 In this lecture, we will talk about the thermodynamic scale of temperature and finally, derive an expression for the efficiency of the Corno engine ok. We start by illustrating the Corno cycle on a PV diagram like this, although we used a slightly different illustration earlier, you may recall that you know we used an illustration like this to show more reversible adiabats and isotherms you know we sort of show it schematically in this in this manner. So, each one of the blue line is a reversible isotherm and the red lines are all reversible adiabats. So, Corno engine will operate between 2 reversible adiabats and 2 reversible isotherms. For instance, we may have a Corno engine that starts here and then goes along this let us say goes up to there and then comes down like this maybe like this and then maybe like this. So, this would be a Corno cycle operating between the isotherms T A and T B and these 2 reversible adiabats. Now, what we want to show here is the following. The efficiency of the Corno engine when it operates in a cycle is 1 minus Q C over Q H and we want to show that this depends only on the reservoir temperatures ok. So, we demonstrated in the following manner. So, we take this cycle some amount of heat Q H is supplied during this cycle and some amount of heat Q C is rejected during this cycle. Now, you can easily imagine operating a Corno cycle which is just shifted between these 2 reversible adiabats, but between the same 2 isotherms ok. So, in that case the Corno cycle would look something like this and notice that there will be no difference between this Corno cycle and this Corno cycle. The heat that is supplied will still be Q H and the heat that is rejected will still be Q C. So, I can move it as long as I maintain the spacing between the 2 reversible adiabats I can move this cycle slide this cycle along this isotherm and the amount of heat supplied and the amount of heat rejected will remain the same only the state points will change. On the other hand, if I actually try to move the cycle slide the cycle vertically down let us see what happens if I slide the cycle vertically down. So, I do something like this. Now, you can easily see that this cycle will be different from the original one in terms of heat interaction because now the heat supplied is Q C. What was the heat rejected in the previous cycle is now the heat supplied and you should remember that Q C is less than Q H. Now, this will reject a different amount of heat Q C prime which would be less than Q C. As we slide down the reversible adiabat and go from one isotherm to another the heat interaction decreases. So, Q H is greater than Q C which is greater than Q C prime. So, you can see that the cycles are different now whereas, if I slide along a reversible isotherm cycle remains the same. The heat interaction remains the same only the state points change but when I slide down the reversible adiabat from one isotherm to another the heat interaction changes which tells me that the heat interaction or 1 minus Q C over Q H is a function only of the reservoir temperature. So, only when I slide down the reversible adiabat from one reversible isotherm to another the heat interaction and other things in the cycle change. So, this tells me that the efficiency of the cycle is dependent only on the reservoir temperatures. So, it makes it clear that it is dependent only on the reservoir temperatures. So, Q C over Q H the efficiency itself is a function psi some unknown function psi of the reservoir temperature T H and T C. So, Q C over Q H may be written as some unknown function phi of T H comma T C that is a most important point. So, you know that the ratio of Q C over Q H is a function only of the reservoir temperatures. In fact, even if you look at the cycle you can see that there is nothing else that the efficiency can depend on the only the reservoir temperatures are available for us to adjust. So, it should be clear that the efficiency will depend only on the reservoir temperatures, but this illustration brings it out even more clearly. Now, how do we make use of this information? So, here is what we do. I have engine 1 and which operates between reservoirs T A and T C. So, T A is greater than T C. So, there is also an intermediate reservoir. So, let us write it like this. So, T A is the hot reservoir, T C is a cold reservoir. Engine 1 operates between T A and T C. It is given an amount of heat Q H 1 from the hot reservoir and it produces an amount of work equal to W 1. Now, I construct another device which is composed of two reversible engines 2 and 3. So, this is the device that I am constructing. So, I can enclose this as a new device. So, that consists of two reversible engines. Engine 2 receives the same amount of heat from the high temperature reservoir as engine 1 and produces certain amount of work. It rejects an amount of heat Q C 2 to this reservoir at an intermediate temperature T B. Engine 3 picks up Q H 3 from this reservoir and then rejects an amount of heat Q C 3 to the cold reservoir at T C. So, I have added one more intermediate reservoir and these two reverse engines operate between these two reservoirs. So, let us go ahead. So, Q C 1 over Q H 1 is a function only of the reservoirs that this engine operates between which is T A and T C. Q C 2 is again a function only of the two reservoirs between which engine 2 operates and those are A and B and Q C 3 similarly is a function only of T B and T C. Now, notice that Q H 2 and Q H 1 are equal. So, Q H 2 and Q H 1 are equal. So, let us just erase this for now. So, Q H 2 and Q H 1 are equal because otherwise the comparison would be unfair. So, we need to give the same amount of heat to the two engines from the same reservoir. Only then the comparison is fair. So, Q H 2 equal to Q H 1. Q C 3 is equal to Q C 1. Let us see why this is Q C 3 which is this one here is equal to Q C 1. Why does this have to be true? Because if Q C 1 is different from Q C 3 then the efficiencies of the two engines will be different. They are both given the same amount of heat from this reservoir and if the amount of heat rejected to this reservoir is different then efficiency of one engine will be different from the efficiency of the other engine. And we have already established that the efficiency of all reversible engines operating between the same reservoirs is the same. So, for that reason and since Q H 1 is equal to Q H 2, Q C 1 must be equal to Q C 3. So, that is what we have written here. And Q H 3 equal to Q C 2. So, Q H 3 is equal to Q C 2. This is because we want this reservoir to operate in a cycle. Remember we want the entire device to operate in a cycle. If during each cycle we reject a certain amount of heat to this reservoir and we pick up a different amount of heat from the reservoir then this reservoir will not actually operate in a cycle. So, how much ever we reject to the reservoir is what we should actually pick up from this reservoir during each cycle. So, if the amount of heat rejected and the amount of heat supplied by the reservoir remains the same and so the reservoir operates in a cycle. So, for that reason we insist that these two should be equal to each other. So, with these constraints I can write the following Q C 1 over Q H 1 may be written as the product of these two quantities. So, this is dependent only on T A and T C, this is dependent only on T B and T C and this ratio is dependent only on T A and T B. Now, this relationship can be satisfied if and only if this function phi unknown function phi is of this form X C of T C divided by X C of T H where X C is again as yet unknown function. So, this is also an unknown function, but now the functional form or functional dependence of the reservoir temperature becomes clear. So, it has to be something like this X C of T C divided by X C of T H. Of course, you must bear in mind that Q C is less than Q H. That means, as the reservoir temperature decreases the heat interaction also decreases provided we are travelling between the same two reversible adiabats provided we are sliding down between the same two reversible adiabats. So, as the reservoir temperature decreases the heat interaction also decreases. So, we have put X C of T C in the numerator X C of T H in the denominator to satisfy this constraint because you may also write this as X C of T H divided by X C of T C and this relationship will still be satisfied, but that would not be consistent with what we just outlined that the heat interaction decreases with decreasing reservoir temperature. So, this is the correct form. There are many, many functions which will satisfy this sort of a relationship. Lord Kelvin initially proposed X C of T to be e raised to T and T could then vary from minus infinity to plus infinity. Supposedly, he did not feel that this was aesthetic. It did not like the fact that temperature could be negative. He did not want temperatures to be negative in this scale. So, he discarded e raised to T and he came up with the even more simple form which is simply that X C of T is equal to T and he said that in this scale the temperatures will go from 0 to infinity. They will not be negative, they will go from 0 to infinity. Now, this scale of temperature has come to be known as the Kelvin absolute or the thermodynamic scale of temperatures. So, X C of T is equal to T which means that Q C over Q H may then be written as T C over T H. So, Q C over Q H may then be written as T C over T H. The only issue that is left unresolved is it is ok to have a new scale like this and then insist that the temperature in this scale should go from 0 to infinity. But how do we relate the values to existing scales? So, by this by the time this was proposed there are already Fahrenheit and centigrade scales. You know that in the centigrade scale water freezes at 0 degree Celsius or water boils at 100 degree Celsius at one atmosphere. So, that is a specific number. How do we come up with the number for this for temperatures in this scale? This means we have to relate the Kelvin scale to the other scales in order to come up with the number. Now, here also the ingenuity of Kelvin comes to the fore. Let us just quickly go back and refresh our memory about what we said about gas thermometers. So, let us go up to gas thermometers. So, you may recall that we discussed this in some detail. So, let us just quickly refresh this. So, there is a bulb which contains a bulb which contains a certain amount of gas up to the level of the mercury. So, initially we fill the bulb with some gas say O2 and measure the pressure by adjusting this sliding scale. And then we remove a certain amount of gas keeping it the same O2. We again take the reading we keep doing this for smaller and smaller amounts of gas. Then we change from O2 to N2, repeat the readings again. Maybe go to a third gas helium or neon and repeat the same experiment and measure the pressure. Now, so now let us see what these readings look like. If you plot the readings on a pressure versus time scale, so this could be for instance O2, this could be N2, this could be say neon. And each one of this dot corresponds to a different amount of mass in the bulb. So, we have different amounts of gas in the bulb. So, each one of these readings refers to that. Note that pressure here is proportional to temperature because it is an ideal gas and PV is equal to m times R times T. And it is a constant volume thermometer. So, V is a constant. So, as I reduce my m, the temperature of the reservoir whose temperature and the reservoir whose temperature is to be measured, that temperature remains constant. So, T remains constant. So, you can see that as I keep going down, the as I keep reducing the mass, the pressure value keeps going down. And what was interesting about this plot is the following. These are very, very accurate temperature measurements using constant volume gas thermometer. As we mentioned earlier, constant volume gas thermometer probably comes closest to being a thermometer whose readings are independent of the working substance, meaning O2, N2 or neon. Still they must behave as an ideal gas, but it is probably the most accurate thermometer that can be fabricated. What was interesting about this was that, all these readings for a particular gas fall on a straight line. And if the straight lines corresponding to different gases are all extended up to the x-axis, which is the temperature axis, they all intersect at the same point. And that is at a temperature of minus 273.15 degree Celsius. Of course, this corresponds to zero pressure because the y-axis is on the, because pressure is on the y-axis. So, they all seem to intersect at minus 273.15 at zero pressure. So, Lord Kelvin then asserted that minus 273.15 was the lowest possible temperature that can be achieved. In fact, he said that this is the lowest possible temperature in the entire universe minus 273.15 degree Celsius. Even in the outermost space where we have a high degree of vacuum, he said this is the lowest temperature that can be achieved. Because this corresponded to P equal to zero, the thinking was that outer space, once you go far beyond, it is as close to being absolute vacuum as possible. So, this must be the lowest possible temperature that can be achieved. So, he then asserted that a temperature of zero Kelvin corresponds to minus 273.15 degree Celsius. What is that? There was no evidence at all for any of these assertions. It was just probably an educated guess based on whatever was known at that time. And the interesting aspect is that this assertion that zero Kelvin corresponds to this has withstood the test of time. That is probably what is most remarkable. So, what comes out of the analysis is a new temperature scale and also a connection between the new temperature scale namely the Kelvin scale and the existing scales that zero Kelvin corresponds to minus 273.15 degree Celsius. So, the efficiency of the Carnot engine may now be written as 1 minus Tc over Th. Of course, a temperature of zero Kelvin is not attainable because if you can attain zero Kelvin, then the efficiency of the Carnot engine will be 100 percent, which would be in violation of the Kelvin-Planck statement. The lowest temperature that has been realized so far is 500 pico Kelvin by a team of scientists from MIT. Now, so in order to have an efficiency of 100 percent, we require a temperature of zero Kelvin or we require a reservoir at a low cold reservoir at a temperature of zero Kelvin. Now, let us just go back to the illustration of the Carnot engine that we had. So, here we have a Carnot engine. Now, let us look at process 23. Process 23 is reversible adiabatic expansion when the cylinder has been taken from the high temperature reservoir and kept on this insulated stand and it continues to expand until the temperature reaches Tc, as we said before. So, the expansion continues until it reaches the temperature of the cold reservoir. Now, let us say that the cold reservoir temperature is zero Kelvin. So, this 23 as we can imagine for a temperature in order, in order for this process to reach a temperature of zero Kelvin, you can easily imagine that it will keep going like this and eventually it will asymptote and it will never be able to reach a temperature of zero Kelvin. Number one, number two, for it to actually reach a temperature of zero Kelvin, the cylinder length must be infinite. Cylinder has to be infinitely long for you to actually expand this until it reaches a temperature of zero Kelvin, both of which are not feasible. So, this shows this argument that I am giving here is actually put forth by Peter Atkins and it is a very interesting argument or a very different way to look at this. That both these things suggest that attaining a temperature of zero Kelvin is not possible. So, in the same manner, we can write for a refrigerator, a corner refrigerator, the COP being 1 over Th over Tc minus 1 and for a Carnot heat pump, we may write the COP like this. So, we may finally answer the question that we posed earlier, what is the maximum possible efficiency of a direct engine or maximum possible COP of a reverse engine. Now, we can actually give some numbers. Let us say that we have thermal power plant, which uses the Rankin cycle, let us say operating between high temperature of 1200 Kelvin and rejecting heat to the ambient at 300 Kelvin. Now, I can plug in these values, these temperatures into this expression and then see that the efficiency of such a power plant cannot be more than 75 percent. So, Kelvin Planck statement said that the efficiency of such a plant cannot be 100 percent. So, we were then speculating whether it could be 95, 90, 85, 80 and so on. Now, it turns out that it can only be 75 and modern power plants which use for example, something like an ultra supercritical cycle, which is a variant of the basic Rankin cycle are able to attain efficiencies between 55 and 60 percent. So, you can see that the efficiencies of actual existing power plants have come quite close to the corner efficiency itself. So, now, we can finally calculate a number. So, if you tell me what the reservoir temperatures are, then we can actually calculate the efficiency. In the same manner, let us say that we have a domestic refrigerator that is operating between minus 10 degrees Celsius and the ambient at 30 degrees Celsius. You can see that the COP of this refrigerator cannot be more than 6.575. Clausius statement says the COP cannot be infinity, but now we can see how what the real numbers are like. So, real number is even for a Carnot engine which itself is a theoretical construct can only be 6.575. So, that is what I alluded to in the beginning of our discussion on second law. Not only will second law tell you what it cannot be, it will also tell you what it can be. So, now, you appreciate the difference between the two. Kelvin Planck statement says that it cannot be 100 percent. Clausius statement says it cannot be infinity. Now, based on our discussion of the Carnot engine, we now know what it can be. For example, in this case, the efficiency of the direct engine is 75 percent, efficiency of this refrigerator is 6.575, which are far from 100 percent and infinity. So, second law answers both these questions very nicely. That is probably one of the most important and fundamental contributions of second law to the field of mechanical engineering and indeed thermodynamics. Now, there are some interesting aspects of the Carnot engine, even though it is a theoretical construct. What is that? The efficiency of a Carnot engine. So, this is the efficiency of any engine. This expression holds for any engine that operates in a cycle. This expression holds only for a Carnot engine. Now, if you look at this expression, notice that it depends only on the reservoir temperatures and does not depend on the nature of the working substance. The working substance inside can be anything. As long as it is a pure substance, we should be alright. So, it can be anything. So, the efficiency of the Carnot engine depends only on the reservoir temperatures and not on the nature of the working substance. Which means that if I actually run a Carnot engine, let us say, let us go back to this illustration. So, let us say that we run a Carnot engine between two reservoirs. Let us say that the temperature of this reservoir is not known. But the temperature of the cold reservoir is known. So, Tc is known, Th is not known. That is the temperature that we wish to measure. So, we measure the amount of heat in principle. We measure the amount of heat that is supplied to the engine. We can determine the work that the engine produces as it executes a cyclic process and then determine the efficiency of the engine. So, basically what we are doing is, we are operating the Carnot engine between two reservoirs, one whose temperature is not known, other one whose temperature is known. So, we measure Qh and by measuring the work that the engine puts out per cycle, we know Qc. So, we know the efficiency of the engine. Now, we know one of these temperatures. So, the other temperature can be evaluated. And the value that we get is independent of the working substance. So, the Carnot engine actually is the only thermometer which can actually give readings which are independent of the working substance, at least in principle. At least in principle. That is very, very important to know.