 Welcome back. Now it will be nice if we do some exercises. Exercise sheet D is being provided separately, you will see it here. It is on the second law of thermodynamics and all the systems which are included here are closed systems or you can model the systems involved as closed systems. Let us look at a few of these and I will provide some approach, some hints, some procedure so that then you can take it upon your own and complete the details. First let us look at the very first exercise here, D01. What we are provided here is some sort of a conductor, a system called a conductor which connects two reservoirs at two different temperatures. Let us sketch it like this. We have one reservoir, we have another reservoir. This reservoir is at T1 which is at 1200 k and this reservoir is at T2 500 and there is some system, it could be a solid metallic block or could be something else which connects these two. Actually in principle no connection is really needed because you know you could consider this reservoir to be our sun, something like 5000 odd Kelvin. This reservoir could be us earth at something like not 500, something like 300 Kelvin and the energy is transferred as heat by the process of thermal radiation. So the intervening space happens to be our system. So this is the so called conductor and let us say that conductor itself or that intervening space is our system and the heat transfer from this reservoir to this reservoir q dot is given to be 115 watts. Let us simplify this clutter and let us look at it only from the system point of view. I will now say that our system is this out here the temperature is at this point we have temperature T1 is 1200 k and the heat which is transferred is let me say q dot 1 the rate of transfer is 150 watt. So we are working here on per unit time basis so that instead of q we have q dot and the units instead of June are watt. Similarly on this surface there is a q dot 2 which is also 150 watt and the temperature here is T2 is 500 Kelvin. We are given that the flow of heat is steady that means 150 watt actually this should be minus 150 watt going out of the system minus 150 watt. So it is given that the flow of heat is steady and because of that the conductor is in steady state that means over time its state does not change and we are asked to determine what is the entropy produced by the conductor or in the conductor. This is entropy production rate. We have seen that entropy produced has the units of Joule per Kelvin or kilo Joule per Kelvin. Here it will be watt per Kelvin or kilo watt per Kelvin. Now since we notice that there is no work interaction involved there is a steady state so our change in energy is 0 there is no work transfer and the amount of heat transfer net amount of heat transfer is 0 because 150 watt is coming in from the top 150 watt is going out from the other end. So hence a steady state is maintained. Now let us apply the second law. What does the second law say? A second law in our normal form would be delta S is integral dQ by T plus SP with the condition that SP should be greater than or equal to 0. Using the time variant mode the change in entropy would become dS by dt for our system. The first term on the right hand side would become integral dQ dot by T plus S dot P with the condition that S dot P must be greater than or equal to 0. Now expanding this term we will get this to be dS by dt equal to this Q dot is at two places one at the top which is Q dot 1 by T 1 the second part would be plus Q dot 2 by T 2 plus S dot P with the second law being S dot P greater than or equal to 0. Now notice that in this relation steady state so dS by dt is 0 the left hand side is 0 and on the right hand side Q dot 1 is 150 watt, Q dot 2 is minus 150 watt, T 1 is 1200 k, T 2 is 500 k. So all that you have to do is substitute and obtain the value of S dot P and after obtaining S dot P we should appreciate that it will be greater than 0 and because it is greater than 0 the process we are studying is a real life possible irreversible process. And what is the cause of irreversibility? The cause of irreversibility is we have a temperature difference a big finite temperature difference between T 1 and T 2 and heat is being transferred across that finite temperature difference. Another way of looking at this problem is not to look at just the system. Another way of looking at this is to consider that our reservoir 1 at T 1 reservoir 2 at T 2 and the intervening conductor has a system all together and if we consider these together then we can model this neglecting or asserting that there is no interaction between any of the reservoirs and any extraneous system that this is our thermodynamic unit and because it is our thermodynamic universe we can say that it being an adiabatic system S dot P for this will be our entropy produced or entropy being produced in the universe and since the universe consists of 3 subsystems this will be the rate of change of entropy of reservoir 1 plus rate of change of entropy of reservoir 2 plus the rate of change of entropy of our system itself. This is what I had written as dS by dt earlier and this because we have a steady state with 0 and here our heat flow rates had shown from the system point of view where q dot 1 equals plus 150 watt and here we have q dot 2 is minus 150 watt. Now these two terms turn out to be equal to S dot reservoir 1 would be q dot absorbed by reservoir 1 by T 1 plus q dot absorbed by reservoir 2 by T 2 and q dot absorbed by reservoir 1 is minus q dot 1 because q dot 1 is absorbed by the system so what is absorbed by the reservoir is minus q dot 1 so this would be minus q dot 1 divided by T 1 and again here you will have minus q dot 2 by T 2. Calculate this and you will find that you get the same answer as earlier. Thank you.