Does Peltier element become a weaker thermo-insulator when taking off power?

Question

Imagine we have two bodies (environments) with different temperatures connected to each other through Peltier element. Say, one body has temprature $T_l$, the other $T_h$. For the sake of experiment, heat exchanges only through Peltier element, which acts as some sort of heat-insulator. Initially, we neither apply power to the element, nor take it away. At some moment, the system will come to equillibrium with some temperature $T_a$, $T_l<T_a<T_h$.

The questions are:

1. If if take off power from Peltier element, will it become a weaker insulator? Or in other words, will the hot body cool faster, as well as cold one warm faster, compared to initial passive scenario? Or vice-versa? Or the answer is not so straightforward?
2. If we take power from from Peltier element, will the new equillibrium temperature of the system $T'_a$ be less then $T_a$?

Unfortunately, can't yet do my own experiments due to lack of equipment, and the only I can do is just do this experiment mentally, but in the end I just come to contradictory conclusions for both questions.

UPD: as @Chemomechanics noted in the comments, I would like to clarify my reasoning which gives contradictory conclusions.

Firstly, the intuitive resasoning described in the comment seems relevant, that the more power we pull out, the faster gradient goes down. And this reasoning answers "yes" to both questions.

But if we look at this experiment from another angle, in analogies, the answers seem to be the opposite to both questions. And I can see at least 2 analogies here:

1. We have 2 communicating vessels with different levels of water. And we have some generator inbetween of them, which converts kinetic energy of waterflow into something. While we don't pull out the power from the generator, the water passivle flows from one vessel to another. But once we start to pull out the power, we add resistance to the flow. But the overall amout of water will stay the samy in the end.
2. Another analogy is electric charges, and acts the same way as in pt. 1. By pulling out the power, we only slow down the process of equalizing potentials by adding resistance, essentially. But the overall charge stays the same in the end.

So, if these two analogies are relevant, then the answers to my questions would be that by pulling out power from Peltier element we actually add "resistance" to normal heat-flow. And final temperature of the system will be the same as in the passive scenario.

Answer 1

I'm glad you updated the question with the analogies, as these can provide much insight into how to precisely express and analyze the original problem.

For example: What is the hydroelectric analogy to the temperatures associated with a thermoelectric engine? Is it water? Temperature again? Energy? Or something else?

What is the thermoelectric analogy to a turbine spinning freely in a hydroelectric plant? Is it an isolated thermoelectric device? Or is it a shorted-out thermoelectric device? Or something else?

To start, note that all three engines in the original example and analogies (thermoelectric, hydroelectric, and electric) operate based on an energy difference: A gradient exists in some intensive property (here, temperature, pressure head, and voltage), and we're going to allow the conjugate extensive property (entropy, matter, and charge, respectively) to shift, using the driving force to extract energy as work by some means. OK, great.

You say, by analogy, that matter is conserved in the hydroelectric engine, and that charge is conserved in the electric engine, and that this suggests that the temperature and temperature difference should not fall more in the thermoelectric system if we extract more work. But the correct analogous term to matter and charge in the thermoelectric engine is not temperature (the intensive variable) but entropy (the extensive variable). And indeed, no entropy is removed from the thermoelectric system when we extract work. Furthermore, the intensive variable gradients (temperature difference, pressure head difference, and voltage difference) all decrease as we run the corresponding engines, with an increasing depletion rate as we extract more power.

So the apparent contradictions may be resolved, at least in part, if the analogies are revised to use the appropriate corresponding thermodynamic variables.

Is the final temperature lower in the thermoelectric system if we collect power from it? It has to be; when we remove energy from closed, single-phase systems in internal equilibrium, they cool down. The temperature drop can be quantified by performing an energy balance in conjunction with the work output and the heat capacities of the thermal reservoirs and device. As discussed above, the existence of charge and matter conservation laws is irrelevant here because no such law exists for temperature—it's energy that's conserved, and we're removing some of it via work.

Does the temperature difference between the reservoirs drop faster if we extract more power? Here, we have an ambiguity depending on whether "Initially, we neither apply power to the element, nor take it away." means that the device is left alone or shorted out to convert any electrical energy immediately to thermal energy. If it's the former, the appropriate analogy appears to be a blocked hydroelectric engine (with slight leakage only); in this case, drawing off power by hooking up an external load decreases the temperature gradient faster. If it's the latter, the appropriate analogy appears to be an open (freely spinning) hydroelectric engine (with the available energy gradient being dissipatively wasted through frictional heating); in this case, drawing off power by replacing the shorted connection with a load can be expected to reduce the rate of temperature change.

Put another way, "no load" is ambiguous when discussing both electrical and mechanical systems and could mean either of the opposite extremes of an absence of a load (i.e., infinite resistance to flow, perhaps more common in electrical contexts) or a load of zero (i.e., zero resistance to flow, perhaps more common in mechanical contexts). Once the ambiguity is removed and the appropriate analogy formulated, the remaining contradictions should be resolved.