 Welcome back. We are now near the end of our studies of Carnot cycle and the Carnot theorem. Let us see where we have reached. What have we achieved so far? So what we have achieved is the following. We now have a thermodynamic basis for temperature scales. We have a thermodynamic temperature scale which does not depend on the property of any material. However it does depend on our ability to be able to create and operate a reverse equality heat engine. But we got rid of that load by showing that the Kelvin scale does not matter whether it is thermodynamic, one based on the Carnot cycle and the Carnot engine, Reversive way to T heat engine or based on the ideal gas. They are the same. It does not matter. So now we have a unique Kelvin scale and instead of two symbols theta and T, let us have just one symbol T for this. So now we have a thermodynamic basis for not only the thermodynamic Kelvin scale but also for the ideal gas Kelvin scale. And remember temperatures on the ideal gas Kelvin scale can be measured reasonably easily compared to temperatures on the thermodynamic Kelvin scale which requires us to implement a reversible machine. And hence because the Celsius scale is now derived from the Kelvin scale and since most other temperature scales say for Fahrenheit example are based on the Kelvin scale directly or indirectly, we have the Celsius scale and other scales with a thermodynamic basis. And finally and indirect consequence of all this is we can determine if we have perfect data or estimate if we do not have perfect data the maximum performance or the best performance of an engine or a refrigerator based on the 2T heat engine or 2T refrigerator. For example we know that assuming that we have the 2T heat engine the efficiency of any such engine which will be less than or at most equal to in the reversible limit but in real life this equal to will hardly ever be applicable less than a maximum efficiency which is based on the reversible case which is given by 1 minus Q2 by Q1 for a reversible case which we know is 1 minus T2 by T1 where T2 by T1 are thermodynamic temperatures so these are on the Kelvin scale. So if we know the temperatures between which our engine works we have an estimate for the maximum efficiency and we know that the real life efficiency is going to be less than that. Similarly the coefficient of performance of any refrigerator based on the 2T refrigerator is going to be less than or in the thermodynamic limit equal to the COP maximum provided thermodynamically which is based on a reversible refrigerator which will be Q2 by W for a reversible machine which is Q2 divided by Q1 minus Q2 for a reversible machine which will be now T2 divided by T1 minus T2 on a Kelvin scale. Just as an illustrative example let us take a typical steam plant these are examples. Today a typical steam plant using steam works with a higher temperature the maximum temperature reached by steam of approximately 600 or so degree Celsius. Let us use round numbers 900 Kelvin. The temperature at which it is rejected to the environment to air or water is near about ambient. Let us take a round number 300 K. This gives us a maximum efficiency of 1 minus 300 by 900 which is roughly 0.67 or 67 percent. Actual efficiency will be less than that significantly less than that. The typical steam plant efficiency today is of the order of about 35 40 percent. Similarly when it comes to refrigerators let us determine that we aim to produce 1 ton 1000 kg of ice in 24 hours a day. How many units of electricity will be needed? One unit of electricity is 1 kilowatt hour. Now let us assume that the ice is produced at roughly 0 degrees C and let us say that 0 degrees C water is available for producing that ice. In that case the latent heat equivalent to 1000 kg of ice formation will have to be extracted. The refrigerator will work at roughly 0 degrees C at the lower temperature and ambient temperature as the upper limit. So for this let us consider a refrigerator. Let us say T2 is approximately 270 Kelvin that is roughly 0 degrees C. T1 is our ambient temperature let it be 300 K. So the maximum COP will be T2 divided by T1 minus T2 which is 270 divided by 300 minus 270 which is 9. Now remember by the definition of COP this is Q2 by W. W represents the amount of electricity required. So the amount of electricity required will be Q2 by 9 and what is Q2? Q2 will be the mass of ice produced 1000 kg multiplied by the latent heat of water that is 334 kilojoule per kilogram. This is the latent heat amount of energy required to convert 1 kg of water to 1 kg of ice at 0 degrees C. Energy difference is 334 kilojoule per kilogram. This divided by 9 and this comes to approximately 37,000 kilojoules. Now to convert this into kilowatt hours we first convert kilojoules into kilowatt seconds. We know that this is equal to this conversion factor 1 kilowatt second per kilojoule. So we have our answer in kilowatt second but we want kilowatt hour. So we multiply this by a factor which is hour per second and we know that one hour equals 3600 seconds. So this gives me approximately 10 kilowatt hour as our answer. Now that means at least 10 units of electricity will be needed to produce 1 ton of ice in one day. Of course the actual amount of electricity needed will be more than this because first this is an approximate calculation. Second this is the maximum COP that we expect. Actual COP will be much less than 9 and of course we have assumed that water is available at 0 degrees C. If it is available at anything higher than that a higher amount of energy will be needed. So this type of you know back of the envelope type of calculations it is something we can very easily do after whatever we have studied so far. The next step is to apply whatever we have studied so far to first any cycle not just a 2T engine or a 2T refrigerator cycle to any cycle and then to apply the principle of second law to any thermodynamic process. That is what we will proceed to do now. Thank you.