striphe said:
The sink has a temp of 10C and the source has a temp of 20C. They both have a heat capacity of one kilojoule per calvin per litre. The source and sink are both one litre in size.
If these were connected together so that heat could be transferred between the two (no heat engine) I would expect that they would reach an equilibrium of 15C with 5 kilojoules moving from the source to the sink.
If i place a heat engine in the place of the medium that allows heat transfer, that doesn't involve any gas or liquid exchanges between the sink and source (e.g. a sterling engine or thermoelectric generator) What equilibrium would result and how much energy would be extracted from the system (using a realistic parameter to refer to the efficiency of that heat engine) ?
As Russ indicated, this is not a trivial problem to solve, and the origin of the problem is the fact that you made your reserviours finite.
If you had two thermal reserviours, one at 10C and the other at 20C, the maximum efficiency that can be obtained to convert heat into work is about 3.5%. If I am reading about Stirling engines correctly, the maximal efficiency is around 40%, which means your Stirling engine will convert, at best, about 1.7% of the thermal energy into work, when operating between the two (infinite) reserviours.
By making the source and sink finite, you have greatly complicated the problem, because as you draw heat from the source, its temperature changes and as you dump heat to the sink, its temperature changes also.
One way around this is to let the heat engine remove very small amounts of heat per cycle, and letting the source temperature T = T(t) and the sink temeprature T' = T'(t) as well. Then, as the engine converts heat into work, the source and sink will both approach the same temperature T_f and the efficiency of the engine approaches zero. This approach preserves the notion of equilibrium (or steady state, at least), which makes the problem easier to solve.
Going to a full non-equilibrium situation, where you attempt to withdraw the entire 10 kJ at once leads to a whole host of difficult problems: for example, how do you assign a temperature during the process?