# Entropy Generation during Heat Transfer Processes

I'm having a hard time trying to understand a few concepts of the second law of thermodynamics.

Let's take the example bellow:

A heat source at 800K loses 2000kJ of heat to a sink at 500K. Determine the total entropy generation.

SOLUTION

The case involve heat transfer through a finite temperature difference, and therefore it is irreversible. The magnitude of the entropy generation (irreversibility) associated with each process can be determined by calculating the total entropy change for each case. The total entropy change for a heat transfer process involving two reservoirs (a source and a sink) is the sum of the entropy changes of each reservoir since the two reservoirs form an adiabatic system.

Or do they?

The problem statement gives the impression that the two reservoirs are in direct contact during the heat transfer process. But this cannot be the case since the temperature at a point can have only one value, and thus it cannot be 800 K on one side of the point of contact and 500 K on the other side. In other words, the temperature function cannot have a jump discontinuity.

Therefore, it is reasonable to assume that the two reservoirs are separated by a partition through which the temperature drops from 800 K on one side to 500 K on the other. In that case, the entropy change of the partition should also be considered when evaluating the total entropy change for this process. However, considering that entropy is a property and the values of properties depend on the state of a system, we can argue that the entropy change of the partition is zero since the partition appears to have undergone a steady process and thus experienced no change in its properties at any point. We base this argument on the fact that the temperature on both sides of the partition and thus throughout remains constant during this process. Therefore, we are justified to assume that the $\Delta S_{partition} = 0$ since the entropy (as well as the energy) content of the partition remains constant during this process. The entropy change for each reservoir can be determined since each reservoir undergoes an internally reversible, isothermal process.

$$\Delta S_{source} = \frac{Q_{source}}{T_{source}}=\frac{-2000kJ}{800K}=-2.5kJ/K$$

$$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}=\frac{2000kJ}{500K}=4.0kJ/K$$

$$S_{gen} = \Delta S_{total} = \Delta S_{source} + \Delta S_{sink} + {\Delta S_{partition}} = 1.5$$

My question is: If both reservoirs are internally reversible, since their state don't change and the "partition" is also internally reversible for the same reason. What is causing the entropy generation (irreversibilities)?

• @BobD It seems to me that, in the case of the two reservoirs, all the generated entropy within the partition is flowing into the cold reservoir along with the entropy flowing from the hot reservoir. In the case of the irreversible isothermal expansion, the reservoir is assumed to be ideal, so all the generated entropy remains in the gas. If you can determine the final state of the system from knowing the external work, then the generated entropy within the gas during the process is equal to the total entropy change minus the entropy flow from the reservoir $Q_{irrev}/T_{res}$ Commented Feb 13, 2020 at 15:29
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