 Welcome back. Let us look at what is the entropy change of a reservoir when there is a heat interaction with the reservoir. This will be needed when we solve some exercises because if reservoir is one of the systems involved we will have to evaluate data s of the reservoir and maybe integrate dq by t of the reservoir. To determine the change in entropy of a reservoir let us go back to our Carnot engine. The Carnot engine is a reversible 2T heat engine. So, going back to our sketch let us say this is our reservoir 1, its temperature is T1. We also have a reservoir 2, temperature of which is T2 and in between we have a reversible engine which absorbs heat q1 from reservoir 1, rejects heat q2 to reservoir 2 and produces work W. Notice that this is a reversible engine, a Carnot engine and hence for every cycle the delta s of this reversible engine per cycle will be 0. But let us evaluate this and we will notice that this is 0 but it should also be equal to, now notice that this q1 is absorbed when the temperature of the system that is the reversible engine is T1. So, this will be the heat absorbed is plus q1, temperature at which that is absorbed is T1 plus there is a change in entropy when heat is rejected at T2. So, here the heat absorbed by our reversible engine is minus q2 divided by the temperature is T2 and we know from our analysis and our definition of the thermodynamic temperature scale that the expression here on the right hand side is indeed 0. Now let us look at it from the point of view of the reservoir. Even this heat transfer, the process of this transfer of heat q1 from the reservoir to our reversible engine is a reversible process. So, it takes place at essentially 0 temperature difference. So, the temperature here is also T1 and hence if we use our definition of change in entropy ds is dq by T, here T is unchanged. So, we can write delta s is q absorbed by T by the reservoir. So, hence for the reservoir 1 we can write delta s reservoir 1 is heat absorbed by reservoir 1 that would be minus q1 divided by T1 and in a similar fashion this heat transfer is also reversible. So, takes place at 0 temperature difference essentially. So, the temperature here is T2 heat absorbed is q2. So, for this reservoir delta s reservoir 2 will be heat absorbed by that reservoir which is plus q2 divided by. Generalizing this we should note this general case that delta s of a TER or a thermal energy reservoir will be useful to define it as the heat absorbed by the thermal energy reservoir divided by temperature of the thermal energy reservoir. This is something to be noted and it will be useful during problem solving sessions. Thank you.