What is this entropy-related thermodynamic concept called? I'm trying to sharpen my understanding of a thermodynamic concept.
Suppose I've got an idealized building whose interior is held at 10 degrees cooler than the outside air by a heat pump that consumes X watts continuously. Suppose next that the situation is reversed including the operation of the heat pump such that the interior is maintained 10 degrees warmer than the outside air by the heat pump consuming Y watts. Isn't there a 2nd-LoT-related reason why X can never be the same as Y, ultimately because heat escaping outward from a thing is not reversible into heat flowing into that same thing? Isn't the outward flow of heat into the cooler environment "easier" in some sense than the inward flow of heat from the warmer environment?
 A: The two scenarios are almost symmetric, i.e. the $X$ is almost the same as $Y$, but there is a very small direction dependence of the rate of heat pumping that one can achieve with a give heat pump power, and the achievable rates are set by the second law. 
We are talking about a room at temperature $T$ and the energy cost of pumping heat:


*

*From the room at temperature $T$ to the outside World at a temperature $\Delta T$ higher, i.e. from temperature $T$ to $T+\Delta T$; and

*From the outside World at a temperature $\Delta T$ lower into the room, i.e. from $T-\Delta T$ to $T$.


Scenario 1: Let's work out what we need to expel heat $Q$ from the room. We take a quantity of heat $T$ from the room, the room's entropy decrease is $\frac{Q}{T}$. We then add that heat to the outside World, with corresponding entropy increase $\frac{Q}{T+\Delta T}$, i.e. $\frac{Q}{T+\Delta T}-\frac{Q}{T}<0$ if $\Delta T > 0$ and so the transfer cannot happen spontaneously and we need to add further entropy to one of the reservoirs for a total system entropy balance. To do this, our system must convert work $W$ to heat and add it to the outside World to increase the latter's entropy, thus:
$$\begin{array}{cl}&\frac{Q}{T+\Delta T}-\frac{Q}{T} + \frac{W}{T+\Delta T} = 0\\\\
\Rightarrow & W = Q\,\frac{\Delta T}{T}\end{array}$$
A "co-efficient of performance" is the ratio of the heat pumped to the work input:
$$\eta = \frac{T}{\Delta T}$$
and it approaches infinity as the temperature difference approaches nought.
Scenario 2: If we make the same reasoning for this scenario, we find:
$$\frac{Q}{T}-\frac{Q}{T-\Delta T} + \frac{W}{T} = 0$$
and our co-efficient of performance would then be:
$$\eta = \frac{T-\Delta T}{\Delta T} = \frac{T}{\Delta T} - 1$$
and is slightly smaller than for the other scenario, and you can see that there is an asymmetry that gets small as the temperature becomes large in comparison with the temperature difference. 
The above analysis is the comparison with the same co-efficient of performance definition in both cases (ratio of heat removed from colder reservoir to the work done). However, from a practical standpoint one could argue that a more meaningful co-efficient in Scenario 2 is the ratio of heat added to the hotter reservoir to the work done: the purpose of the heat pump in this case is to be a heater, after all. I believe this is almost certainly the co-efficient of performance manufacturers would use (since it is higher and therefore more markettable). If we use the revised definition (ratio of heat added to the hotter reservoir to the work done) we get:
$$\eta^\prime=\frac{T}{\Delta T}$$
and this is the same co-efficient as for Scenario 1. But note that we are using a subtly different definition here.
So to answer your title's question, the concept you are groping for is likely the co-efficient of performance for a heat pump, but its definition is subtly different depending on whether your talking about a heat pump as a cooler or as a heater.
