 7th problem, the heat engine X operates between reservoir at 1000 degrees centigrade. So, 1000 and we will say add 273, so 1000, 273 Kelvin and a body A. So, this is the heat engine X and a body A at 600, that is, this is in 873 Kelvin, okay. Heat transfer to the engine is 1000 kilojoules and the work output of this is 220 kilojoules. Another engine Y, okay, operates between body A, that means this is Y here, operates between body A and atmosphere at 27 degrees, 27 means 300 Kelvin, this is atmosphere, so this is rejecting heat here. Heat rejected by Y to the atmosphere is 400 kilojoules, that is given, okay. And this does some work, I will say this is W Y, okay. And I will say this is Q C X and this is Q H Y, okay. The body A maintains its temperature steadily at 600 degrees centigrade, that is 800 Kelvin, it maintains steadily, that means the heat, how can it maintain, this is the body, a finite mass and heat capacity, so some heat is coming in, so temperature will tend to rise, some heat is rejected by this or received by Y, that means the temperature will increase and decrease, correct. Now, if the heat transfer to the block or body A and the heat transfer from the body A if they are equal, then only the temperature can be, it is 73 Kelvin or 600 degrees centigrade. And this body has heat interactions only with the two engines, okay. Now, you can say that that may be some heat addition. So, if for example, if the heat given to the engine and taken by Y, okay, given from the engine X and taken by the engine Y, they are different, the rest can be from the atmosphere, but it is clearly given that the heat interaction for the body A is only with the engines X and Y. So, calculate the work output and thermal efficiency of, that is I want to calculate W Y and thermal efficiency of Y, which is equal to W Y divided by Q H Y. These are the two things I want to calculate. Compare the efficiency with the Carnot cycle, operate with the same heat source and sink, that we can do. For, see, compare the efficiency of Y with the Carnot cycle, Carnot cycle operating between 873 and 300, we can calculate and do it. So, this is the given data here. So, let us say for X, W equal to Q H X minus Q C X. So, this is 200 W X. So, 220 is given. 220 equal to 1000 minus Q C X or I can say Q C X equal to 780 kilo joules, that is God. Now, since the body A maintains its temperature at 600 degree centigrade or 87 degree Kelvin, only by interacting with the engines X and Y, the heat interaction between the engines X and rejected by X has to be transferred to Y. So, I can say Q H Y should be equal to Q C X equal to 780 kilo joules. So, that is done. So, we can easily calculate the, so this is the important point here. So, the body A, which is interacting only with X and Y and maintaining its temperature at 600 degrees, it is possible, only when the heat received by A should be rejected by A, that is, received by A from X should be rejected by A to Y. This is the important key. So, once you know this, then I can write this statement, Q H Y equal to Q C X. So, without that, we cannot solve this problem. So, what happens now? Q C Y is given here, Q C Y is given as 400 kilo joules. So, that implies W Y will be equal to 780, sorry, I will write first, Q H Y minus Q C Y equal to 780 minus 400 equal to 380 kilo joules. So, that is the work. First one is work, then thermal efficiency. Second is thermal efficiency of Y will be equal to what? I will write in the next page. Thermal efficiency of Y will be equal to W Y divided by Q H Y. So, that is equal to 380, 380 divided by 780 equal to 0.4871. Now, for a Carnot cycle operating between these two temperatures, that is what? 873 Kelvin and the ambient temperature of 300 Kelvin, the efficiency of Carnot can be found as 1 minus 300 by 873, which is equal to 0.670 kilo joules. So, that is equal to 6563. So, now we, so it is clear that the efficiency of the real cycle, real or reversible cycle is less than that of the Carnot cycle. So, here, if you find the Carnot efficiency to be lesser than what you get here in the actual cycle, then the cycle is not possible. Please understand this, okay. So, the efficiency of the cycle can match. So, for example, Carnot cycle efficiency also can be 0.4871, then it is fine. All it is, it can be greater, like 0.6563, it is fine. But Carnot efficiency cannot be 0.45. Then the cycle, what we have worked on will be impossible, okay. So, that is very important to understand. So, all the reversible cycles will have more efficiency than the corresponding irreversible cycles operating between the same temperature, source and sink temperatures. That is what is in this problem. So, you can see that these are problems where the engines are connected in series and there is an intermediate source or sink, okay. So, for example, for x, the body A becomes sink and for y, the body A becomes source. Now, different heat transfers, the QGX and QHY can be different. In that case, if this is, A is not a reservoir here, please understand that if A is a reservoir, that means it has huge thermal capacity, mass into specific heat, then the temperature will not change. But if A is a finite body, like a body, if you say it is a finite body of finite mass and thermal capacity, then the heat transfer from the engine X will tend to increase the temperature of A and the heat rejected by A to engine Y will cause the temperature to decrease. So, that is the important point to find here. So, this 1000, 2000 degrees, that is 1000 degrees and the ambient are infinite temperature reservoirs. Infinite temperature, that means that the thermal capacity is very high, the temperature will not change by the heat transfer given to X or heat transfer from Y to the ambient. But on the other hand, for a finite body, which has a smaller mass and a specific heat, then the corresponding heat transfers to the body will tend to increase the temperature of the body or from the body will tend to decrease. In this case, it is clearly given that the body interacts only with X and Y as well as its temperature is maintained constant. That is possible only when the heat received by the body from X is equal to the heat rejected by the body to Y. So, this is very important to understand. Then only we can write that Q Hy equal to Q C X. So, if A is a reservoir, then this need not be true. Like we cannot say that Q C X should be equal to Q Hy become a different. Like in the previous problem here, this may be a single reservoir. So, we can see that the heat rejected by the engine Q C E and heat taken by the heat pump Q C P, they are different. Not only they are different, that is two times this, this is possible. So, the eighth problem, a house is to be maintained at 27 degrees centigrade when the outside ambient temperature is 5 degrees. So, it will be maintained. That means 300 Kelvin. Ambient temperature is 5 degrees. That is 278 Kelvin. Now, so we can see this. This is a space which has to be maintained at 300 Kelvin here and ambient is 278 Kelvin. Two methods or strategies are considered here. Two strategies which are, first one is operate a space heater. So, in the room put a space heater at 1 kilowatt power and operate it for 1 hour. That means, see, what is the heater? It has some coils and electric current passes through that, it becomes hotter and that will heat. So, the capacity or the power supplied to the heater is 1 kilowatt and it is operated for 1 hour. That is one. So, normally which we use. Second is supply is the same work. That is 1 kilowatt work to a reversible heat pump. Reversible heat pump. That means, I will say there is a reversible heat pump which will receive Q C P and reject to the room, that is Q H P to maintain the temperature 300. Now, this obviously will receive W H P which is equal to 1 kilowatt. That is what, because same power. Now, this is also, the technique again, if you do this, what will be the, we want to know what will be the heat rejected to the room by this heat pump. So, supply is the same work that is W H P to a reversible heat pump that operates between ambient and house, like this. In each case, determine the amount of heat that has been transferred to the house. So, this is the problem. So, how to do this? First is, cold temperature reservoir is at minus 5 degrees centigrade. So, minus 5 degrees. So, that is actually 268 minus 5. It is minus here. Minus 50. That is 268 Kelvin. And ambient temperature, which is, sorry, the room temperature, I will say T R will be at 27 degrees centigrade equal to 300 Kelvin. So, this is, T C is ambient temperature. Now, this is what is given. So, first the, when you operate the space heater, okay, so, when 1 kilowatt heater is operated for 1 hour, the amount of heat transferred to the room, room is how much? 1 kilowatt into 3600 Kelvin. So, this is the amount of heat transferred per 100 seconds, or I can say 1 kilo joule per second into 3600 seconds, which is equal to 3600 kilo joules. That is the amount of heat which is transferred. So, what is the work supplied? 1 kilowatt is the work supplied. And for operating 1 hour, 3600 kilo joules has been supplied. So, that will be the heat which is supplied. What is C O P for this C O P? In this case, it will be equal to, again, the, whatever you supply as work has resulted as the heat supplied to the room. You have to consider the system properly. System which actually has the coils of the heater excluded, that will give you the delta T. So, based upon temperature difference. So, here it will be 3600 divided by 3600, which is equal to 1. So, C O P will be 1 in this case. First case where you consider the space heater at 1 kilowatt operating for 1 hour. The B, this case, the, when you find use a divisible heat pump. So, in this case, C O P will be equal to, I will say heat pump will be equal to T H divided by T, because reversible, correct? T H minus T C, since reversible. Now, that will be equal to what? That will be equal to 300 divided by, so how we get this basically is because C O P of the heat pump will be equal to Q H P divided by W H P, correct? Or we can say Q H P divided by Q C P, sorry Q H P minus Q C P. So, now I can write this as 1 divided by 1 minus Q C P divided by Q H P or I can say 1 divided by 1 minus T C by T H. So, that is what I am getting here. So, T H is 300 and this is 268. So, that will be 9.375. So, compare this. For, if you just use the space heater, you are supplying work of 1 kilowatt for 1, 1 kilowatt work for 1 hour. So, what is the work supplied? 1 kilowatt, that is 1 kilo joule per second into 3600 again. So, that is getting transferred as the heat to the room, that is it. Now, if you consider this heat pump, which works in a thermodynamic cycle, it has a C O P of 9.373. So, which is about more than 9 times. Now, what is the heat supplied for the same amount of work transfer? That is 1 kilowatt. What will be the, so 1 kilowatt means what? That is for 1 hour. So, I will say 1 kilowatt into 1 hour. So, which is equal to 3600 kilo joules, the Q H. So, here what is the heat supplied to the room? That is Q H P, correct? So, Q H P will be equal to C O P into W, because C O P equal to Q H P, sorry, Q H P divided by W. So, this means, since this is correct, so this means C O P is what? 9.375 into 3600, which will be equal to 33750 kilo joules. So, when you just put the heat, heater, based upon its power rating, if it is 1 kilowatt running for 1 hour, 3600 kilo joules of work is supplied and same amount of heat also will be supplied. Assuming there is no heat loss, obviously heat loss should be there, not be there in the case where you operate the reversible heat engine also. Now, in the case of heat pump, your C O P is about 9 times more, more than 9 times. So, that means, you are supplying 9.375 times the heat which is supplied in case of the space heater. So, obviously, the better method or a better thing to do is to run a heat pump. Okay, one more point is, this is reversible heat pump, which we have considered here, so the C O P is very high. But even if you use a heat pump, which is basically not ideal or not reversible, then also C O P of the heat pump can be, say, 3 or 4. Even in that case, you are getting 3 to 4 times the heat which is transferred with the help of a space heater. Okay, so always it is better to do a, run a heat pump and do the heating of the room instead of just using a space heater. A small room, it is fine. For bigger room or if you want to maintain this, this is a one-hour, it is fine. If you want to maintain this for several, like several hours or 24 by 7, then base upon the temperature of the ambient, etcetera, proper heat pump should be chosen and the room should be heated.