 Hello and welcome to lecture 8 of this lecture series on Introduction to Aerospace Propulsion. In the last few lectures, lectures 4 onwards, we were discussing about some of the very fundamental aspects of thermodynamics. We came across several terms that we are going to use very frequently in this course. So, I hope you have understood the significance of many of these terms like what is the system and what is the boundary of a system, or the different types of system like closed system, open system and so on. What is meant by internal energy, what is enthalpy and so on. So, these are some of the terms that we are going to use very often in the during the rest of this course. So, it is very essential that you understand the significance and implications of each of these terms. So, in today's lecture what we are going to discuss about is a very fundamental law of thermodynamics and that is known as the first law of thermodynamics. We have already discussed about the zeroth law of thermodynamics. So, if you recall during one of the earlier lectures, we had discussed about the zeroth law of thermodynamics and what was zeroth law of thermodynamics. It was essentially stating that if two bodies are in thermal equilibrium with a third body, then all these three bodies are in thermal equilibrium with each other. And as we had discussed zeroth law of thermodynamics forms a very basic law of thermodynamics. And because it was formulated only after the first and second laws of thermodynamics were formulated, it was termed as zeroth law of thermodynamics. Because it is a much more fundamental law than the first and second laws of thermodynamics. So, after the zeroth law as I had mentioned during the first lecture, we shall also be discussing about the first law of thermodynamics, the second law and the third law of thermodynamics. So, in today's lecture what we are going to talk about is about the first law of thermodynamics applied to a closed system. If you recall a closed system is one wherein you can only have energy transfer across the system boundaries and no mass transfer. So, we shall be discussing the application or the implication of first law of thermodynamics as applied to a closed system. So, let us look at what are the different topics that we are going to discuss in today's lecture. So, in today's lecture we are going to discuss about these following topics. We shall primarily be discussing about the first law of thermodynamics applied to closed systems. We shall be discussing about energy balance, then we shall talk about energy change for a system. Then what are the different energy transfer mechanisms that are involved and then we shall formulate the first law of thermodynamics as applied for a cycle. And again if you recall a cycle is one wherein the initial and final states of the system are the same. Subsequently we shall formulate the first law of thermodynamics for a system which is undergoing a change of state. And towards the end of the lecture we shall also discuss some very interesting aspects known as perpetual motion machines of the first kind. These are basically devices which have been proposed which apparently well which obviously violate the first law of thermodynamics and therefore these are devices which never worked. So, we shall discuss these interesting devices towards the end of the lecture. Now we shall start our lecture with discussion on a very important experiment which was carried out more than 200 almost 250 years ago. In the 1800s in mid 1800s Joule carried out a series of experiments which laid the foundations for the first law of thermodynamics. And these experiments were very interesting in the sense that till the time Joule carried out the experiments people considered heat to be a form of an invisible fluid which they called as caloric. And caloric as people believed in those days was a fluid which is basically heat. And so heat transfer takes place between a body which has a higher caloric to a body which has a lower caloric and that was known as the caloric theory of heat. And so Joule was not very convinced with this statement of this concept of heat being a type of invisible fluid. So, he decided to carry out some experiments to find out whether heat is indeed a type of fluid. So, what he did was in his experiments which we shall discuss in detail that in his experiments he basically transferred work across an adiabatic system. And then he observed that transfer of work across the adiabatic boundaries caused an increase in temperature of the system boundaries. And then after you remove the insulation what happens is whatever temperature the system had acquired will be transferred back to the surroundings in the form of heat. And so using this experiment Joule was able to prove that essentially heat is not any fluid which flows from one body to another, but it is just a form of energy. And so Joule's experiment forms the basis for well it laid the foundations for defining what is known as the first law of thermodynamics. So, if you will if you look at what Joule did during 1840s to 1849 he decided to investigate the equivalence of heat and work. Well during that time it was only considered that heat is a form of a fluid will not work. Well Joule decided to carry out experiments which can prove that basically heat and work are just two different forms of energy transfer. And therefore, thermodynamically they are the same. And as I already mentioned before Joule's experiment heat was considered to be an invisible fluid. And it was known as caloric and it flows from a body of higher caloric to one which has a lower caloric value. The same way as we as we discussed today that heat is actually a form of energy transfer which takes place from a body which has a higher temperature to a body which has a lower temperature. So, heat is a mode of energy interaction with between a system and its surroundings or between two systems simply by the virtue of temperature difference. And so this concept of flowing of caloric from a higher body with higher caloric to one with lower caloric was known as the caloric theory of heat. So, in Joule's experiment basically it was a kind of an experiment which proved that basically heat and work are identical and they are just two different forms of energy. So, this is the basic experiment which Joule carried out. And it is a very simple experiment as it look like looks like, but the important thing to remember here is that it was this experiment which basically proved that heat is not basically it is essentially not a form of fluid which flows from one body to another, but it is a form of energy. So, let us look at what are the different constituents of this experiment. The experiment consists of an adiabatic vessel as you can see here. This adiabatic vessel contains a fluid it could be water or any other fluid. And there is a paddle wheel which is immersed in the fluid. And then this paddle wheel can be driven by a pulley on which a weight is suspended. So, as the weight is dropped it causes rotation of the paddle wheel work paddle wheel. So, across the system boundaries you can see that there is a work transfer which is taking place if there is a work done on the system. And as a result of this Joule observed that there is a change in the temperature of the system from what it was originally. Let us say the system temperature was initially T 1 and as the weight is dropped it causes the paddle wheel to rotate. And as the paddle wheel rotates it is basically doing work on the system. And so therefore, that energy has to appear in some form of the other on the system itself. So, where does that energy appear? Well that energy appears in the form of increase in temperature of the system. So, the temperature of the system which was initially T 1 would now become T 2 let us say which is where T 1 is less than T 2. And this temperature was measured using the thermometer which was also immersed in the fluid. So, this is a very ingenious experiment which Joule devised to show that it is possible to convert work into heat. And therefore, heat and work thermodynamically are equivalent in the sense that both of them are just two different forms of energy transfers or energy interactions. So, what Joule did was basically to take a certain fluid in an adiabatic vessel that is insulated and then do work on the system through a paddle wheel. That is as you drop the weight basically what happens is when the weight is dropped it rotates the paddle wheel. And therefore, that paddle wheel does a work on the fluid resulting in increase in temperature of the fluid. Now, after the paddle wheel after you have done work on the system now if you remove the insulation what happens is that now the system has a temperature T 1 well T 2 which is greater than the ambient temperature. And therefore, there is a heat transfer taking place across the system boundaries because there is no more insulation now. And so eventually the system will come back to its position where it was before the experiment began. So, now you can see that what you have basically done is a process which is cyclic in the sense that the system has returned to its initial state and this was possible by two processes. One of the process is work transfer into the system or work done on the system and the second process involves heat transfer from the system. And so the net change in energy of there is no change in internal energy of the system after the cycle is completed. We have already discussed this that for a cyclic process the change in property is cyclic integral of a property is 0. And since internal energy is a property of the system cyclic integral of internal energy should be 0. And therefore, for this particular system at the end of the cycle there is no change in internal energy of the system. So, what it means is that you have done a certain amount of work on the system which appears in the form of increase in temperature of the system. Now, if you were to transfer that much amount of energy back to the surroundings by heat transfer then essentially there is no change in the internal energy or net change in energy of the system. So, therefore, this means that work done on the system is now equal to the heat transfer from the system. So, this is what Joule was trying to prove and therefore, work done is basically not equal to, but it is proportional to the heat transfer from the system. And the constant of proportionality Joule was able to find out through a series of experiments on different fluids. So, instead of water he then used certain some other fluid. In fact, he also did experiments on solids that is he generated he did work on the solid system containing a set of solid blocks. And that work was done by essentially that work done on the system essentially caused friction between the different solids inside the system leading to an increase in temperature. But what he found was at the end of all these different experiments that the proportionality constant between work done on the system and heat transfer from the system is the same irrespective of what material or what substance you use in the system. And that proportionality constant is known as the Joule's constant. And so that basically, so it is after these set of experiments that the SI unit of energy has been termed or has been named after Joule. And therefore, the SI unit of energy is basically Joule as a mark of respect to the person who did these very fundamental experiments which defines one of the fundamental laws of thermodynamics the first law of thermodynamics. So, let us look at some more details about the experiments which Joule carried out. Well in Joule's experiment let us assume that we had work done between states 1 and 2. And therefore, the work done w subscript 1 2 is the work done on the system. And you can easily measure this work because it is just a paddle wheel work which is implemented by dropping of a weight or displacement of a weight. So, as you displace the weight you know how much the distance the weight has moved and therefore, you can determine the work done on the system. And so this work done on the system manifests itself in the form of an increase in temperature from initial state T 1 2 at state T 2. So, the system temperature rises from state T 1 temperature T 1 2 temperature T 2 at the end of the work done process. Now, if you remove the insulation what happens is basically that the system comes back to its initial state by heat transfer across the system boundaries. And therefore, the work done is basically proportional to this heat transferred. And as we have already discussed this constant of proportionality is known as the Joule's equivalent. So, Joule's equivalent is a constant of proportionality which relates the work done on the system to the heat transfer from the system. So, Joule found that if you if you were to carry out an experiment of this fashion it does not really matter what is the fluid which is held inside the system boundaries. The equivalence of work done and the heat transfer from the system is basically the Joule's constant of proportionality. And so if you were to look at the system particular system which we are discussing, we have work done between states 1 and 2. Let us say the process path is shown by this particular process as shown here between states 1 to 2. And so the work done on the system is W subscript 1 2. So, Joule showed that the work done on the system can let us say be plotted in terms of some properties x and y. And at the end of this process as we discussed if you remove the insulation of the system then that causes heat transfer well it results in heat transfer from the system to the surroundings. And therefore, that heat transfer process is shown by this second process as shown in this process diagram from state 2 to state 1. So, basically this represents a cycle wherein the system initially was at state 1. The first process was work done on the system by means of the paddle wheel work which resulted in change of state from state 1 to state 2. And from state 2 by heat transfer the system comes back to its initial state. So, the first process of energy interaction is a work transfer process or work done process and second is a heat transfer process. So, what it means is that for a cyclic process the net work done the sum total of all work done on the system is equal to some constant J times the sum total of all heat transfer that has taken place during this cycle. And this constant of proportionality is what we had discussed as the Joule's constant of Joule's equivalent essentially. Or we can also express the summation in the form of a cyclic integral which is cyclic integral of W is equal to J times cyclic integral of Q. That is for a cyclic process it is possible to equate the work done and the heat transfer through the proportionality constant which is the Joule's equivalent ratio. So, as I mentioned this Joule's experiment forms the basis for the first law of thermodynamics. So, first law of thermodynamics is essentially the conservation of energy as you have perhaps aware that there is something known as a conservation of energy principle. And that is essentially the first law of thermodynamics that energy can neither be created nor be destroyed it can only be converted from one form to another. So, this is the essence of the first law of thermodynamics that you can neither create energy nor destroyed, but you can transfer it or convert it from one form to another. And for all adiabatic. So, as a result of this first law of thermodynamics for all adiabatic processes between two specified states of a closed system the net work done will be the same. And it is regardless of the nature of the closed system or the details of the process because it is essentially an adiabatic process and the process is between two specified states of a closed system. And since we are discussing first law for a closed system today we will only talk about closed systems. The next lecture we will be talking about first law for an open system or flow processes. Now, so the first law of thermodynamics is essentially stating that energy is something which you cannot create and there is a net energy of the universe which does not change. And so there can only be energy transfer between two systems or between a system and its surroundings and it is not possible either to create energy or destroy energy. And we shall see towards the end of the lecture that there are some devices which people have proposed which essentially violate the first law of thermodynamics because they either create energy in some form or destroy energy in some form. And therefore, these devices violate a fundamental law of nature and therefore, these devices are infeasible. And it is for this reason that such devices have never been demonstrated though there are numerous examples of devices which have been proposed they have never been demonstrated because they violate a very fundamental law of nature. Now, let us look at how we can understand the first law of thermodynamics in little more detail. So, there is a simple example given here. So, this example is about change of energy from one form to another. So, here what you see is a cliff at the end of which we place a rock which has a certain mass m and by virtue of the height of this rock above the date the baseline of the datum it has a certain potential energy. Let us say it is 12 kilo joules and because the rock is at rest its kinetic energy is 0. So, remember we are looking at the macroscopic forms of energy. So, the potential energy of this particular mass is let us say 12 kilo joules it kinetic energy is 0 because it is at rest. Now, if you drop this rock from this from the top of from the edge of this cliff as the rock falls down there is a change in its energy. So, you know that potential energy is a function of the height of the particular system from the datum. So, as the rock falls down potential energy decreases because the height of the rock with reference to the datum is reducing. So, its potential energy reduces and it becomes let us say 8 kilo joules at this instant and now because the rock has a certain velocity it also has a certain kinetic energy. But the sum total of potential energy plus the kinetic energy will still be equal to what it was originally that was 12 kilo joules. And therefore, in even as it is falling the net change or net energy of the system will still be the same as it was earlier that is 12 kilo joules. Now, you may wonder what happens after the rock has reached the datum after the rock reaches its bottom then potential energy should become 0 kinetic energy also becomes 0. So, you may wonder that you have actually destroyed energy where has that energy gone well there is no destruction of energy here. What happens is after the rock falls on the ground whatever energy was contained with the rock the initial state that was 12 kilo joules gets converted to heat and work that is that much energy gets dissipated in the form of heat and sound. So, you know that as the rock falls down you would hear a sound. So, where does that sound energy come from sound is also a form of energy and that is basically the energy which the system had initially that was let us say 12 kilo joules in this case gets converted to sound. And there is also some amount of heat which is generated as the rock falls on the ground. So, this is just a very simple example of how you can use first law of thermodynamics for any system. So, as you can see in this case that you are not either creating energy nor is energy destroyed it is only converted from one form to another. And in this case it was converted from potential energy to kinetic energy and finally, it got dissipated in the form of heat and sound. So, there is only conversion of energy taking place between different forms there is neither creation nor destruction of energy. So, the first law of thermodynamics is what explains this kind of energy interaction between a system and the surroundings. Let us look at some more examples and so we have some examples here which describe application of first law to different types of systems. So, first example we have let us take an example of a potato which we are trying to boil or to cook. So, we place the if you place the potato in an oven you know that there is heat transferred into the potato which is why it gets cooked. So, the increase in energy of the potato in this example is equal to the amount of heat that is transferred to the system that is the potato in this case. So, if you were to transfer let us say 10 kilo joules of heat on to the potato the energy of the system that is potato increases by the same amount that is 10 kilo joules. So, there is no work interaction here there is just heat interaction taking place between the system and the surroundings. And in the absence of any work interaction well the net change in energy of the system will be equal to the net heat transfer. On the right hand side you can see another example of another system wherein there is heat transfer into the system as well as the heat transfer out of the system. Let us say we are transferring 25 kilo joules of heat into the system that causes and change in net energy of the system. And there is also heat transfer out of the system and let us say in this example it is 5 kilo joules. 25 kilo joules is the heat transfer into the system 5 kilo joules is the heat transfer from the system. Therefore, the net change in energy of the system which is delta e is equal to q net should be equal to q n minus q out which is 20 kilo joules. So, in the absence of any work interaction there is no work interaction here the energy change of this particular system is equal to the net heat transfer that is heat transfer in minus heat transfer out. So, in this example we have seen that if you look at a system where there is heat transfer into the system as well as heat transfer from the system and if there are no work interactions we shall look at work interactions in the next example. The net change in energy of the system will just be equal to heat the difference of the heat transfer in and heat transfer out of the system. So, this is again a consequence of the first law of thermodynamics applied for closed systems. Now, we will now look at some examples where there is also work transfer into the system across the system boundaries and one of the examples we shall look at is electrical work done on the system. We have seen in the previous lecture that there are different forms of a work done one such form is electrical work which involves heating of a resistor and that causes work done across the system boundaries in the form of heating of the resistor. So, if you consider a system let us say which is adiabatic which means there is no more heat transfer across the system boundaries, but you can still have work transfer across the system boundaries by means of in one example we shall look at a resistor. So, if you have a resistor in a inside a system which is adiabatic or which is insulated therefore, there is no heat transfer from the system boundaries. Now, you are able to do work across the system boundaries using two modes one is electrical resistance and other example we shall see is shaft work or paddle wheel work. So, in this example in the left hand side you can see the first one that is electrical heating of the system. So, using a battery if you were to heat a resistor then the net change in work done on the system will be essentially reflected in the form of increase in energy of the system. So, let us say in this case W in was 10 kilo joules that will be seen in the form of increase in energy of the system by the same amount that is 10 kilo joules if the system is adiabatic. On the right hand side we have a similar example, but instead of electrical work we have paddle wheel work wherein you rotate a paddle wheel using some work. So, in this case again if the work input was 10 kilo joules that will result in an increase in energy of the system by the same amount of 10 kilo joules and this is again applicable as long as the system is adiabatic. So, if it is not adiabatic one also needs to consider the heat transfer from the system boundaries. So, there is heat well the example we shall see in the next few slides will be systems which have both heat and work interactions and then we will see what is the net change in energy of the system. And so we shall apply first law of thermodynamics to two well different types of systems one is if it is a cyclic process which we have already seen some hint through the joules experiment wherein he proved that the net work done on a system is equal to is proportional to the net heat transfer for cyclic processes well that is essentially first law for a cycle we will see that little more detail little later. We will apply again first law of thermodynamics for system which is undergoing only a change of state but not a cyclic process then also the first law of thermodynamics for isolated system. So, we will see that as we progress in the structure. Now, let us look at how you can carry out an energy balance for any particular system which is undergoing some energy interaction with its surroundings. So, the net change which could be either increase or decrease in total energy of a system during a process is equal to the difference between the total energy entering the system and total energy leaving the system. In other words total energy entering the system minus total energy leaving the system is equal to change in total energy of the system. This again is obviously a consequence of the first law of thermodynamics that is that is e in a minus e out is equal to delta e system that is e in refers to the total energy entering the system e out is the total energy leaving the system delta e is the net change in energy. So, if you are to consider energy per unit mass this would be total energy by mass for each of them that would be small e in minus small e out is equal to delta small e system. Now, so energy change is basically energy at the final state minus energy at the initial state. Now, we have already seen this that in the absence of any electrical magnetic or surface tension effects the net change in energy delta e is actually equal to delta u plus delta k e plus delta p e where delta k e is kinetic energy change delta p e is the potential energy change delta u is the internal energy change. And therefore, for stationary systems the net change in energy of a system will be equal to the net change in internal energy of the system. Now, to explain this further if you look at a system which has both heat interaction as well as work interaction across the system boundaries. So, we have an example here of a system which is no longer adiabatic and also there is work done across the system boundaries. So, let us see what is the net change in this energy of the particular system. So, here there is a system where there is some heat transfer into the system there is some heat transfer out of the system and there is a work done on the system. So, how do you calculate the net change in energy of a system? So, you know that if there were no work done on the system then delta e would have simply been equal to q in minus q out. So, that is 25 minus 5 20 kilo joules, but in addition to this you also have a work done on the system which is 8 kilo joules here. Therefore, the net change in energy of this particular system is equal to q in minus q out plus w in which is 25 minus 5 which is corresponding to net heat transfer plus 8 which is work done. Therefore, the net change in energy of this particular system is 28 kilo joules. Therefore, this is an example which shows that you can calculate the net change in energy of a particular system which has both heat interaction as well as work interaction across the system boundaries. By looking at what is the net work done on the system and what is the net heat transfer across the system boundaries. So, that will determine what is the net change in energy of this particular system. And so, what we shall see next is that what are the different modes of energy transfer? I think we already discussed that energy interaction for a closed system can take place only in two modes. One is heat and the other is work that is you can transfer energy across the system boundaries. If it is a closed system it can be done only in two modes that is heat and work. Now, what is what if the system also has well if it is an open system. So, in addition to heat and work transfer for an open system it is also possible to have energy transfer through mass flow. So, if this were the case which we will discuss in much detail in the next lecture that is energy transfer for open systems. So, energy transfer in in general will be equal to E in minus E out that is net change in energy is equal to net heat transfer which is Q in minus Q out plus net work that is W in minus W out work done on the system minus work done by the system plus the energy due to mass flow. So, it could be E mass in minus E mass out this will be equal to net change in energy of a system. Now, that we are dealing only with closed system this particular term will be 0 there is no mass flow across the system boundaries there is only heat transfer. Therefore, delta E system as we have seen in the previous slide should be equal to just delta Q and delta W and. So, if we were to express the previous expression E in minus E out which could be net energy transfer it could be by either heat work or if it is an open system by mass is equal to delta E of the system which could essentially be reflected in terms of change in internal energy kinetic energy potential energy etcetera. And we can also express this in the form of rate that is in the rate form that is rate of net energy transfer by heat work or mass E in which is there is also usually rates are expressed by a dot at the top of the symbol. So, E in dot means rate of energy in minus E out dot which means rate of energy out is equal to d E system by d t that is rate of change of internal kinetic potential energy etcetera. And so for if the rates were constant then the net heat transfer or work done or energy can simply be found by multiplying the rate by the corresponding time interval that is Q will be equal to Q dot times delta T W should be equal to W dot times delta T and delta E is equal to d E by d T times delta T. So, this is how you find the net work or heat or energy if you know the rates and the time interval during which these interactions were taking place. So, what we shall look at next is how do you apply the first law of thermodynamics for a cycle. And we were discussing about Joules experiment which was essentially a cyclic process wherein work was done on the system and then subsequently heat was transferred from the system causing the system to come back to its initial state and therefore, that qualifies as a cyclic process. And so Joules had proved that for such processes delta W is proportional to delta Q. So, that is essentially what a cyclic process first law for a cyclic process means. So, if you look at a closed system which is undergoing a cycle the initial and the final states are the same right. We are already aware that for a cyclic process the cyclic integral of a property should be 0 which means since in energy is a property of a system cyclic integral of that should be 0. So, E in well E 1 and E 2 if you were to look consider that as the initial and final states they should be same. Therefore, delta E of a system will be equal to E 2 minus E 1 which is essentially 0 because E 2 and E 1 are same as a system has come back to its initial state after the process. So, delta E of a cyclic process will be equal to 0. So, if delta E were 0 for a closed system if you apply the first law of thermodynamics what we shall see is delta W and delta Q will be the same because delta E is equal to 0. So, that is what is explained in the slide here. Now, closed system which is undergoing a cycle we know that initial and final states are identical. Therefore, delta E of the system is equal to E 2 minus E 1 which is 0. So, for energy balance it simplifies that E in minus E out is 0 or it means that E in is equal to E out. So, energy in and energy out are the same. So, if you extend the first law of thermodynamics then if it is a closed system which does not obviously involve any mass flow across the system boundaries the energy balance for this cyclic process can simply be expressed in terms of heat and work interactions that is W net out is equal to Q net in or in the form of rate W net out dot is equal to Q net in dot that is rate of work output should be equal to rate of heat transfer input. And so, this is basically the first law of thermodynamics as applied to a cycle. Well, this just explained in terms of an illustration here. We have here a cyclic process which means that the system let us say was initially at a state here at the end of the process it comes back to it is initial state. So, as per the first law delta since delta E is equal to 0 for a cyclic process the net heat transfer during the cyclic process should be equal to the net work done during the cyclic process. So, this is the first law as applied to a cycle that W net is equal to Q net. We shall now look at how you can calculate or determine or apply the first law of thermodynamics for system which is undergoing a change of state which means that it is not undergoing a cyclic process, but it is just undergoing a process wherein there is a change of state. Therefore, the initial and final states will be different. So, in this process obviously delta E will not be 0 because there is a change of state of the system and so we shall see what or how we can apply the first law for a system which is undergoing a change of state. So, in a processes which involve a change of state you could have a scenario where the heat and work interactions are unknown that you do not know whether heat is transferred to the system or from the system or work is done by the system or to the system. So, normally what we do is to assume the direction of heat and work interactions. So, the normal practice is that we usually assume that heat is transferred into the system or usually there is heat input from the system into the system and there is work output from the system. So, this is just a normal practice it is not compulsory that all systems will definitely have such a property there could be systems where there is heat transfer or heat output and there is work input. But, in general it is a normal practice to assume that heat is transferred into the system let us say in the amount q and work is done by the system by amount w. So, if that were the case then if you apply the first law of thermodynamics we get q net in because heat is transferred into the system minus w net out work done by the system is equal to delta E of the system or in other words q minus w is equal to delta E. Now, here q essentially is q net in which means q in minus q out which corresponds to the net heat input and w is the w net out which is w out minus w in that is the net work output of the system. Now, if in a particular example or in some problem which you are trying to solve if you apply this particular sign convention and end up getting negative quantity of q or w it just means that the assume direction for that quantity was not correct and it has to be reversed. Magnitudes would not change just that you may get a negative sign when you try to solve a problem, but that only means that the assumption was initially incorrect and it should be just reversed. So, for a system which is undergoing a change of state the first law of thermodynamics comes to q minus w is delta E that is the difference between the net heat transfer and the net work done will manifest itself as the net energy change of the system that is q minus w will be equal to delta E. So, basically first law of thermodynamics boils down to q minus w is equal to delta E which means that the net heat transfer and the difference between the net heat transfer and the net work output will be equal to or will be showing up as the net energy of the system. So, that is basically what q minus w is equal to delta E means and so that is the first law of thermodynamics as applied to a system which is undergoing a change of state. And if you look at a system which has multiple inputs and multiple outputs in the sense that there are multiple modes of heat transfer and multiple forms of work interactions then let us look at what would be the first law for a system. And remember you can actually sum up the net heat transfers and net work outputs and therefore, get the delta E. Now, on the left hand side is a system which is simply undergoing a change of state there is some heat transfer into the system and the system does some work out. Therefore, the first law states that q minus w will be equal to delta E of the system. Now, what if you had multiple number of q and multiple number of w that is there are multiple modes of heat transfers either to the system or from the system and if there are multiple modes of either work done on the system or by the system then it is basically adding up all these quantities. So, in this case we have heat transfer in the form of q 1 to the system q 2 to the system and q 3 from the system there is. So, q 3 is heat transfer from the system similarly, we have w 1 which is work done by the system w 2 is work done on the system w 3 is work done again by the system w 4 is also work done by the system. So, if you were to apply the sign conventions which we had discussed in the last lecture then heat transfer to the system is usually taken as positive and heat transfer from the system is negative similarly, work done by the system is positive work done on the system is negative. So, if you apply that particular sign convention you get q 1 plus q 2 because both of these are heat transfer to the system minus q 3 because q 3 is heat transfer from the system minus w 1 which is work done by the system and therefore, positive minus w 2 which is work done on the system and hence negative w 3 and w 4 are positive because they are work done by the system. So, this net effect is equal to delta e. So, here we have net heat transfer and net work done is equal to net change in energy of the system. So, this is how you would apply first law for a system which has multiple energy interactions. So, let us summarize first law for different first law expressed in different forms or different ways for closed systems. So, we have seen that in general you can express q minus w as delta e for a stationary system because delta kinetic energy delta potential energy is 0 q minus w is delta u per unit mass it becomes q minus w small letters is equal to delta e small and in differential form delta q minus delta w is equal to d e. So, this is how you would express a first law in general for closed systems and if it is a cyclic process the right hand side naturally becomes 0 because there is no net change in energy of the system and therefore, that right hand side becomes 0 and therefore, q will be equal to w for cyclic processes for closed systems and what if the system is isolated. So, for an isolated system because there is no interaction between the system and its surroundings there is absolutely no energy interaction between the system and the surroundings delta q is 0 because there is its adiabatic delta w is also equal to 0 because it is isolated and therefore, first law gives d e should be equal to 0 or e should be equal to constant. That means, the energy of an isolated system is always constant. So, if a system is isolated that means that it has no energy interaction between the system and the surroundings the net energy of the system is the same or the energy of the system does not change because there is no way in which the system can interact with the surroundings and since it cannot interact with the surroundings it cannot interact or there cannot be any energy interaction and therefore, as per the first law energy should of that particular system will be a constant. Therefore, energy of an isolated system is always a constant. Now, there are some interesting observations based on the first law. No first law is something which you cannot mathematically prove, but just that there are no processes in nature which have known to violated to have violated the first law. Therefore, we consider that that is a sufficient proof in itself that first law cannot be violated because there have not been any proof of violations of the first law of thermodynamics. And so therefore, it is not necessary to prove first law mathematically because the fact that there are no processes which can actually violate first law itself is a proof in it that first law does exist and it is a fundamental law. And first law of thermodynamics is a very fundamental law, physical law in itself and first law the fact that first law is something which you cannot mathematically prove or that it cannot be derived from any other law qualifies first law to be what is known as a fundamental law of nature like Newton's law of nature. You cannot really prove Newton's law from some other law. Therefore, that is why Newton's laws are fundamental laws of nature. Similarly, first law of thermodynamics is also a fundamental law of nature. Now, in as per first law of thermodynamics we have seen that as per first law heat and work interactions are identical. First law does not distinguish between heat and work it does it only considers heat and work as two different forms of energy interaction. But what we shall see later on in few lectures from now is that as per second law there is definitely a great difference between heat and work and that will become clear as we understand the second law of thermodynamics. But as per first law there is absolutely no difference between heat and work there is simply two different forms of energy interactions. And what we shall see now are some interesting aspect of thermodynamics first law of thermodynamics that as we have seen it is a fundamental law and you cannot really violate the first law of thermodynamics. But over the years there have been several such concepts or proposals by different so called inventors who have come up with device such which basically violate the first law of thermodynamics. And therefore, no such device has actually been proved to be working and because if they are actually working they would be violating the first law of thermodynamics. And but there are so many such devices which have been proposed over the years and so many inventors have kind of file patents for their so called inventions which basically violate a fundamental law of nature. And so those devices which violate the first law of thermodynamics are known as perpetual motion machines of the first kind. So, any device which will violate the first law of thermodynamics is known as the are known as perpetual motion machines of the first kind. And such a device obviously will create energy and therefore, such devices cannot exist. And therefore, these devices were never actually demonstrated these were just concepts which were proposed, but they were never demonstrated. Let us look at one interesting example here. Now, here we have an example where we are the inventor so called inventor proposes that we have here a power generating unit where instead of a boiler in a conventional boiler where you add fuel and generate energy. Here the inventor claims that you can use a resistance heater by using part of the power which is generated by the generator to generate to heat up water to generate steam which can be expanded in a turbine. And this turbine drives both the pump as well as the generator. And the exhaust from the turbine goes through a condenser which condenses water and then is pumped into the boiler back. So, there is you can see from the condenser there is heat transfer out of the system and this is the system boundary. And obviously, the generator generates electrical work output. So, though this device so the inventor claim that once you start this device that it can run forever without having to use any fuel. So, it looks like a very interesting innovative concept which will solve the energy crisis forever because you do not need any fuel to run such a system. But if you look at it from the fundamental thermodynamics point of view it violates the first law of thermodynamics. And why does it violate? It violates because it generates a net work output which is queued out out and W net out without any energy input. So, there is no energy input into the system, but it always generates work output. And therefore, it is obviously violating the first law of thermodynamics. And such a device can never exist and can never work which is the reason why such a concept has never been proven though there were proposals and so called inventions, but they have never been demonstrated. So, such devices are known as perpetual motion machines of the first kind or PMM 1. There are also devices which are converse of PMM 1 that is those devices which will consume work output without any heat transfer from the system. So, such devices are known as converse of PMM 1, but essentially they are also perpetual motion machines of the first kind. So, let me quickly recap what we had looked at in this lecture. We had discussed the first law of thermodynamics for closed systems. We discussed about energy balance and energy change of a system and also the energy transfer mechanisms for a system in general. Then we applied the first law of thermodynamics for a cycle and then first law of a system undergoing a change of state. Towards the end of the lecture, we discussed about the perpetual motion machines of the first kind which are devices which basically violate the first law of thermodynamics. And in the next lecture what we shall discuss is first law of thermodynamics applied for open systems or flow processes. And we shall discuss about flow work and energy of a flowing fluid and the total energy associated with the flowing fluid and how is energy transferred or transported through mass. We shall then carry out energy analysis for steady flow systems or steady flow processes or it is also known as steady flow energy equation. And then we shall apply the steady flow energy equation for certain engineering devices which involve steady flow processes. So, that brings us to the end of this lecture and in the next lecture we shall take up first law for open systems.