 Welcome back. Now, let us look at exercise D02. We have a cylinder piston arrangement contains 1 kilogram of water, initially in the saturated liquid state at some pressure that absorbs it from a reservoir, the temperature of the reservoir is provided. During the process the piston moves in such a way that the pressure remains constant. So it is an isobaric process. At the end of the process water has completely evaporated into steam that means the end state is dry saturated steam. We have to determine the heat transferred for heat absorbed by our system, change in entropy of the system, change in entropy of the reservoir and change in entropy of the universe. Universe here means thermodynamic universe. So let us sketch the system first. We have our system with cylinder piston arrangement initially at state 1. In this is our system mass is 1 kg and state 1 is given to be saturated liquid in our symbolism dryness fraction 0 at P1 is 0.8 MPa. So the temperature would be the saturation temperature at 0.8 MPa approximately I think 170 degree C. You can look it up in the steam depth and all other properties as needed. Then we have a reservoir I will show it separately just for clarity and the temperature of the reservoir is 250 degree C. Remember that whenever we have the second law of thermodynamics we should always be working with temperatures in Kelvin, thermodynamic temperature. But as a general tradition we have steam tables and all other information usually in degree Celsius. So be careful not to blindly substitute degree Celsius in the second law, convert it into Kelvin and then substitute. So convert this into Kelvin by adding 273.15. What is the process? Heat is absorbed. Let us call it Q12. The piston moves in such a way that we have a state 2 some work is done and the pressure is constant. So state 2 we are told is dry saturated liquid that is x equal to 1 at again P2 is 0.8 MPa and the process 1, 2 is isobar. If we sketch this diagram on the TS plane, it will start becoming comfortable with the TS plane now. Suppose this is the saturated liquid line and this is the dry saturated vapour line then our process suppose this is the isobar 0.8 MPa, the isobar is likely to go something like this but our state 1 is dry saturated liquid, state 2 is dry saturated vapour. Process can be shown like this. So what is the assumption that we make? It is already given that it is a closed system. Let us proceed. We apply first law. First law is Q equals delta E plus W. So Q12 will be delta E12 plus W12. First thing we do is expand. Q12 is an unknown. We have to determine that. So we write Q12 is delta U12 plus delta E other plus W expansion plus W other. Since there is absolutely no hint of a stirrer or an electric connection or anything like that, we can assume this to be 0 or negligible. It is an assumption. Similarly there is no mention of the system being accelerated, decelerated or moving up and down. So let us assume that change in other components of energy other than the thermal energy delta U12 are also 0. So we finally end up with delta U12 plus W expansion 1, 2 on the right hand side. Then we write W expansion 1, 2 as integral E dV from 1 to 2. We say that this integral is possible. For that we will assume that the process is quasi-static and in that we are told that the pressure is constant. So this particular term becomes equal to P delta V12. And because P is constant, I can write this as delta of P V from 1 to 2. So Q12 now becomes delta U12 plus delta of P V12 which is delta of U plus P V which is delta of H. Expanding this, this is mass of the system into change in the specific enthalpy. And since state 1 is saturated liquid, state 2 is dry saturated vapor, this delta H12 will be HFG as tabulated in our steam tables at 0.8 MPa. So this will be M into HFG whereas HFG is to be evaluated at 0.8 MPa. We are given that M is 1 kilogram, HFG is read off and hence we can determine Q12. That is one part of the solution. Now we are also asked to determine the change in entropy of the system and the change in entropy of the reservoir. Again we notice here that the initial state is saturated liquid, the final state is dry saturated vapor. So if we determine the change in entropy delta S12, this will be M into delta specific entropy 12 which will be M into SFG. Why SFG? That is only because 2 is dry saturated vapor and 1 is saturated liquid. And again mass is given to be 1 kg, SFG is to be read off at 0.8 MPa. That gives us delta S12. Now let us look at the reservoir. We have to determine the entropy change of the reservoir, delta S reservoir. The reservoir R is at T R is 250 degree C. Let it absorb heat, Q rest. What is this heat absorb? According to our previous nomenclature, this will be minus Q12. And hence delta S reservoir will be heat absorbed by the reservoir divided by T reservoir which will be minus Q12 divided by T reservoir. Remember that this is already obtained whereas T reservoir substitute in Kelvin. That gives us delta S13. And finally if we assume that the thermodynamic universe is made up of system plus reservoir, assuming that thermally the system does not interact with any other system except the reservoir. And the reservoir also interacts only with our system and with nothing else. Then this assumption is right. And then delta S of the thermodynamic universe will be delta S of the system, delta S12 plus delta S of the reservoir. When you substitute and calculate this check, is this greater than 0? You will find that it is greater than 0 and I will leave it to you as a guess to determine what is the cause of this change in the entropy of the universe. If it is greater than 0 that means it is a possible irreversible process. Thank you.