 Let us come to the next problem now, question C, water at P1 equal to 100 degree centigrade and P is equal to 0.4 MPa is supplied Q heat energy is equal to 1500 kilojoule per kg of heat at isobaric conditions in a closed system. Determine the final state of the system depending on the final state find out the change in enthalpy drainage fraction if required temperatures etc at point number 2. So, the state 1 is given at 0.4 MPa pressure and 100 degree centigrade as T1. The pressure is not atmospheric pressure here therefore, this is not the saturation temperature in this case point to be noted. So, this is a basically again a constant pressure forces and Q has been supplied some heat energy supplied to this and therefore, there will be some kind of expansion alright and because of which the state 2 will be different from state 1. Let us now again we can understand we can possibly have a cylinder piston arrangement have state 1 in that and state 2 can be found out. So, always try to draw some kind of a cylinder piston arrangement or a schematic to show the system first. Try to show your system also on a PT diagram if you are sure about the state and that will actually clarify lot of doubts for understanding purposes those things are always essential ok. So, let us come to question C solution now. So, question C I can again show a cylinder piston kind of a arrangement here and again my system could be as shown over here alright and this could be my state 1 initially and we can find out the state 2 later ok. The first law of thermodynamics tells me that Q is equal to delta E plus W change in energy and work done from the system. Assuming that it is a standard assumption assuming that change in kinetic energy and potential energy and any other energy are 0 and also W others other than the expansion space other than the expansion over is 0. We can say that Q is equal to delta U plus P delta V simple assumption and therefore, we can say that state 1 and state 2 we can say U 2 minus U 1 plus P times V 2 minus V 1 because the process is at constant pressure and this is nothing but U 2 plus P V 2 minus U 1 plus P V 1 V 1 P V 1 and therefore, this will come equal to H 2 minus H 1 change in enthalpy which is equal to delta H. So, we we say that whatever Q goes during this constant pressure process if all these assumptions are ok then we say that Q is equal to delta H. Now, we see that the state 1 is at P is equal to 4 bar pressure is 4 bar and T is equal to 100 degree centigrade and we also see that Q upon M is equal to delta H. If I just write capital H is equal to small H specific enthalpy change that we are writing as and Q is equal to 1500 kilo joule per kg. So, we can say that 1500 kilo joule per kg is equal to H 2 minus H 1. Now, if I want to locate the state 1 the state 1 is very important that I should know what is the P sat at 100 degree centigrade. We know P sat at 100 degree centigrade is atmospheric pressure. So, I say that 0.10142 MPa or 1.014 bar is the P sat value which is standard value and my pressure is 4 bar here that means P sat T 1 is less than P 1 what does it mean? If P sat T 1 is less than P 1 we are in sub cooled liquid region we are in sub cooled region or compressed liquid region alright. Now, if I go to table 3 I can get these values the enthalpy values at 4 bar and 100 degree centigrade because I mean sub cooled region I will go to table 3, is not it? And from here I will get now state 1 the enthalpy related or whatever values I want I can get it from now table 3. So, can I go to table 3 now alright. So, this is table 3 let us say 0.4 MPa that we are talking about approximately 4 bar and here I can locate the point of 100 degree centigrade at this point alright. So, I can get the values related to enthalpy here which is my state 1 in this case alright. So, my H 1 here can be read from here as 419.39 kilojoule per kg alright. So, my H 1 which is what I want because I am interested in delta H. So, H 1 is equal to 419.39. So, my H 1 value is equal to 419.39 alright. And as we understand from here Q is equal to delta H. So, my H 2 is equal to H 1 plus Q by m is not it? From here I have got Q is equal to delta H which is H 2 minus H 1. I can get the value of H 2 from here and therefore, my H 2 will be 419.39 plus 1500 kilojoule per kg which is equal to 199.39 kilojoule per kg. Now this is my state 2 and now I have to understand when I know state 2 what I know is enthalpy of state 2. I will have to again locate now whether the state 2 lie is it also in the sub-cooled region? Is it inside the dome or is it in a superheated region? That can be understood now when I again go back to the steam table and locate the value of H f and H g at 4 bar. Because the process happens at constant pressure I will go to 4 bar now which is where I will refer to now other table and get the values of H f and H g compare those values with this H 2 and locate the state ok. So, let us go to table 2 now and find out those values. So, here we can see that for 0.4 MPa we can locate the values of H f and H g. So, H f is 604 you can read from here kilojoule per kg and H g is 2738.1 kilojoule per kg while our value of H 2 is around 199.39. It means that it lies between H f and H g it also means that therefore that this steam is wet steam or the 0.2 lies inside the dome all right. Had it been more than H g it would have been superheated region, had it been less than H f it would have been in a sub-cooled region. But because this is between H f and H g the state 2 lies for sure inside the dome and therefore the state 2 is wet steam. The state 1 was sub-cooled liquid and because it has been given some Q value it has been given some heat energy it has become now wet steam it is some vapor plus some liquid liquid plus vapor to get that all right. So, let us come back to the solution. So from table 2 we understand that H f is less than H 2 is less than H g at 4 bar and therefore we say that state 2 is wet steam all right. The moment I say wet steam I want to know what is the value of X and therefore I know other formulation now that H 2 is equal to H f plus X time H f g. I know the value of H 2 which is 1919.39 is equal to H f we know that it is equal to 604.65 plus X time 2133.4 from here we can calculate the value of X time to be equal to 0.616. So, we understood that the value of X is now 616.616 I could also see from the table that the temperature at this value the T set value at this temperature because the state 2 being the two phase region the temperature is going to be T set value all right and that is equal to 143.608. So, T set value at 4 bar is going to be 143.608 degree centigrade this is a temperature and if I want to plot this process now on a PV diagram all right my 0.1 was somewhere at this point in a sub cooled region this is state 1 and 0.2 has this is the process that is going to happen at constant pressure and 0.2 will be somewhere at this point because my X happens to be 0.6 at the center if I say midpoint of this will have 0.5 as a X dryness fraction this is 0.6. So, I can approximately say that this is the process I am talking about from 1 to 2 a constant pressure process for which I will say at state 2 we have got H 2 is equal to 1919.39 kilojoule per kg I found out that T 2 is equal to 143.6 degree centigrade and I also found out that X is equal to 0.616 and this will be the answer of this question all right. So, this was our problem the answer will be the enthalpy at 0.2 is 1919 the X 2 the dryness fraction at 0.2 being a wet steam is 0.616 and the temperature at that point is 143.6 degree centigrade and this is the answer of this question from the simple formulation of the first law depending on the pressure and the temperature given at 0.1 and because it has been supplied with Q the state changes from subcooled liquid to a two phase mixture or liquid plus vapor filter. Thank you.