 Carnot introduced several important ideas, first he introduced the idea of a reservoir, he did not use these words many of these words come to us, a reservoir is simply a system that is so large that addition of taking subtraction of heat from the reservoir does not make a difference to its temperature. So he introduced the concept of reservoir, he assumed that there was a TH a high temperature and a low temperature reservoir, you can use T as small t if you like to indicate, you still do not know what absolute temperature is, now Carnot introduced the idea of absolute temperature although the idea of ideal gas absolute temperature had been introduced by Gailusack earlier, now he showed have to go back and say here, if you measure the volume of gases and your temperature by any empirical temperature small t, Gailusack showed that the volume can be expressed in this fashion and he showed and a turned out to be 1 by this temperature increase volume increase 1 by 273.15 or whatever, so the numerical value if T was measured in degree C, right this is experimentally measured and so Gailusack essentially the conclusion was that if you went to minus 273.15 degrees there will be no volume and therefore there cannot be temperatures below that, so the empirical temperature scale had already been introduced, T ideal gas was simply T plus 273.15, okay this was known but the ideal gas temperature is not the same as the absolute temperature that Carnot came up with, you can show that the two have to be proportional, luckily for you the proportionality constant was chosen as 1, so you can completely confuse yourself about the ideal gas thermodynamic absolute temperature and the ideal gas absolute temperature and make no mistake, in which they have chosen some 1.23 then we would have had one more question in the exam to trick you, if you use 1.21 then we can give you minus 2 or do something like that, but luckily you have been spared that, so what Carnot said was that he is going to take heat from the higher temperature, reject heat to a body at lower temperature and perform a certain amount of work ?w, he said this engine could in principle he called this a reversible engine, he simply assumed that an engine could be run backwards, so he takes another engine which is also a reversible engine, he can take here a certain amount of heat say ?ql' from the lower temperature, reject heat ?qh' to the higher temperature and put in work ?w', the key idea was reversible, earlier this is around the time when everybody was working on various engines, you know you had diesel engines, you had the internal combustion engine was there, everybody was making a slightly more efficient engine every other week or at least claimed that was a more efficient engine and Carnot asked the theoretical question is there a limit to all this madness, can I tell them when to stop and he assumed one thing, he made a universal law which is still the law, he said you cannot take a body, you cannot take heat from a body and convert it completely to work without rejecting some heat to a body at a lower temperature, this is an assumption he made because of the nature he found I mean people knew that frictional heat generation was there when you did work, you could convert work to heat but converting heat to work had never been achieved with a single body, it is simply human experience to be able to see it and generalize it takes a genius, we are discussing the Carnot cycle, the key concept in it what you take for granted is the concept of a reversible engine, there is a reservoir there are several concepts, the word reservoir has been used or source and sink and the word sink is now used so widely for example they will say research is a sink, this is from the finance secretary, he says you take up any amount of money and nothing comes out of it and no change in state as far as he is concerned. So these words have become common parlance now, so you are talk of a machine that takes heat Q ? Q let us say does the amount of work ? W and rejects heat this is ? QH or lower temperature to a sink at a lower temperature, both the sink and the source are reservoirs, Carnot argued that any useful engine has to work in a cycle because it will have to come back to the same state and then go around cyclical changes so that you can repeat it and get reasonable amount of work if it is a one-shot affair it is no use, the second thing he said was that any introduce the concept of a reversible engine, this engine works forwards this was known you could convert heat to work by that time auto diesel had all actually demonstrated it, he showed that you could take heat reject heat and convert the balance into work this was also essentially possible they had already had the refrigeration kind of cycles in very crude form you could take heat from a lower source put in work and reject heat at a higher temperature this is possible this is what you do in refrigeration so the point is that he looked at an engine and said I am going to set up an engine and its reverse and ask what is the most efficient way of converting heat to work the he made a statement there is you have to make a hypothesis all the time now this is sink this is source he made a statement that it is impossible on the basis of just experience he said it is impossible to take heat from a source go through a process and return the heat to the same source that mean take a net heat return less heat this net heat can be completely converted to work and vice versa actually vice versa he did not make the statement he made the statement only one way vice versa is actually possible he said you cannot take that is you cannot leave the state of the universe the same go through a cyclic process take heat convert it to work completely so that is the assumption with that assumption he is going to give you a proof and the proof is trivial let us say you have ? Qh taken up so net heat if you if these both work in cycles okay let us say this works over n cycles that is you operate this engine over n cycles in this engine over n prime cycles n and n prime are integers but you can choose integers large enough so that the ratio of n by n prime can be any number any reasonable but net heat taken from the source is n times ? Qh minus n prime n prime times ? Qh prime net heat rejected to sink is ? Ql n times ? Ql minus n prime times ? Ql prime then network this work done by the first engine is n times ? w minus n prime times ? w prime when you finish an integral number of cycles this is left unchanged it is an engine and the remarkable thing about Carnot is he refused to be drawn into details of what the engine was it is not as if people did not pester him if it said what engine are you talking about and already there were local advertising campaigns Otto said are you talking about mine these are said are you talking about mine and so on he said I am completely contemptuous of the type of engine which is first thing you have to do you have to when you do generalizations you have to take out all the non-essentials so he said I do not care what the engine is it works in a cycle that means when this once the cycle is over it comes back to its original state these two of course are by definition sinks and sources so they do not change at all nothing happens to them you add a bit of heat it does not matter as long as it is finite so this is the net process and the process tells you that this should be equal to this minus this okay so n ? w now he said first of all choose n n ? such that n ? ql is equal to n ? ql you can always do that you just running this n times this n ? times you have control over it you simply decide to choose this n and n ? such that this is these two are equal so this term will vanish if there is a reversible engine and if there is so you have the engine and the reversible engine and the two operate this has to be satisfied and if this has to be satisfied and this is 0 then you have got conversion of a certain amount of heat into work completely with no change in state of anybody so this is impossible unless easy 0 it is a bit tricky this is why the thermodynamics is difficult good it sort of it looks like tautology but the point is this I cannot take heat let us say concretely from the reservoir net amount of heat and convert it completely to work that is the principle but that is what is happening here because I have chosen the key concept was the reversible engine it can run backwards and the reversible engine can be chosen so that at this stage I have not said anything about the efficiencies of the engines so I have got by efficiency is meant ?w by ?qh the fraction of heat that is converted to work if you like I have not said anything about it right now this can be set to 0 if I set this to 0 I get a contradiction if this term is positive if this term is negative I will simply reverse both engines so that this term is positive that is I will run this engine forwards and this engine backwards so that is a second part of it but it is possible for me to choose this to be positive and this to be 0 if I do that I have conversion of heat to work without change in state of anything in the universe and this is impossible right the first law tells you that these this quantity has to be equal to these two together that is all the first law tells you it does not say anything about whether you can convert heat completely to work or not the second law tells you not only can you do that each has to be equal to 0 by hypothesis if you like the hypothesis becomes a law the hypothesis says that you cannot take heat from a body converted completely to work without leaving the rest of the universe and I mean leaving the rest of the universe unchanged that is all the argument is a beautiful argument therefore if this has to be equal to 0 one possibility is you can rearrange this equation and ask this is possible that is this implies ?w by ?qh is equal to ?w' by ?qh' because I can set these two equal these two equal it is one way of satisfying it of course if this is actually 0 then you are not doing anything right you are not taking any heat out of the system out of the reservoir source then you are not doing anything therefore you cannot say anything that is the trivial case the non-trivial cases in this term this you are taking net heat out but this is equal to this and this is equal to this right these two are equal or this by this is equal to this by this so the non-trivial case implies in the non-trivial case or ? this is one ?w is ?qh – ?ql by ?qh is equal to ?qh' – ?ql' by ?qh' this implies again the ?ql then comes the beauty of his assumption he said I do not care what the engines are I do not care how they work one I mean you can think of any kind of engine because it is independent of the type of engines and depends only on ql and qh l and h therefore it must depend only on the temperatures of the source and the sink because the only characteristic of the source and the sink the only characteristic of each is its temperature is its degree of hotness so this is equal to function of th ?ql I do not know the form of the function yet this is 1780s feel a little humble you know you can do anything no but 1780 here is a guy equipped with practically nothing even the first law was not stated in explicit form we had an intuitive feel for it and he goes through a whole book called reflexions he just is reflecting on the nature of heat and work to be able to just sit in a corner and think about heat and work for one hour you must be given a price he has a guy thinking about just heat and work for two years is supposed to be an army guy it is amazing so having said this he did not stop here he said no not only is this a function of th and pl he went further and said he knows what form it is what he is going to show is actually a function of th divided by the same function of pl it is not only of this form it is not an arbitrary function of th and pl it is also actually a ratio of a function of th divided by the same function of pl and to do that he did another clever thing I am very bad at drawing which is one of the reasons I took to thermodynamics because here you have to only draw circles lines in some nothing else so let us say I have a P1 for convenience then I have a reservoir at T2 so it is great fun to draw now if you have let us say I take ? Q1 reject ? Q2 and then from this we will call this ? W12 between 1 and 2 this is between 2 and 3 I do some work I reject heat ? Q3 so I have now three reservoirs since it was great fun playing with two reservoirs now you play with three I mean you can try playing with four you won't get any new result so it takes wisdom to stop with the number of car now had the wisdom to decide it was worth only playing with three then he said why not work between these two so he had an engine or put in engine work here ? W13 so here you are rejecting heat here to T1 again you can put primes if you like the argument can be written mathematically but you have to try and say the words yourself in order to get this right what Karma said was simply this ? Q1 by ? Q2 or Q3 if you like mathematically it is just this this is sort of clear you can just write it and multiply and divide by Q2 but this according to car now is f of T1 T3 this is f of T1 T2 and this is f of T2 T3 remember as long as I have reversible engines this ratio has to be the same function because this has to be equal to this you know the primed engine is the same as the non-primed engine as long as both are reversible and both have to be reversible because otherwise I would have just taken work converted it to heat which is possible I can't convert heat to work therefore I have to be able to run it in reverse run both in reverse so this is the result and this from mathematics once you have function of T1 T2 etc. And although the word state variable was not defined at that time car now recognize the temperature is itself a state variable so here I have this and he said what is the form of f so you take logarithms log of f of T1 T3 is equal to log of f of T1 T2 plus log of f all I do is take this thing and differentiate with respect to T 1 can differentiate with respect to T1 so you get ? log f of I will call this f 1 3 okay we will call we will symbolize this by f 1 3 that means f of T1 T3 T1, T3 with respect to T1 ? log f of 1 2 with respect to T1 see this is a function of T1 and T3 right after I differentiate this is a function of T1 and T2 and if the differential of this these two have to be equal each has to be a function only of T1 it can't be a function of T2 because then this will be invalid it can't be a function of T3 because this will be invalid so this is equal to I will say function of T1 T3 this is function of T1 T2 so each can be equal to only a function of we will call it a G of T1 the best it can be is function of T1 because T1 T2 T3 are completely independent temperatures I can choose them exactly as I please for arbitrary temperatures if a function of T1 T3 has to be equal to a function of T1 T2 each can at best be a function of T1 alone now this is typical of calculus you differentiate then you integrate and you do not get the same answer you get slightly different because I have put in an argument that this is G of T1 if I now integrate this partial equation this implies that F13 or log F13 if you like is G of T1 DT1 integral at constant T3 this is at constant T3 right so if I integrate I get integral G of T1 DT1 right plus because this is of T3 is held constant plus some function of T3 it also implies this equation also implies I am running out of time for you to ask questions F12 is equal to again integral of G of T1 DT1 plus function of T2 so I get ln F13 let us call this integral H okay integral G of T DT is H of T and this is three lines means definition if I put three lines there instead of equal to sign definition so I get H of T1 plus some function of T3 let us call it since it is an integration constant H what shall we call it capital F surprising how we are constrained in notation L of F12 is again H of T1 plus F of T2 notice if I had differentiated this equation instead of differentiating with respect to T1 suppose I had differentiated with respect to T3 then I got this term F12 F23 it would have been equal to a function of T3 which I would have integrated I would have got H of T3 the point is that this capital F has to be the same as this really in or suppose it is a logarithm so it could be plus or minus this because the integration constant can be of different signs to exponential of this F12 is equal to exponential of this let us find another e power H of T1 we will call it H of T1 times e power F of T3 call this G of T3 and this is H of T1 again in G of T2 so look at this all he said was the only statement he made was it cannot be extracted from a body and converted to work without leaving without a change in the rest of the universe seems like a harmless statement and you would think some fool is say making some statement let him get away with it from that he will put all kinds of constraints on the efficiency of an engine that you can derive. So let us look at this if you take this F13 I have that expression there so I get H of T1 G of T3 is equal to H of T1 G of T2 then you have H of T2 and G of T3 so clearly H of T1 G of T3 come back and then you have this quantity here which therefore has to be 1 if it has to be 1 H is 1 by G so finally the glorious statement of Carnot that if you have an engine working between T1 and T2 then Q1 by Q2 for a reversible engine is equal to G of T1 is it G or is it H H H of T1 by H of T2 now I have further arguments I have to show you that H of T is in fact a monotonic function of T the challenge here is if it is not a monotonic function of T you can go back set up this cluster of engines and show by rearranging the engines you can convert heat to work without a change in the rest of the universe so that I leave the words to you because any number of times I say the words you have to say it yourself so you just have to say it yourself if you do not feel like a fool go in front of a mirror and try and convince yourself that that is true I am going to simply say H of T has to be monotonic function I think I will have to stop there I will give you 5 minutes for question so this is a crucial argument because if it is not a monotonic function at some point if H of T can switch signs and so on I will choose that interval and operate my engines in that interval so within that interval I will produce a conflict with the first statement you have to look at this carefully because after this it is all everything else is deductive once you prove it is a monotonic function then what function should it be it turns out if you apply the same Carnot cycle to an ideal gas a particular chosen cycle it does not matter which one you choose but if you choose an intelligent cycle then you can come to the conclusion that this H of T is actually the absolute ideal gas temperature it is actually equal to T if T is centigrade then H of T will be T plus 273.15 actually times any alpha you like luckily for you in history chose alpha to be one I will stop there I will discuss this again next class if you have any doubts you have to come back but I suggest you read Feynman and then put it away and say the arguments for yourself because what appears obvious is not obvious when you close the book then you discover what is obvious is that you know it is obvious to you that Feynman understand physics but that may have been obvious earlier so I will stop there I think Gibbons said teaching is futile except in those cases where it is unnecessary in that the same thing Feynman says in his professor in this original okay you have questions yeah sit down sit down because one is a function of one and three the other is a function of one no it will be T2 the other one will be T3 T2 cannot be equal to T3 so it cannot be of the form T1 T2 no no if log of f of 1 2 is you are saying if that is T1 times T2 I am saying it cannot be if log of f of 1 2 is T1 T2 if I differentiate with respect to one I will get T2 but it is also equal to log of f of T1 T3 with respect to T2 T1 which will give you T3 these arguments are fairly foolproof I may make mistakes here but if you do it carefully incidentally I always have one line added in all my note in all my I am allowed this is called subject to silly mistakes it does not take away from thermodynamics I can make silly mistakes I may make a sign error I may put a one instead of a two and all that but you have to convince yourself the basic arguments are not invalid these are two time tested and if you find a fault in them be sure you will get an over price I mean not a fault in what I do that may be any number of mistakes I am talking about the subject itself in fact the famous statement of Eddington is if all the laws were to disappear where to be found wrong he would not be surprised except if we found the second law of thermodynamics okay it started discussing the Carnot cycle we said two things first is I should have told you also an elementary another elementary concept that heat flows from higher temperature to lower temperature this is also elementary concept simply a uncontradicted experience so given that we came to this conclusion you had TH we will call this T1 this temperature T2 less than T1 your heat we showed that ? Q1 by ? Q2 is equal to some function you know what we called it just right now as f of T1 by f of T2 I suppose I called it H yesterday I said H of T is a monotonically increasing function it has to be a monotonically increasing function of T that sort of obvious because if it was a decreasing function then at T2 that function would be greater than at T1 so I can do this work and transfer this amount of heat back from here to here the net result would have been that I would have taken out heat from a body at temperature T2 converted it completely to work without any change in state of any other body so that is not possible so H of T has to be monotonically increasing it just detailed arguments you just have to sit down with it and do it but you do it only once because afterwards you take it for granted you go through all your calculations as if everything was taken for granted but I think you should see it once incidentally this will never come in the exam so for those of you who are looking for a question in the exam you do not have to worry but the real fact is that the arguments are very subtle and very simple but the sum of those arguments leads to very profound results otherwise you could not have done the rest of thermodynamics so fundamentally this has to be an increasing function this suggestion was then apply this to a specific cycle I will not go into details of why the cycle was chosen for an ideal gas in the specific cycle was to adiabatics into isothermals that is you go through to adiabatics into isothermals and you have done this before simply a matter of calculus if you do this it turns out that H of T this gives you the result H of T is identical with T ideal gas absolute is equal to T plus 273.15 where T is in degree C in this it turns out this is called actually it is alpha times T ideal gas and this was chosen as alpha times T absolute here after we use capital T for absolute temperature which is the Carnot temperature Carnot absolute temperature it is called degree Kelvin of Lord Kelvin who really refined the whole thing I mean what Clausius and Lord Kelvin did was to rewrite the whole thing in language that was acceptable but this you should say this is not alpha okay this is equal to T ideal gas sorry this is equal to alpha T this alpha was chosen to be one what we need is simply T absolute temperature in the ideal gas scale to be proportional to the absolute temperature thermodynamic temperature is just called absolute thermodynamic temperature here after we would not make a distinction and having said that next question was does this have other implications and Clausius did this at this stage Carnot had proved that the maximum efficiency you can get in any thermodynamic engine was simply dependent on the absolute temperatures the source in the sink and the maximum efficiency in terms of work would be simply T1- T2 by T1 so that is the so it a Carnot he was looking at was simply delta W by delta Q1 which was equal to T1- T2 by T1 this has developed a point where everybody believes it so much that the patent office will reject any heat engine if you submit a process for patent which shows that the efficiency is greater than this patent office will simply reject you that but more importantly what Clausius said was the following this also implies delta Q1 by T2 is equal to delta Q2 by T1 is equal to delta Q2 by T2 rearranging the same equation and here was a discovery because Joule had just discovered that you the internal energy which is whose difference differential du is represented by delta Q- delta W delta Q is a function of path delta W is a function of path but the difference is independent of the path similarly Clausius observed that you had another quantity that was independent of the path delta Q depends on the path but there is exactly one function which is the absolute thermodynamic temperature which can convert the differential quantity delta Q which is dependent on the path to a quantity that is independent of the path and this is what this was called DS if you like delta Q by T I am sorry and say he said DS is the entropy change is equal to delta Q by T the integral of DS over the whole cycle would be delta Q1 by T1- delta Q2 by T2 because when heat is absorbed it is positive when heat is given off it is negative as far as the body is concerned and over the cycle the integral is 0 this implied that the integral cyclic integral over delta Q by T was equal to 0 for the process you are always looking at the engine that is your body as far as the body that is operating in a cycle is concerned delta Q by T integral cyclic integral is 0 which implies that there exists a DS is there exists in S such that you have to say that this implies there exists in S which is a function of state such that DS can be defined as delta Q by T and because corner was dealing with closed systems there is no mass transfer in this engine from the engine to this reservoir or this thing this definition is what clause is adopted for a closed system I will say here brackets losses any engine could be less efficient than this which implied that the entropy of the system plus the entropy of the surrounding the net change in the entropy could be greater than 0 and that is the origin of the statement by Clausius that the entropy of the universe always increases.