0\in\mathcal C$, $c+d\in\mathcal C$ (run two devices side by side), and $r\cdot c\in\mathcal C$ for $r\ge 0$ (scale a device). A cycle is reversible when $c\in\mathcal C$ and $-c\in\mathcal C$. A cycle operates between $h$ and $k$ when $c(t)=0$ for every $t\notin\{h,k\}$. A Carnot engine between a hot reservoir $h$ and a cold reservoir $k$ is a reversible cycle operating between $h$ and $k$ with $c(h)>0$, $c(k)<0$ and $W(c)>0$. The efficiency relative to the hot reservoir is $\eta=W(c)/c(h)$.

Statement. Let $T:\Theta\to\mathbb R$ be a scale for which every Carnot engine $e$ between reservoirs $k<h$ satisfies $|e(k)|/|e(h)|=T(k)/T(h)$. Then every such Carnot engine has efficiency

This combines equation (1) with the definition of the thermodynamic temperature scale: on that scale the efficiency of a reversible engine depends on the two reservoirs only through the ratio of their thermodynamic temperatures.