power output divided by the power input. For the heat-engine, the power input is the heat current out of the hotter reservoir (reservoir $L$), $J_L$, and the power output is the electrical power generated $P_{\rm gen}$. Thus, the heat-engine (eng) efficiency is $$\begin{aligned}
\eta_{\rm eng} = P_{\rm gen}\big/ J_L.
\label{Eq:eff-eng}\end{aligned}$$ This efficiency can never exceed Carnot’s limit, $$\begin{aligned}
\eta_{\rm eng}^{\rm Carnot} &=& 1-T_R/ T_L,
\label{Eq:Carnot-eng}\end{aligned}$$ where we recall that we have $T_L > T_R$.

For the refrigerator the situation is reversed, the load is replaced by a power supply, and the power input is the electrical power that the circuit absorbs from the power supply, $P_{\rm abs}$. The power output is the heat current out of the colder reservoir (reservoir $L$), $J_L$. This is called the cooling power, because it is the rate at which the circuit removes heat energy from reservoir $L$. Thus, the refrigerator (fri) efficiency is $$\begin{aligned}
\eta_{\rm fri} = J_L \big/ P_{\rm abs}.
\label{Eq:eff-fri}\end{aligned}$$ This efficiency is often called the coefficient of performance or COP. This efficiency can never exceed Carnot’s limit, $$\begin{aligned}
\eta_{\rm fri}^{\rm Carnot} &=& (T_R/T_L -1)^{-1}, \ \  
\label{Eq:Carnot-fri}\end{aligned}$$ where we recall that $T_L < T_R$ (opposite of heat-engine).

Strangely, the laws of 