I was considering some theoretical limits on transferring heat from one system at to another to another. It is well known that if you have two systems at temperatures $T_1$ and $T_2$, respectively, with $T_2> T_1$ then you need to provide work to transfer heat to the system at temperature $T_2$. Now, that work is then coming from some different thermodynamic process that we choose not to describe. But leaving that out is actually compromising the ideal solution as extracting work first and then having to dump that at the system at $T_2$ is not necessarily optimal.
This led me to consider three systems at temperatures $T_1$, $T_2$, and $T_3$, respectively, with $T_1<T_2<T_3$ where we want to pump heat to the system at temperature $T_3$ but now there is no external source of work. In the limit of infinite heat capacities, we then find that ratio between the heat supplied to the system at $T_3$ and the heat extracted from the system at $T_2$ in the optimal reversible case will be:
$$\eta = \frac{T_3}{T_2}\frac{T_2 - T_1}{T_3 - T_1}$$
See below for the derivation. So, in principle at least, one can construct a heat pump that heats a house using outside environments at two different temperatures without the need for work to be provided. E.g. in winter the outside temperature near the ground could be -10 °C while at some depth below the ground it could be 10 °C.
The question is then how to realize this in practice. Presumably one would need some sort of a Stirling machine. What can one expect about the amount of heat that can be extracted from the environment per unit time in some realistic setting?
To derive the efficiency formula, one can consider a Carnot process that extracts work from the systems at temperature $T_1$ and $T_2$ and another Carnot process that acts as a heat pump between the systems at temperatures $T_2$ and $T_3$ using the work provided for by the first Carnot process. However, a simpler derivation is to consider any arbitrary process that adds an amount of heat of $q_i$ to system $i$, the First law then implies that:
$$\sum_{i=1}^3 q_i = 0$$
Reversibility implies:
$$\sum_{i=1}^3 \frac{q_i}{T_i} = 0$$
Solving for the $\eta \equiv -\frac{q_3}{q_2}$ then leads to the formula given above.
