 Welcome back. Now it is time for us to look at the Carnot cycle. In the previous discussions we noticed that to measure temperature on the thermodynamic temperature scale, we need a reversible 2T engine or a reversible 2T refrigerator. One method of implementing such a reversible 2T machine is by the Carnot cycle. Carnot cycle is a cycle that means a set of processes which can be used for implementing a 2T reversible heat engine or 2T reversible refrigerator. Let us look at the specification of the cycle as an engine. We will sketch the cycle on a state space of the working fluid or the system which undergoes the cyclic process. We will see it on the TV diagram. The temperature is important because the cycle has to work between two temperature levels. The higher temperature, let me call it TA and the lower temperature, let me call it TB on the ideal gas Kelvin scale. And let us say that on the thermodynamic Kelvin scale, let the two temperatures be theta A and theta B respectively. Our final aim is to show that TA equals theta A and TB equals theta B. But we will have to follow a rather scenic route, long-winded scenic route till we come to that conclusion. Let me first sketch the cycle and then explain the processes. Of course, the way the cycle looks depends on the working fluid. But generally, this is the way it will be seen. The requirements for this to be a Carnot cycle are as follows. The first and most important requirement is that all processes must be reversible. Even if a small part of the process is irreversible, it is not a Carnot cycle. There are four processes involved, 4, 1, 1, 2, 2, 3, 3, 4. Let us look at the specification of the four processes. Process 4, 1, remember it is a cycle. So, I can start analyzing it from one point. Complete the analysis and I return to that point. From which point you start, it is left to you. Let us start with the process 4, 1. This is an isothermal process and heat is absorbed by the system from the high temperature reservoir. Let me show that the heat absorbed is this, Q absorbed. Then the second process 1 to 2 is adiabatic and it is of course an expansion. In 4, 1 also there is an expansion involved. You can see the volume increasing from 4 to 1. But important thing in this adiabatic expansion is that the temperature reduces from T1 to T2. And remember that T1 equals Ta, T4 equals Ta also. The third process 2 to 3 is an isothermal process and heat is rejected, this process 2 to 3. And the fourth process which completes the cycle is 3 to 4 is adiabatic, but it is a compression process. And during this compression process, the temperature rises from T3 to T4. So, notice that there are two processes 1 to 2 and 3 to 4 which are adiabatic. And in the first process 1 to 2, the temperature of the system drops from T1 which equals T4, which equals Ta, 2, T2 which equals T3, which equals TB. And in the second process, the temperature rises from T3 to T4. So, there are just two temperature levels and two adiabatic processes. One in which temperature drops from the higher level to the lower level 1, 2. And the second from 3 to 4, again adiabatic in which the temperature rises from the lower level to the higher level. During the process 4-1, there is heat absorption from the high temperature reservoir. During the process 1-3, there is heat rejection to the low temperature reservoir. The corresponding symbolic diagram for this would be something like this. Let this be the engine. Let me call it the Carnot cycle engine. It works between two reservoirs. The temperature of this reservoir is Ta on the ideal gas Kelvin scale, Theta A on the thermodynamic Kelvin scale. The temperature of the reservoir is TB on the ideal gas Kelvin scale and Theta B on the thermodynamic Kelvin scale. There is some work done but that is something which we need not measure. The heat absorbed from the high temperature reservoir is Q absorbed as shown here. And this is the heat rejected which is Q rejected as shown here. Thank you.