 So, in the last lecture, we actually looked at the definition of efficiency for a cyclic process and also definition of coefficient of performance for a reverse engine. So, this efficiency is applicable for a direct engine and the COP is applicable for a reverse engine. So, we saw these definitions and so you can see from these expressions that in principle efficiency can go up to 1 and again in principle the COP can go up to infinity. However, the question that we asked at the beginning namely what is the maximum value for the efficiency in principle and what is the maximum value for COP. So, in fact, if we answer one of the questions, the other one is automatically answered. So, basically the question that we are asking is what is the maximum theoretical possible value for efficiency, can it be one under some circumstances? Let us say we have an ideal device, is it possible that such a device will have 100% efficiency meaning it will convert all of the input heat into work. So, that is the question that we would like to answer now. So, we take it up like this. So, we start with the Kelvin Planck statement which reads like this. It is impossible to construct a device that operates in a cycle and produces no effect other than the raising of a weight and exchange of heat with a single reservoir. So, basically what this statement asserts is that it is not possible for a device to have 100% efficiency for a direct heat engine to have 100% efficiency. Note that there are certain very important operative words in this statement. The device must operate in a cycle which means we are looking at a heat engine. And you may recall that we defined a heat engine to be a thermodynamic system which operates continuously and has work and heat exchanges with the reservoir. So, we understood the term operating continuously to mean that it operates in a cyclic process. So, what this says is that it is impossible to construct a device that operates in a cycle produces useful work. Remember, our definition of work was that if the interaction of the system with the surroundings is such that it results in the lifting of a weight or raising of a weight in a gravitational field then the system is doing positive work. So, that is where this comes from. So, it is impossible to construct a device that operates in a cycle produces positive work but exchanges heat only with one reservoir. Which means that it is receiving heat from reservoir, it is producing positive work and it has heat interaction with only with one reservoir. That means that it is not rejecting any of the heat that is being given to it. If it rejected any of the heat that means it will be interacting with more than one reservoir. So, that is where the single reservoir comes from. So, basically what it says is that if the device operates in a cycle then it cannot convert all of the heat into work. So, eta max is not equal to 100 percent is what this statement says. Again, this statement is based on experience there is no proof for the statement. It is a statement that is based on experience and we saw the reason for it that the heat accesses the internal energy of the working substance which is a disordered mode and when you access the disordered mode not all of the heat can be converted to work. So, this is a statement that is made based on that. So, it answers the question that we raised just now. What is eta max? Can it be 100 percent? So, right now we have ruled out eta max equal to 100 percent. Let us look at the second statement then we will summarize this. Second statement is called the Clausius statement. Again, it says that it is impossible to construct a device that operates in a cycle again. The cyclic operation is important here and produces no effect other than the transfer of heat from a coal to a hot body. Heat automatically does not flow from coal to a hot body. So, if you want to move heat from a coal to a hot body we have to do some work. So, what the Clausius statement says is that it is impossible to construct this device which operates in a cycle and moves heat from a coal body to a hot body without any work input. No effect other than what this suggests is that it operates in a cycle and continuously moves heat from coal to a hot body without any other interactions with the surroundings which means no work input to the reverse engine. That is what this says. In other words, the COP of reverse heat engine cannot be equal to infinity. So, COP max cannot be equal to infinity, eta max cannot be equal to 1, COP max cannot be equal to infinity. You need to put in some work to move heat from coal body to a hot body is what the Clausius statement says. Informal and quite popular version of the statement is very obvious a refrigerator does not work until it is turned on. So, if you do not turn on the refrigerator then whatever you put inside if you take something at room temperature put it inside the refrigerator assuming that it has not been turned on then it will remain at the same temperature no heat will be removed from this and you know cool down. So, refrigerator does not work until it is turned on. So, COP max cannot be equal to infinity. So, we have two statements which are eta max not equal to cannot be equal to 1 COP max cannot be equal to infinity. So, the next question that naturally arises from this is the following if eta max cannot be equal to 100 percent what can it be is it 95 percent is it 90 percent or how do we calculate that. Similarly, for COP max if it is not equal to infinity what is it is it some large number 10 raise to 6 10 raise to 5 whatever it is. So, how do we calculate that those are the that is the next question or these are the two questions that arise naturally from these two statements. Now, before we address these two questions we also need to consider one very important aspect we have given two statements here one is the Kelvin Planck statement on the face of it it appears that this statement is applicable to direct engines on the face of it that is what it that is what it appears that it seems to be applicable to direct engines and on the face of it it seems that the Clausius statement is applicable to reverse engines. So, when we are given two such statements we are met with the disconcerting possibility that there may be a device which violates one of these two statements but not the other. So, in other words is it possible that someone can construct a device which violates the Kp statement but not the Clausius statement or vice versa. Now, if that were to happen then you know we would be in very deep trouble because that would cut out the foundation you know of the second law of thermodynamics. So, we need to ensure that there is two statements although on the face of it they appear to be different one is for direct engine the other one is for reverse engine we need to make sure that the two statements are equivalent. In other words we need to show that any engine that violates one will also violate the other which is what we will try to do next. So, we will try to establish the equivalence of the two statements. So, let us start with an engine direct engine that violates the Kelvin Planck statement. So, here is a direct engine it takes an amount of heat QH from the hot reservoir the reservoir we consider to be an infinite reservoir. So, no matter how much heat we take from the reservoir its temperature does not change. So, the engine is supplied with an amount of heat QH during every cycle. Remember we went from a steady flow to per cycle basis. So, this operates in a cyclic process and during every cycle it receives an amount of heat QH and it produces an amount of work QH without rejecting any of the heat in clear violation of the Kelvin Planck statement. Now, if you give me a device like this and here is what I can do. So, this violates Kelvin Planck statement. So, now I have a reverse engine E inverse which is a legitimate reverse engine. So, this is a legitimate reverse engine it moves an amount of heat QC from the cold reservoir. Again the cold reservoir is also an infinitely large reservoir. So, no matter how much heat we put into it the temperature remains the same. So, it moves an amount of heat QC from the cold reservoir and dumps it into the hot reservoir. The work that is required for this is supplied by the direct engine. So, the direct engine produces an amount of work QH that is used to move an amount of heat QC from here and transfer an amount of heat QH plus QC to the hot reservoir. This is a legitimate engine. Now, if you look at the combined device. So, if you look at the combined device which is given in this gray box. So, whatever is inside the combined device operates in a cycle because that was what we were given that engine E operates in a cycle takes a certain amount of heat from the hot reservoir converts it entirely to work. So, whatever is inside the box both the direct engine and the reverse engine operate in a cycle and during each cycle the engine as you can see moves an amount of heat QC from the cold reservoir and puts in a net amount of heat QC into the hot reservoir. Notice that during every cycle the combined device takes an amount of heat QH from the hot reservoir but it also puts back or rejects an amount of heat QH plus QC to the hot reservoir. So, the net heat rejected here is QC. So, this device moves an amount of heat QC from the cold reservoir during every cycle and puts an amount of heat QC same amount of heat QC in the hot reservoir without taking in any input work which is in clear violation of the Clausius statement. So, if you give me a device that violates the Kelvin Planck statement then I can use it to construct a device which violates the Clausius statement but not otherwise I can do that only if I have a device which violates the Kelvin Planck statement. Now, let us look at the second scenario. So, here is a reverse engine which clearly violates the Clausius statement it moves an amount of heat QC from the cold reservoir it operates in a cycle moves an amount of heat QC from the cold reservoir during every cycle and rejects the same amount of heat to the hot reservoir and there is no work input to this device in clear violation of Clausius statement. Now, if you give me such a device here is what I can do with this. So, here is a direct engine this is a legitimate direct engine it receives an amount of heat QH from the hot reservoir and rejects an amount of heat QC to the cold reservoir notice that it rejects the same amount of heat to the cold reservoir as the amount of heat that is moved by the reverse engine. So, this engine is constructed to operate in a cycle take an amount of heat QH from the hot reservoir and reject an amount of heat QC to the cold reservoir same as this. So, this engine puts out an amount of work equal to QH minus QC. Now, if I combine these two devices then if you look at the device inside the gray box this operates in a cycle. So, this device operates in a cycle because everything inside it operates in a cycle this engine operates in a cycle this engine is stipulated to operate in a cycle. It operates in a cycle and if you look at the heat interaction of this combined device with the reservoirs notice that it has a net positive interaction with the hot reservoir. So, the net positive interaction with the hot reservoir is QH minus QC the net interaction heat interaction with the cold reservoir is 0 because it rejects an amount of heat QC during every cycle and it also is supplied with an amount of heat QC during every cycle from the cold reservoir. So, the net heat interaction with the cold reservoir is 0 and it generates an amount of work equal to QH minus QC which is positive. So, the combined device operates in a cycle has heat interaction only with one reservoir and produces a positive amount of work which is in clear violation of the Kelvin Planck statement. So, if you give me a device which violates the Clausius statement I can then use it to construct a device which violates the Kelvin Planck statement and vice versa which shows that we are not going to end up in a situation where a device violates one but not the other. If it violates one then it violates the other statement also in principle. So, we are not going to end up in that kind of a dilemma where we have a device which violates one but not the other. So, this establishes the equivalence of the two statements. Note that I do not really have to show that if a device does not violate the Kelvin Planck statement then it also does not violate the Clausius statement that I do not have to show. I only need to be concerned about devices that violate one of these two statements and establish the equivalence in this manner. So, this establishes that the two statements are equivalent. Now, we so having done this we now turn to the question of what is eta max or what is COP max? It is not equal to eta max is not equal to 100 percent, COP max is not equal to infinity, what is it then? So, we now turn to this question. So, in order to answer this question what we are going to do is look at an ideal device a theoretical construct. So, we have the best possible device and it executes the best possible cyclic process. The combination of the two we hope or we hypothesize will give us the best possible efficiency in the case of a direct engine and best possible COP in the case of a reverse engine. So, we need two things best possible engine to operate in best possible cycle. Let us take a look at the process. You may recall this illustration from before when we discussed fully registered process we showed this in connection with path I mean sorry state path and process. So, we said that you know this was our system and we said that if I remove the mass all at once then it is more or less an unrestrained process and the amount of work that I developed in this is very very minimal if not 0. If I divide this into four equal pieces then it is a partially registered process some amount of work can be realized we may not be able to actually calculate it perhaps it can be measured but we cannot calculate it because our PDV integral PDV is applicable only for a fully registered process but some amount some non-zero amount of work may be realized because it is a partially registered process. So, this is unrestrained this is partially restrained. We argued then that if I divide the mass into an infinite number of pieces and start taking them off one at a time then I will have a fully registered process and I get the maximum possible amount of work from this expansion process. In addition we also said that the mechanical disequilibrium is very very minimal or infinitesimally small so that we go from one equilibrium state to another to another and so on. So, we then can trace the process undergone by the system like this. The most important thing is that the system is always in thermodynamic equilibrium when it goes from one state to another the departure from equilibrium is infinitesimally small. In this case mechanical equilibrium is infinitesimally small. For such a process of course you know we can plot the process on the diagram it is fully registered and more importantly than that for such a process I can stop the process at any intermediate state. For instance I can stop the process here and then if I start putting the weights back on one at a time the system will retrace the same path back to state one. So, a fully registered process is a reversible process. So, we realize the maximum work in a fully registered process and now we are saying that the process is also reversible. So, any process with work interaction will give you maximum work if it is a reversible process. Now, what if it is a process where we are putting in work let us say we are trying to compress this gas the same argument holds we will put in a minimum amount of work to compress the system from let us say state two to state one if the process is a reversible process. So, let us say that you know this is our initial state for compression that is label this x just to prevent any confusion. And now if I want to compress this to a final state let us say y, then I start putting on this small this infinitesimally small mass one at a time then I trace the process. So, then I can define the process and I can complete this process and this is what the process looks like and this is the process for which the work that I put in will be the least minimum. So, if it is a work producing process then a reversible process gives me the maximum amount of work. If it is a work absorbing process then a reversible process absorbs the minimum or least amount of work that is best possible for work absorbing or work producing processes. Now, what about heat interaction? The basic idea is the same basically what we are saying is that the departure from equilibrium must be infinitesimally small. The source of the departure from equilibrium may be mechanical it may be thermal. For instance, instead of the way I can actually add heat to the system. If I add a lot of heat very, very quickly then the system expands rapidly it undergoes an under strain expansion and the beneficial effect of adding heat is lost. I am not able to realize that in the form of work as much as I would like to. Now, if I add the heat in small increments you may recall that we said that heat transfer occurs by virtue of two things. So, we said heat interaction of a system with the surrounding occurs by virtue of two things both of which must be present. One is difference in temperature other one means to communicate. This is what we wrote down earlier so means of communication. So, if I make this difference in temperature infinitesimally small. So, for instance when I am supplying heat to the system I make sure that the system is at a temperature that is infinitesimally less than the temperature of the reservoir. If the reservoir is at a temperature th then when the heat transfer takes place the system is at a temperature of th minus d th infinitesimally small. Or if it is rejecting heat then I make sure that the system is if it rejects if it is rejecting heat to a reservoir at a temperature of Tc then I must make sure that the system is at a temperature of Tc plus dTc when it rejects heat. In both cases the departure from equilibrium will be minimal. In this case the origin of the departure from minimal from equilibrium is thermal. In the case of the mass the origin of the departure from equilibrium is mechanical. But irrespective of the origin the departure from equilibrium must be as must be infinitesimally small that is the most important thing. So, when I have process with work interaction I must make sure that it is a fully resisted reversible process because that is a tautology it is fully resisted or it is a reversible process. Whenever there is heat interaction I must make sure that the system is at a temperature that is infinitesimally different from that of the source or the sink. Which means that essentially the system will be executing an isothermal process whenever we supply heat it is at a temperature that is infinitesimally different from the reservoir which means that it is actually in a sense executing a reversible and isothermal process. Remember just like when the departure from equilibrium is very small we end up with a reversible process. Regardless of the source of the departure from equilibrium it may be mechanical or thermal regardless of that the process will be a reversible process. And in the case of heat supply or heat rejection it is a reversible isothermal process because the temperature is just infinitesimally different from the that of the reservoir. So, let us sum up what we have said so far. So, long as the departure from mechanical equilibrium is small work developed during the expansion during expansion is the highest possible and work required in the case of a compression process is the smallest possible for a fully resisted process. In other words processes with the nonzero work interaction in a cycle must be reversible processes. Remember that we are trying to put together the best possible cycle. So, in such a cycle any process with the nonzero work interaction must be a reversible process. Similarly any process with the nonzero heat interaction must be very close to being in thermal equilibrium for it to be reversible. So, we want the heat interaction process also to be reversible. And since the temperature difference between the system and the reservoir is infinitesimally small it is obvious that the system needs to execute a reversible isothermal process in case of nonzero heat interaction. So, the ideal cycle will consist of reversible processes and in case of work interaction it will again be a reversible process. In case of heat interaction it will be a reversible isothermal process. So, the best possible cycle one with the highest efficiency or COP must be composed entirely of reversible isothermal and reversible adiabatic. Where did we get this adiabatic from? Remember we said that whenever there is a nonzero heat interaction it must be a reversible isothermal process which means that any other process in the cycle must not have any heat interaction. It can have a work interaction but it cannot have a heat interaction. So, we have to be very careful about what we are saying here whenever there is work interaction we are saying it must be a reversible process. We have said nothing about the heat interaction. So, if for instance we supply, so let us go back to the previous example. So, let us say we are supplying heat to the system mass is not there we are supplying heat and again we are making sure that it is a reversible isothermal process which means that heat is being supplied. So, there is heat interaction and the piston moves up the gas expands. So, there is also work interaction. So, all we need from a work interaction perspective is that it must be a reversible process. Heat interaction may or may not be present. It must be reversible that is all. Whereas, in the case of heat interaction we are saying that it must be a reversible isothermal process. So, the restriction there is an additional restriction in the case of the heat interaction process. So, any other process in the cycle then must be adiabatic or there must be no heat interaction. So, that is where we are getting this from. So, the best possible cycle must be composed of reversible isothermal processes for heat interaction and reversible adiabatic process for everything else. So, for processes with only work interaction we should have reversible adiabatic and for the other processes we should have reversible isothermal. Now, this is not a unique choice. We can always argue since it is a theoretical construct. We can always argue that I can construct some other reversible cycle which is composed entirely of reversible processes. We will address that later on. For now, we will say that the best possible cycle is composed of reversible isothermal and reversible adiabatic. Such a cycle as most of you are probably aware is called a corno cycle. So, that is the best possible cycle. Now, what is the best possible engine? Obviously, the best possible engine is a corno engine. So, let us see what that looks like. So, here we have a piston cylinder mechanism containing a certain amount of a working substance. It is insulated except at the bottom. So, initially let us say that we place the cylinder on a reservoir at temperature T H. Now, the system absorbs an amount of heat Q H from the reservoir and as it absorbs heat it also expands slowly in a reversible isothermal process because it is a heat interaction process. So, this is reversible isothermal at temperature T H. Once that amount of heat Q H has been supplied, we take the engine, I am sorry, we take the cylinder from this bath and then put it on an insulated stand. So, the bottom is also now insulated and the gas is allowed to expand further. Now, it expands until the temperature reaches T C and this must be a reversible adiabatic expansion. So, the cylinder is entirely adiabatic. All the surfaces are adiabatic and it has to be reversible. So, it is a reversible adiabatic expansion until the temperature becomes equal to T C, which is the temperature of the cold reservoir. So, once the temperature reaches T C, we remove it from the insulated stand and we keep the cylinder on this reservoir, which is at a temperature T C. Notice that T C is less than T H. Now, we compress the gas until a certain amount of heat Q C is rejected to the reservoir. So, we compress the gas. So, there is work interaction, there is also heat interaction, but this is a reversible isothermal process at temperature T C. So, we continue to compress it until a certain amount of heat Q C is rejected to the reservoir. Now, we take it from the reservoir, again keep it on the insulated stand and compress it further until its temperature becomes equal to T H. So, we started with a system being at a temperature T H. Remember, it is a reversible isothermal process at T H. So, that means the system was at a temperature T H at the beginning. And then we supplied an amount of heat Q H, then we put it on the stand and allowed it to continue to expand until the temperature reached T C. Then we put it on the reservoir at T C and we compressed it until a certain amount of heat Q C was rejected to the reservoir. Then we put it on the insulated stand and continue to compress it until its temperature reaches T H and then these processes are executed again. So, the engine executes a cyclic process. There is no friction, there is no dissipation, there is no other mechanism, it is an ideal construct and such an engine is called a corner engine. So, the engine is ideal and the cycle is composed of processes which are the best possible from a work perspective, work and heat perspective. So, we expect that this engine will have the highest possible efficiency or if it is operated in the reverse as a reverse engine, it will have the highest possible COP. Let us illustrate this on a PV diagram. So, the PV diagram of the corner cycle looks like this. Notice that 1, 2 is a reversible isothermal process at T equal to T H, 2, 3 is a reversible adiabatic process, 3, 4 is reversible isothermal at T equal to T C, 4, 1 is reversible adiabatic. Because all the processes are reversible, I can run the engine in the reverse direction. So, I can go from 1, 4, 3, 2, 1, that is also possible. The magnitude of the heat interactions and work interaction will remain the same. Only thing that would change is the sign. If it is a fully reversible process, the magnitudes of heat and work interaction remain the same in the forward and the reverse direction, the sign alone changes. We will come back to this later on when we talk about temperature of 0 Kelvin. What is that? 2, 3 is the reversible adiabatic process where we allow the gas to expand until its temperature becomes equal to T C. So, we will come back to this later on. But for now, this is what the corner cycle looks like on a PV diagram. Later on, we will draw the corner cycle on a TS diagram where temperature specific entropy. So, here we have pressure specific volume. Of course, since this is a PV diagram, the area enclosed in the cycle is the network. So, the area enclosed inside the cycle is the network. And from first law of thermodynamics, that is also equal to the net heat interaction of the system with the surroundings. Now, we tried to answer the question or at least quantify the statement or sorry, answer the question that we posed earlier. Since we are saying that the corner cycle and the corner engine are theoretical constructs. If as an inventor, let us say, I claim that I have developed an engine which operates in a special cycle, not necessarily the corner cycle, it is a fully reversible cycle. One that produces more work than the corner engine given the same amount of heat from the same reservoir. So, we need to make sure that the comparison is fair. So, we have a high temperature reservoir at TH and we give the same amount of heat to these two engines. So, I have two engines X and Y, both are reversible engines. Engine X executes a corner cycle while engine Y which is what I have developed executes a reversible cycle, not necessarily the corner cycle. So, both of them are supplied with the same amount of heat QH from the same reservoir at TH. Is it possible for I claim that engine Y develops more work than engine X? Is this possible? Is what we need to answer next? And that is a very fundamental question. So, we need to make sure that we are able to address this effectively. Let us see how we do this. So, since X and Y are both reversible engines, since particularly since X is a reversible engine, I can run the engine in reverse which is what I am going to do next. So, I run the engine X in reverse. So, it takes in an amount of work WX, moves an amount of heat QCX from the high temperature reservoir and rejects an amount of heat QH to the I am sorry, moves an amount of heat QCX from the cold or low temperature reservoir, rejects an amount of heat QH to the high temperature reservoir. Now, what I am going to do is I am going to supply the work that is required for engine X from engine Y. Remember, engine Y is also reversible, but in the direct mode, engine Y produces an amount of work equal to WY. So, that is what I am going to do next. So, the work that is required for engine Y, engine X is supplied by engine Y like this. So, engine Y runs engine X. Now, if my claim is true that the efficiency of engine Y is greater than engine X, then that means that engine Y produces more work than engine X. In other words, since they are both given the same amount of heat QH, engine Y rejects less heat than engine X to the cold to the low temperature reservoir. That means that engine Y is able to do more with the heat that is supplied to it than engine X, which means that it rejects less heat to the reservoir at temperature Tc, cold reservoir temperature Tc. Since, WY is greater than WX, this is positive, this is greater than zero. So, what I am saying is my claim is that the efficiency of engine Y is greater than that of engine X, which is actually a Kornow engine. So, this is a reversible engine and it executes a reversible cycle different from a Kornow engine and I claim that its efficiency is higher. So, in this scenario, you can see that I am driving engine X using engine Y, it produces a positive amount of work. Let us see. So, if I look at the the device, the combined device in the shaded box, the net heat interaction with the high temperature reservoir is equal to zero because during every cycle, remember X and Y, both engines operate in a cyclical manner. So, both the device inside the box operates in a cycle, the net heat interaction of this device with the high temperature reservoir is zero because during every cycle, it receives an amount of heat QH, it also rejects an amount of heat QH, same amount of heat QH. So, that means the net heat interaction with the reservoir is zero. And you must recall that since the efficiency of engine Y is claimed to be higher than that of engine X, QCY is less than QCX. So, QCY is less than QCX, which means that QC net of this engine is positive because it is supplied with this combined device is supplied with QCX during every cycle and it rejects QCY during every cycle, which means that QC net is positive. And the work output is also positive since WY is claimed to be greater than WX. So, if you look at the device in the box, it operates in a cycle, produces a positive amount of work and has heat interaction with only one reservoir, receives heat only from one reservoir, which is in clear violation of Kelvin Planck statement. Hence, eta Y has to be equal to eta X, it is not possible that eta Y be greater than eta X, eta Y has to be equal to eta X. So, what we are saying here is a very powerful statement. The Carnot cycle which is entirely composed of reversible isothermal and reversible adiabatic processes has the highest efficiency. Any other engine which operates in a reversible cycle between the same reservoirs and receives the same amount of heat QH will have the same efficiency as a Carnot engine operating between these two reservoirs. So, that is very, very important. It can be any reversible cycle. You can construct any reversible cycle, but it will have the same efficiency as a Carnot engine operating between the same reservoirs. That is a very, very powerful statement to make because it is very, very general. When I say any reversible cycle, that is precisely what it means, it can be any reversible cycle. So, it is a very sweeping statement and one which is very, very important. So, now in our quest to answer the question, what is the maximum possible efficiency for a direct engine? We said that it is not 100 percent. Then we said if it is not 100 percent, what is it going to be? Then we said that it is going to be the efficiency of an engine which operates in a Carnot cycle or efficiency of a Carnot engine will be the maximum possible. So, what we need to do next is quantify this. What is if it is not 100 percent, what is it? Can we calculate the efficiency of such a Carnot engine? That is what we are going to look at next. Before we do that, let us see what happens in real life. The Carnot engine is an ideal construct and can never be built in practice. It is sort of like a thought experiment. It is a theoretical construct. Its usefulness lies in the fact that the efficiency or the COP of such a engine is the highest possible. That is where its usefulness lies. In real life, there are many factors which cause actual engines to depart from this ideal engine and suffer a degradation in performance. So, actual engine we are labeling as irreversible. So, the ideal engine or Carnot engine is composed entirely of reversible processes. We said that all the processes in the ideal cycle must be reversible. So, in reality, they depart from being reversible. So, they are actually irreversible. What are some of the factors which render the cycle irreversible? These are friction, dissipation. It can be a partially resisted process or it can be a thermal disequilibrium, meaning heat transfer across a finite temperature difference. We said that the temperature difference between the reservoir and the system must be only infinitesimally small. In most cases, it will be there will be a finite temperature difference. So, that can also cause thermal disequilibrium and an irreversibility or it can also be mixing and chemical reactions and so on. Now, some of these irreversibilities may be classified as internal to the system, some may be classified as external to the system. We need to be able to understand which are external, which are internal, so that later on when we actually do or evaluate the entropy that is generated in the universe as a result of these devices, you will be able to see where the entropy generation comes from. Is it due to external irreversibilities or is it due to internal irreversibilities? As engineers, that is very important for us to understand. And you may recall that this was one of the learning outcomes that we mentioned in the beginning of the course. How to calculate entropy, not only efficiencies of devices, but entropy generation in the universe as a result of operation of these devices. So, internal irreversibilities as the name suggests are internal to the engine itself. So, this is friction in the engine or dissipation in the engine, mixing inside the engine and possible chemical reactions that are occurring inside the engine. So, inside the cylinder for example, if it is a piston-cylinder mechanism, these are things that can be happening inside the piston-cylinder mechanism, these are internal irreversibilities. External irreversibilities, partially resisted process or unrestrained processes and heat transfer across a finite temperature difference, which is thermal disequilibrium. As we mentioned earlier, this is thermal disequilibrium. These are external irreversibilities. Notice that it is possible for us to have an internally reversible engine, but with external irreversibilities. So, the engine may be free of all these irreversibilities, but these irreversibilities may be present, those are external to the engine. Or we can have internal irreversibilities, but no external irreversibilities, both are possible. That is why we are classifying them properly, so that we can account for entropy generation in the universe as a result of these sorts of irreversibilities, which are inevitable in nature. So, if let us develop this point a little bit more. So, we said for example, so if you revisit the corno engine, for example here, we may have a corno engine that is internally reversible, meaning there is no friction, there is no dissipation, there is no mixing, there is no chemical reaction. It is internally reversible, but let us say that there is an external irreversibility, meaning there is a temperature difference between the reservoir and the system. In such a case, the corno engine will still execute the same cycle as here. The only thing is, this TH will no longer be the reservoir temperature, but temperature of the system. So, the system receives the heat at temperature TH and it executes the reversible isothermal process. So, it remains at the same temperature as it executes the process. So, this will not be the temperature of the reservoir, it will be the temperature of the system. Similarly, this also will not be the temperature of the reservoir, it will be the temperature of the system. If it is an internally reversible engine, but with external irreversibilities. So, we can easily account for external irreversibilities in this type of scenario. Internal irreversibilities are somewhat more difficult to account for friction and other things. So, the only thing we can say is as a result of internal irreversibilities, the work produced by the engine decreases by a certain amount. That is all we can do and that is how we will handle internal irreversibilities. But external irreversibilities can be handled in a somewhat more elegant manner. So, what we will try to do in the next lecture is answer the question finally, what is the efficiency of a corno engine? How do we evaluate the efficiency of a corno engine? Once we are able to evaluate the efficiency of a corno engine that answers the question that we started out with namely, what is the maximum possible efficiency? How much, what is the maximum amount of heat that can be converted to work? So, once we do this, we will be able to answer that question. So, we will take it up in the next lecture.