 Now, the fourth problem, the Carnot engine uses hot and cold reservoirs at 1684 Kelvin and 842 Kelvin respectively. Ok, so now we can draw this 1684 Kelvin and the another reservoir is at 842 and a Carnot engine. So, this reversible engine operates between these two reservoirs. The entire, the heat input for this engine is QH and let us say this is QC. Ok, now the entire work delivered by this engine is used to operate a Carnot heat pump, the entire work W of the engine is supplied here. The pump removes heat from the 842 Kelvin reservoir. So, for example, I can draw this like this, 842 Kelvin reservoir, pump removes heat from the 842 Kelvin reservoir. So, let us say this is the heat removed by the pump, I will say P, ok, pH, then and puts it into the hot reservoir at another temperature of T dash, ok. So, now this is P, P, I will say this is pH and this is PCA, I will say, useful notation. This is the scenario. The amount of heat removed from the 842 reservoir is also QH. So, this QPC is equal to QH, to understand. So, the heat engine receives QH and puts QC to 842 Kelvin reservoir. As told to reservoir temperature, you add heat or remove it, the temperature will not change. It will be maintained at 842 Kelvin. So, now the heat pump removes heat or receives heat from the 842 Kelvin reservoir and rejects it to another temperature reservoir or another temperature T dash. So, we want to find the temperature T dash. What is also given is, the amount of heat removed from 842, that is, removed means what? This QPC is equal to QH, that is, the heat received by the engine. So, this is what is given. Again, this heat pump is corner heat pump, please understand. So, both the engine and the heat pump are reversible. So, that is the main problem, I think, here. So, we can solve this problem. So, let us find the work done by the engine that will be equal to efficiency of the engine into QH. What is the efficiency of the engine? Given the temperatures, we can find efficiency as 842, this is 1 minus 842 divided by 1684, that will be equal to 0.5. So, that efficiency of the engine. So, that means W of the engine will be equal to 0.5 times QH. That is the one, equation one. Now, this entire work is delivered to the heat pump, corner heat pump. Now, for the pump, we can write the first law, that is, minus WR equal to QPC minus QPH or WR equal to QPH minus QPC, which is equal to QPH minus QPC is given as QH itself. So, we can write this. Now, we know that this WR will be equal to W, sorry, here WV, WV equal to WR. So, because entire work is given. So, now you can see this. What is WR? 0.5 QH, correct, from this. So, this is expression, expression one, from one, I can write WR equal to 0.5 QH, which is equal to QPH minus QH, which implies QPH equal to QPH. That is the heat, which is rejected by the heat pump to a reservoir at the temperature of T dash. That will be equal to 1.5 times QH. So, this is the equation I get. Now, what is COP of the heat pump will be equal to, for a heat pump, what is the objective heat transfer? The heat which is rejected to the high temperature reservoir will be the objective energy transfer. So, we can say that will be equal to QPH divided by W. So, now I know this. What is this QPH? 1.5 times QH divided by WR, WR, WR is 0.5 times QH, 0.5 times QH. So, that will be equal to, so QH cancels basically. So, I will get the COP as 3, COP of the heat pump as 3. Now, COP is defined as what? For the heat pump, QPH divided by QPH or say WR, we can say WR, I will say QPH minus QPC. So, in terms of temperature, because this is Carnot, in terms of temperature I can say this is, QPH is received, rejected by the heat pump to the temperature reservoir temperature of T dash. So, I can say this T dash divided by T dash minus 842. So, that will be equal to 3, ok. So, from this I can find T dash as 1263 Kelvin, do you understand? So, here the interesting problem, the, there is a common reservoir 842 Kelvin. Engine rejects heat to this reservoir and a heat pump takes heat from this reservoir and rejects to a higher temperature reservoir at T dash. Now, the entire work developed by the engine is supplied to the heat pump, where the engine receives an amount of heat equal to QH from a reservoir at 1684. And the heat received by the heat pump from 842 Kelvin reservoir is equal to the heat which is received by the engine. So, under this context, considering that both the engine as well as the heat pump or Carnot that is reversible, we can write the efficiency in COP in terms of temperature in Kelvin and get the solution. Fifth problem, a reversible heat engine operates between two systems at constant temperatures of 600 degrees centigrade, that is 873 Kelvin and 313 Kelvin. A reversible heat engine. So, we can say that engine is reversible and it does some work also W. And let it, the heat, ok, QHE, I can say, similarly this is QCE, I can say, that is work by the engine. The engine drives a reversible refrigerator or which operates between systems at constant temperatures of 40 and minus 20, ok. So, that means I can say this is 40 and minus 20 is 253. So, now that means this receives heat from the low temperature, that is QCR and rejects to high temperature QHR, ok. This is the CRU here. Now, the heat transferred to the engine is 2000 kilojoules. So, that is, this is given as 2000 kilojoules. The net out, the net work output of the combined engine and refrigerator is 350. That means the entire work developed by the engine is not given. That is, some work is coming out also, that is this. So, that means I can say work to the refrigerator is WR, ok. The engine is also reversible here and refrigerator also reversible. We have to evaluate the heat transfers involving the refrigerator. That is, what is QCR and QHR? That is what we want to find. So, now, what is the efficiency of the engine? Because of the reversible engine, reversible engine, ok. I can say, efficiency is equal to 1 minus Tc by Th, which is equal to 1 minus 313 divided by 873. So, that will be equal to 0.6415. Now, what is the work developed by the engine? That will be equal to efficiency of the engine into QHE, which is equal to 0.6415 into 2000 kilojoules per cycle, ok. So, that will be equal to 1283 kilojoules, ok. That is the work. Now, net work output of the engine and refrigeration combination from the system, taking these two systems is 350. That means work for refrigerator will be equal to work developed by the engine minus the work net, that is 350 kilojoules. So, that will be equal to 933 kilojoules, ok. So, that means the net work output, see refrigeration, refrigerator only will receive the work. It is not going to give any output. So, the work developed by the engine partly given to refrigerator and partly given to the surroundings. So, the surroundings value is given here, 350 kilojoules. So, from that we have, we can find the work supplied to the refrigerator. So, that is WE, the total work developed by the engine is 1283 kilojoules. So, the work sent to the or shared with the refrigerator will be WE minus 350, ok. 350 is the net work done by the combination of engine and refrigerator. So, considering these two as a system, ok. So, now this is done. So, by first law we can say WR, WR equal to minus WR equal to QCR minus QHR, or we can say QHR minus QCR equal to WR, correct. So, this is the equation. Now I know the value of WR, 933 kilojoules. So, this will be equation one for me, ok. Now, COP of the refrigerator, COP of the refrigerator, because I know the temperatures now. What is that? COP by definition it is QCR divided by WR. So, now this is QCR divided by WR is QHR minus QCR. Now, in terms of temperatures I can write 253 divided by 313 minus 253. So, that will be equal to 4.2167. That will be the COP. Now, COP is calculated using the temperature because of the fact that the refrigerator is reversible in nature. So, we can see for example here QCR if it is taken to the denominator, it will be 1 divided by QHR minus QCR minus 1. QHR divided by QCR can be written as corresponding TH and TC for the refrigerators. So, that will be the answer. So, now from this I can find QCR equal to WR into COP of the refrigerator. So, that will be equal to 4.2167 into 933, ok. So, that will be 3934.15 kilo joules per cycle. Everything is per cycle. So, now from this I can get. So, one of the heat transfer is got. Now, apply equation one. From equation one, I can get what? QHR equal to WR plus QCR. So, I know now WR is 933. QCR, now only we have found, that is 3934.15. So, QHR will be equal to 4867.11 kilo joules. So, these are the two heat transfers. So, again a problem where the reversible heat engine and reversible refrigerator are involved. Temperatures are given. So, when temperatures are given, of course you have to convert that into Kelvin. So, if you use temperature and degrees centigrade, all the answers will be wrong. So, main thing is you have to convert the temperatures into Kelvin. And now the ratio of heat transfers can be written as the ratio of the temperatures which are expressed in Kelvin. So, this is the problem. The sixth one, a heat engine operating between two reservoirs, 1000 Kelvin and 300 Kelvin is used to drive. So, there is a heat engine operating between these two. So, engine used to, so this engine basically is not given as reversible. We have to note every point here. So, there is some work developed for the engine. So, let us say this is QHEWEQCEU should notations. This is used to drive the heat pump, heat pump HP which extracts heat from reservoir. So, this is 300 Kelvin, extracts heat that is like this. So, I can say QCP at a rate twice that of, that at which the engine rejects heat to it. That means this will be equal to 2 times QCE. That is, engine rejects an amount of QCE to 300. The HP takes 2 times of that. So, if it is QCP, then the QCP can be written as 2 times QCE. Now, this has to reject heat, that is QHP let us say, and to some temperature here, some temperature of the reservoir, let us say that is T. Now, if the efficiency of the engine is 40 percent of the maximum possible, that is, efficiency of the engine equal to 40 percent of the efficiency of the car not engine. Maximum possible is what? Maximum possible efficiency will be developed by the car not engine. Maximum possible efficiency is developed by a reversible or a car not engine. So, I can write. What is efficiency car not? I can find this. How? Because I know the temperatures. That will be simple, 1 minus the lower temperature reservoir 300, which is in Kelvin divided by the high temperature reservoir value that is 1000. So, this will be equal to 0.7. So, efficiency of the engine will be equal to 0.4 into 0.7 equal to 0.28. That is it. Similarly, the COP of the heat pump, so obviously this engine and heat pump are not reversible. They are not car not cycles. Yes, heat pump is 50 percent of the maximum possible. So, that is COP of the heat pump equal to 50 by 100 into COP of the car not operating between two temperature reservoirs. What is the temperature of the reservoir to which the heat pump rejects heat? That is, what is this T? What is the value of T? Next, what is the rate of heat rejection from the heat pump? If the rate of supply of heat to the engine is 50 kilojoules per second, that is, this is 50 kilojoules per second. So, what is the rate of rejection from the heat pump? That is, what is QP? QHP. If the rate of supply of heat to the engine, that is, from the 1000 Kelvin, engine receives the heat, that is, if that is 50 kilojoules per second, what will be QHP value? Okay, now this is what is given. So, now please see that the car not efficiency, COP can be calculated once I know the temperature. Okay, and now we can just write the expression. For heat pump, the objective heat transfer is QHP, QHP divided by the work transfer We, We is equal to QHP minus QCP. So, I can write this as 1 divided by 1 minus QCP divided by QHP or this is equal to 1 divided by 1 minus TCP. TCP is 300 by T. So, I can write this as T by T minus 300. That is the car not. Okay, so we can write COP of the heat pump as 50 percent of the COP of the car not, which is equal to 0.5 times T divided by T minus 300. So, that is done. Now, I know the efficiency of the engine. From that, I can calculate the work. Correct. So, what is work developed by the engine? That will be the efficiency of the engine into Q. That is the heat received by the engine. That will be equal to 0.28 times 50, 0.28 into 50. That is not. So, now from first law, for the engine, W E equal to Q H E minus Q C E or I can say 0.28 into 50 equal to Q H equal to 50 minus Q C E. So, which implies Q C E will be equal to 36, 36 kilo joule per cycle. Okay, so now this is W or will be equal to W E. Okay, now from the first law for the heat pump. So, this is the heat pump actually. So, W H P equal to W E. So, W H P equal to Q H P minus Q C P. Q H P minus Q C P. So, this is equal to Q H P E minus Q C P can be written as 2 into Q C E. This is given again. If we go back, we will see that Q C P is given as 2 times Q C E, correct. So, it can pump, extracts heat from the reservoir at 300 Kelvin at the rate twice that of the value that is rejected by the heat, that is Q C E, 2 times of that. So, this you know. Now, I know this also. So, from this, W H P is W E. So, I know everything. So, I can calculate Q H P from this substituting the value as 86 kilo joule per cycle. Now, how will you calculate the CO P of the heat pump is given as 0.5 T divided by T minus 300. So, from this, I can calculate T, but CO P should be known. What is CO P? That is Q H P divided by W H P. Now, we know both. So, which implies T will be equal to 326.6 Kelvin. We can also do in other way. First, we can see that for the pump, Q H P minus Q C P equal to W H P. So, this is equal to W E. So, now, I can write this as Q H P minus 2 times Q C E equal to W H P. Now, I know W E is equal to Q H E minus Q C E. Correct. So, now, efficiency of the engine is equal to 1 minus Q C E divided by Q H E. So, efficiency of the engine is 0.28, 0.28 equal to 1 minus Q C E divided by Q H E. So, from this, I can find Q C E equal to, in terms of Q H E, I can substitute. Correct. So, 1 minus 0.28, that is 0.72 into Q H E. So, W E can be calculated as point, sorry, Q H, Q H E minus 0.72 times Q H E. Do you understand? So, that will be equal to, so I will actually write in the other way, Q C E I will substitute here. So, this means Q H E I will remove from this. So, what I will do is, Q H E equal to Q C E divided by 0.72 minus Q C E. You will write like this. So, now, using this Q H P minus 2 Q C E equal to W E is, W H P equal to W E, it will be Q C E into 1 divided by 0.72 minus 1. So, that will give you, so you can take the other side. So, Q H P can be written as 2 plus 1 by 0.72 minus 1 times Q C E. So, now, we can say, C O P of the heat pump equal to Q H P by W. So, now, W also can be written in terms of Q C E. Correct. So, substituting that I can say 2 plus 1 by 0.72 minus 1 into Q C E divided by W can be written as 0.389 Q C E. Now, this cancels. So, now, I will get a value of 6.14138. So, that means, I can say, this is the actual. So, from that I can directly find. So, this is equal to what? 0.5 times T divided by T minus 300. From this also, I can find T as 326.6 Kelvin. So, there are two ways to do this. You need not calculate, take into account of Q H E. Directly, you can do this by connecting the engines, engine and the heat pump. The work actually developed by the engine is totally supplied to the heat pump. From that we can do.