 So, to simplify matters, we will define what we mean by a reservoir or thermal reservoir or a constant temperature energy reservoir. This is defined for convenience and to reduce the complexity of derivations. A reservoir is a system such that any finite amount of heat interaction does not change its temperature. And we write finite amount because if we say infinite amount it is difficult to set up such reservoirs. But if we say finite amount which is good enough for our purpose because everything we do in real life is finite to some extent. So if we say finite amount of Q interaction does not change its temperature, it is possible to set up such reservoirs using either something like a liquid and vapor in equilibrium at constant temperature or a body with a large thermal mass like a large block of copper. In real life the large amount of water in our oceans or the large amount of air our surroundings are excellent approximations to such reservoirs. And a reservoir is characterized by its temperature and if you want to know the ice junction which we use in our heat transfer lap as the reference temperature that also is an excellent example of a small capacity thermal reservoir. Similarly, when somebody is running temperature is fever, we use an ice pack and hold it on his head to maintain the brain temperature and keep it under control. That is another excellent example of a small capacity thermal reservoir till the ice melts completely the temperature is maintained to near 0 degree C. So a reservoir is characterized by its temperature and will be represented by something like this a tray or a bracket with the temperature written in it. So, this is reservoir at temperature T, this is reservoir at temperature T 1, this is reservoir at temperature T 2, this is reservoir at temperature T 3, this is reservoir at temperature T 4 and so on. Every temperature you want you can set up a reservoir for that. Why? The advantage is if you have any heat interaction with the reservoir a finite amount its temperature is not going to change. Other properties may change the energy of the reservoir will change its internal composition will change. Other properties like enthalpy entropy may all undergo a change. All that is needed is its temperature does not change do not make a mistake saying its state does not change. State does not change that means it is not having any interaction that is not true. It is capable of having heat interaction. Other parts of the reservoir may have other interactions we do not care, but this heat interaction does not change its temperature. Now after having define the heat reservoir and the efficiency of an engine we reduce the Kelvin Planck statement to this format that if I have a reservoir at some temperature T then if I have an engine and if I propose that the engine produces work by absorbing heat from just a single temperature reservoir that means an engine with an efficiency of 1 W equals Q so efficiency of 1 or 100 percent this naturally violates the Kelvin Planck statement. So Kp statement is equivalent to saying that the efficiency of an engine which interacts thermally only with one reservoir cannot be 100 percent such an engine is not possible. Now the next question which we have to ask is suppose instead of one reservoir I have two reservoirs. Absorb heat from two reservoirs so Q1 greater than 0, Q2 greater than 0 W equals Q1 plus Q2 greater than 0. Is this possible? Of course I have to say while asking this that T1 is not equal to T2. Mind you it is perfectly okay to assert that T1 is not equal to T2 for the simple reason that zeroth law helps us check that. I just allow these two reservoirs to interact with each other across a diathermic wall. If there is a heat interaction then I know T1 is not equal to T2. If there is no heat interaction I know T1 equals T2 and T1 equals T2 would make this perfectly equivalent to this because if these two reservoirs are at the same temperature I might as well combine them and say that I have just one reservoir. If these two reservoirs have the same temperature I can combine them and say that I have just one reservoir. So that question has already been answered by Kelvin-Plank statement that this is not possible. Is this possible? At this stage because I cannot directly interact with you I cannot ask the question. I ask my students this question and some people say yes some people say no but now we will proceed like this and this is the typical way we will be proceeding. We are given that T1 is not equal to T2. So what does it mean? It means that because T1 is not equal to T2 it means that heat interaction between the two reservoirs directly across a diathermic wall is possible and it will take place either from reservoir 1 to reservoir 2 or from reservoir 2 to reservoir 1. We still do not know in which direction it takes less. So what we will do is we let the engine work if it is possible for it to work. This also has an efficiency of 100%. Now let us say that it is possible to have a heat interaction such that heat flows from T1 to T2 reservoir 1 to reservoir 2, one of the two possibilities. Then all that we do is we adjust the behavior of the intervening diathermic wall and other intervening systems, other intervening boundaries such that this heat interaction is matched to Q2 and then what we claim is that look at what is happening. Out here I have a system called a reservoir which is at T2 and all that it does is absorb heat Q2 from this system at T1 and provides it to the engine. So I can consider my engine along with this reservoir T2 as an extended engine. This does not undergo any change of state and since this reservoir has absorbed Q2 and rejected Q2 and its temperature is not changing, net interaction is 0. So its state will also not be changing and hence this is perfectly equivalent to a reservoir at T1 providing to my extended engine Q1 as well as Q2 both positive and providing me with work output which is Q1 plus Q2 again positive giving me an efficiency of my extended engine E plus reservoir to be 100%. Now this by Kelvin Planck statement is not possible. Then you can argue that look it may not be possible to transfer heat from T1 to T2 but it may be possible to transfer heat from say T2 to T1. Again you can have the same argument now you extend the engine by adjusting first adjust this equal to Q1 and then extend the engine to include T1 in it and you will come to the same conclusion. In either way we show that if you have an engine which absorbs heat from two distinct reservoirs, two reservoirs at two distinct temperatures, absorbs from each of them and produces the power output making its efficiency 100%. Even this is not possible. So what is possible if we have two distinct reservoirs? Let us see. Let us say that we have two reservoirs T1 not equal to T2 and notice we are never claiming so far. We have not so far claimed that any temperature is higher than any other temperature because 0th law and 1st law does not allow us to do that. We are only able to say two temperatures are equal or two temperatures are unequal and let us say that I have an engine E absorbs from one reservoir say an amount of heat Q1 provides an amount of heat at is rejected to another reservoir at Q2. Notice the direction Q1 is greater than 0 as shown here Q2 is greater than 0 in the direction shown here and produces a work output which is now Q1 minus Q2 but because it is an engine it has to be greater than 0 which indicates that we must have Q1 greater than Q2 in magnitude then is this possible. This definitely has an efficiency of our engine less than 1. This seems to be possible. Now let us see. Now we go into the realm of consistency. We know from our day to day observations that given two systems at two distinct temperatures say T1 and T2 we know from the normal observations that energy flow across a diatomic partition between the two systems that is energy flow in the form of heat flow. We take place only in one direction the so called from a high temperature system to a so called low temperature system. We our idea of high and low temperature may be arbitrary but we know that if you have a system containing steam at one bar and another system containing ice at one bar the heat flow will be always from the steam containing a steam at one bar system to the ice at one bar system. It will not be from the ice to the steam system. Let us check whether this idea that this is possible is consistent with that. So what we do is we try to see whether this is consistent with that idea Let us do the following thing now. We have been given that T1 is not equal to T2 that means heat transfer is possible from T1 to T2 or from T2 to T1, one of those two ways. Let us say let us have the possibility of heat transfer from T1 to T2. Let that quantity be some Q3 adjusted. You will notice and I leave it to as an exercise that you will notice that whatever the amount of quantity you adjust 0 to Q1 to Q2 whatever the efficiency of this engine will never be made exceeding 1 or even equal to 1. All that happens is efficiency is defined as W divided by Q supplied and even if you extend this engine to include this all that happens is Q supplied will go up, Q rejected will go up. In fact, your efficiency effective efficiency will only reduce. This means that such an engine is consistent with direct heat flow from the reservoir at T1 to reservoir at T2. What about the other way round? Let us argue this out. Let again sketch our system to reservoirs at T1 and T2. Let us have our engine E working by absorbing Q1 from this reservoir and then rejecting Q2 to this reservoir and providing W equal to Q1 minus Q2 greater than 0. Let us say that because T1 is not equal to T2, let us say that let us try this. Maybe it is possible let us see whether heat can be transferred from T2 to T1 and let that quantity be Q3. Now, it is always possible for us to adjust Q3 and let us adjust Q3 equal to Q2. Now, what happens? Now, we notice that this reservoir T2 absorbs Q2 from the engine and provides Q2 to the reservoir at T1. Let us extend our engine to include the reservoir at T2. Now, what do you have? This is now equivalent to just one reservoir and extended engine absorption of Q1 minus Q2 from this reservoir and producing work equal to Q1 minus Q2 leading to an efficiency of the engine equal to 1, which Kelvin Planck statement says not possible. So, what is the conclusion? The conclusion from this is the following. The conclusion is if an engine works such that if you allow direct transfer of heat is possible is not possible. Now, I will leave it to you as an exercise to show the other way round. We have shown that this means this is possible this is not possible. You can show it the other way round so that we can go from implications to equivalence. Show that I will leave it to you as an exercise if we have two reservoirs T1 not equal to T2. Did I write T1 not equal to T2 here? Yes, I should write T1 not equal to T2. That is a most important or very important requirement. If T1 and T2 are two reservoirs at distinct temperatures and the two temperatures are such that when allowed to directly transfer heat between them heat flows from T1 to T2 then if you try to work an engine between them the engine will work by absorbing energy or absorbing heat from the reservoir at T1 and rejecting it to reservoir at T2 all quantities here shown are positive, but not note the directions. You have to show that homework. So what does this equivalence means? This equivalence means that maybe I should as a consequence the fact that this is equivalent to all quantities positive in the directions shown. That means if this is possible this is possible and if this is possible I should forgot to write an engineer. If this is possible then this is possible and not the other way around and if this is possible then this is possible and not the other way around. These two are perfectly equivalent. Now the next step here we have seen the equivalence of an engine working between two temperatures to distinct temperatures in a particular way and the direction in which heat is transferred between the two reservoirs at two distinct temperatures. Now what happens if we have three distinct reservoirs, three reservoirs at three distinct temperatures. Later on you can extend it to four reservoirs, five reservoirs till you get bored or till you run out of time or pen and paper you can extend this, but the next assignment next homework is this. What we have let T 1, T 2 and T 3 or I should not 1, 2, 3 always gives you an idea of order. Let us say we have T A, T B and T C three distinct temperatures. I should even write not equal to T A because just writing only up to this T A not equal to T B not equal to T C does not assert that T A is not equal to T C. So better ways to say is that T A, T B, T C are all distinct. No pair is equal. Show that before then there is an if. Show that T 1 and T 2 are such that T A and T B are such that T A and T B are such that Q is transferred, heat is transferable from A to B and T B and T C are such that heat is transferable from B to C. Then you should show that if T A and T C are allowed to interact through a diatomic partition, then heat will be transferred from A to C the other way. Because if you assume this and this, then this would show that this violates Kelvin Planck statement. This is what you have to show, just the way we have done earlier. Now what is the importance of this? The importance of this is that and of course you can have the equivalence of this because we have shown that this is equivalent to this. You can extend this by saying now whatever I have shown in black are only heat interactions. I will show it in green. Another part of this is that if you have or maybe I should use another page like this, you can also show that T C are all distinct. When you try to run an engine between T A and T B, it runs by absorbing some heat from A, rejecting some heat to B and producing some work W. If T B and T C are such that an engine works by absorbing some heat Q B or Q B prime, rejecting Q C prime to T C and producing some work W prime, then if you try to run an engine between reservoir at T A and reservoir at T C, then the engine will run by absorbing the double prime. All quantities are shown are positive and not this way, the other way. You can show that these two are consistent with this, but these two together with this violates the Kelvin Planck statement. Now, I think, did I miss defining a term? I do not think I mentioned that. When we said that such a thing is possible absorbing from one and rejecting it to other, such an engine is known as a 2 T heat engine. I think I forgot to mention that. So, that definition standard textbook definition we will be using. We do not need to use it, but that is called 2 T heat engine, something which absorbs heat from one reservoir, rejects heat to another reservoir at a distinctly different temperatures and produces a positive amount of work. Now, after showing this, the next step is important. The next step is the following T A, T B, T C, T D whatever are distinct temperatures, such that if I allow them to heat interaction between T A and T B, heat interaction takes place in the given direction at shown. If I allow heat interactions between T B and T C, heat interaction takes place in this direction at shown. If I allow heat interaction between T C and T D, it allows heat interaction only in this direction, T C to T D. Similarly, I can have T E, T F, they are arranged in this way. This is given. Now, the question is, if allow heat exchange between this and this, which direction will it go? You should be able to show that if I allow between T A and T D, the heat interaction will take place in this direction only and not in this direction. Similarly, if I allow heat exchange between say T B and T E or T F, which direction will it take place? You can show that it will take place only in this direction and it will not take place in the opposite direction. Because if you assume that it takes place in the opposite direction, then you can combine this with the other reservoirs and show that the Kelvin Planck statement is violated. This is only for heat transfer. Instead of heat transfer, you can replace them by two T heat engines and the same procedure can be demonstrated. The same idea can be demonstrated. What does this mean? This means that a set of reservoirs, distinctly different temperatures can always be arranged in order. For example, from L to R, such that heat transfer always take place reservoirs say R. I will use R for reservoir. Take place from R at left and R at left to N R at right. If you show this, then you will agree to this statement. What does it mean? That means all these distinct temperatures as represented by the reservoirs are in some order. That is the order is important. If I arrange it from left to right, such that the reservoir at left is able to transfer heat to the reservoir at its immediate right, that is the way if we arrange. Then if I allow any two reservoirs to interchange or exchange heat through a diathermic wall, then we can show that the direction can be determined by finding which is the reservoir which is at the left. From that reservoir, heat will be transferred to the reservoir at the right. So, given two distinct temperatures say T x and T y, in which direction will heat flow? You find out in this hierarchy, where does T x lie? If T x lies to the left of T y, then the heat transfer will take place from reservoir at left that is T x to the reservoir at right that is T y. If T y happens to be at the left, then the heat will be transferred from T y, which happens to be at the left to T x, which happens to be at the right. That means, this implies that there is a hierarchy or order of temperature and this order is illustrated by the fact that we can arrange the corresponding reservoirs in one direction, say from left to right and the advantage of this would be, if we are asked the question, take two arbitrary reservoirs, allow them to exchange it directly or take two arbitrary reservoirs, try to run a heat exchange, two T heat engine between the two. Then we can definitely say that if a direct heat transfer is allowed, heat transfer will always be from the reservoir at the left to the reservoir at the right. If a two T heat engine is to work, that will work only by absorbing heat from the reservoir at the left and rejecting heat to the reservoir at right. Anything other than this will not occur because that occurrence will be shown to be equivalent to violating the Kelvin Planck statement of the second law of thermodynamics. Now, we come to our definition and convention. The definition and convention is this. This hierarchy is known as the hierarchy of temperatures. We label the temperatures in such a way that temperatures at left have a higher numerical value than temperatures at the right. Hence, we say that these temperatures represent the one at the left represent higher or hotter temperatures, the one at the right represent lower or cooler temperatures. Now, our definition would be, if I have two reservoirs and if I find that a direct heat transfer takes place in this direction and not in the opposite direction and if I run a heat engine between the two, it will work by absorbing heat from this reservoir and rejecting heat to this reservoir while producing a positive amount of work. If this is so, then we define T 1 to be higher than T 2. This is now our definition of a higher temperature and a lower temperature. Remember that for this, we require the Kelvin Planck statement and hence we require the second law of thermodynamics. Now, we can go back and say that these reservoirs in this slide are arranged from higher temperature reservoir at the left to lower temperature reservoirs at the right. Using the same definition, we can say that our 2 T heat engine, the original heat engine will work only when it absorbs it from a higher temperature reservoir and rejects heat to a lower temperature reservoir. And heat will be transferred from a higher temperature reservoir to a lower temperature reservoir, but that is the definition of higher temperature and lower temperature. I strongly recommend that you do these exercises of showing that and the exercises all depend on the same trick. If I want to show that x implies y, where x and y are statements, assume that x implies not y, the opposite of y and then show that x and the opposite of y together lead to a violation of the Kelvin Planck statement of the second law of thermodynamics. And the Kelvin Planck statement can never be violated, so that means x and not y can never be together and that means if x is true, not y should be false and hence y should be true. This is the standard mathematical trick used in algebra, but very often in geometry called reductio at epsilon. We might have not used it after our may be tenth or eleventh standard, but thermodynamics is the place where we use it again and you start doing it and you will enjoy using it for various cases. Now, we have reached a stage in thermodynamics where we have defined a hierarchy of temperature, so at this stage there should be no confusion between what is a higher temperature and a lower temperature. Now, the next thing that we have to do is define some more terms. This thing we will do first thing tomorrow morning after our afternoon's discussion session. We will define the term what is meant by a reversible process and then we will talk of a reversible cycle and a reversible engine and then state and prove properly the Carnot's theorem. From there it is just standard derivations to thermodynamic scales of temperature, the Clausius inequality and the definition of entropy and relation between entropy and other properties. So, it is time for lunch. Thank you. Let us break for lunch.