 So typically we could just simply call this as reference temperature it does not have to be 298 and as an example I have written a couple of things here one is the atomic hydrogen formation reaction so we noted that H2 is like the reference element for H and therefore a half H2 giving H is the formation reaction for atomic hydrogen and so there is a HF0 298 for H so you look at the notation here so in fact you could improve on the notation by saying 298, script MI on the subscript there where script MI refers to the species for which you are looking at the formation standard heat of formation. So here for the formation reaction of atomic hydrogen you get a number that looks like this but on the other hand if you now look at the formation reaction for the high H2 molecule the formation reaction is H2 gives H2 and obviously you are now having no change across this reaction therefore it makes sense for us to take the standard heat of formation as 0. The last class one of the questions that came about for a definition of a heat of reaction and more so for the formation reaction was if you now take the initial state a certain pressure and temperature and then you now wait until you come back to the same pressure and temperature for the final state and look at the heat that has been removed or added to the system during this process what about the expansion work okay because we now keep the pressure constant that means we could allow for the volume to change and therefore there will be a pressure work that the system does because of the volume change that is allowed so what we have to recognize now is note that all heats or enthalpies okay. That means and it is not for nothing that we chose a symbol H that would imply that the heat is an enthalpy or the reason why H is associated the symbol H is associated with enthalpy is looking at the other way it is actually a heat okay so this is to say all the heats that we are looking at are not internal energies but enthalpies okay. So these are enthalpies that means that is they include the expansion work so for example if you now say suppose we now have a container and a piston and a weight W indicates constant pressure right and suppose initially we have a mixture of H2 and ½ mole of oxygen molecule and suppose we now have a heat interaction across the system boundary and finally we now have a different volume right at which you now get a H2O for the final. So here we have a T equals 298 K P equals 1 bar and here we have 2 T equals 298 K P equals 1 bar right so the convention that we use the convention is so heat transfer is positive positive when into the vessel so positive when into the vessel right so we now take Q to be positive if it is going inside and so for here here is in this example where we are taking a H2 and ½ to initially and then becomes H2O the heat is liberated liberated that is leaving the vessel leaving out of the vessel okay and that is that is we will also try to now make it like Q P Q P refers to the subscript P here for the Q refers to the heat transfer at constant pressure right so that is Q P equal to negative 240 kilojoules per mole right. So when you are saying per mole it is actually per mole of what is formed so in this case it is 1 mole of atomic hydrogen that is formed incurring 217.999 kilojoules here you are looking at 1 mole of water that is formed liberating 240 kilojoules per mole right so this is this means negative means exothermic right positive means endothermic right and so what is happening here is first law of thermodynamics applied to the system means that a ? E which is the internal energy change is equal to Q P minus the work done if you could write it the other way that is the heat is equal to ? E plus the work done or ? E is equal to Q P minus the work done and this means E2 minus E1 is equal to Q P minus P times let us say V2 minus V1 or you can also now rearrange things so you can say E2 take this PV2 to the left hand side so it becomes a plus PV2 and then you have a minus of minus PV1 so you have a plus PV1 on the right hand side which can be taken to the left hand side becomes you now try to group it with E1 with a negative sign that is equal to Q P so from here you can see that this is actually H2 minus H1 equal to Q P that is equal to the ? H so this is this is actually for any heat of formation or any I am sorry any heat of reaction okay so for whatever you mean by heat of reaction as defined with respect to having a constant pressure and temperature at which you start out and coming back to the same temperature and pressure in the end will basically give rise to this because we now keep the pressure constant right and you get back to this thing so effectively what it means is the heat that you are looking at is essentially an enthalpy right so it is what you are or an enthalpy change in the in the reaction in the case of heats of formation you will find that you are looking at reference elements to start with okay which are assigned a zero heat of formation therefore you can directly say the heat of formation of a what should I say of any any species is essentially its enthalpy okay because it is not really an enthalpy change anymore because it is changing from zero for the reference element all right therefore you can simply say that all heats are enthalpies right now in the case of a constant volume process of course the case of a constant volume process we will have e2-e1 equal to Qe which means we do not keep the pressure fixed the pressure will vary but the volume will be fixed okay so I will leave it to you to actually find out whether Qe will always be greater than Qp or not okay that is something that you can try to find out for yourself and you will also look at some puzzles in the end between Qe and Qp that I will try to point out if you get some time but let us just continue at the moment okay and the other point I want to make is constant typically most experiences or with constant pressure right example or let us say gas turbine rockets or stars etc. Farley different ones like gas turbine and rocket may be like aerospace but stars is not necessarily aerospace furnaces also all these things you typically have a constant pressure situation in things like piston engines the pressure keeps the pressure keeps varying the volume keeps changing okay so it needs a constant pressure not constant volume but let us not bother about it at the moment for many many applications constant pressure combustion is a good approximation okay the other thing is so standard tables now standard tables such as what is called as the JANF JANF thermochemical tables catalog or give simply can say simply simply give ? Hf0 298 for many species in fact this tells you why we need to actually have a standard heat deformation okay so it is possible now if you standardize the conditions at which you want to catalog or tabulate the heat deformation then it is possible for you to actually have tables like the JANF thermochemical tables which will give this for many many species okay so it is available so if you now look at the heat that is released Qp as essentially a ? H which is nothing but H2- H1 right that is what we had here just coming back to that after these notes so we now have this can be written as then ? i equals 1 to n ni double prime let us say Hf Hfmi at P2, T2-sigma i equals 1 to n ni single prime Hfmi at P1, T1 which is to say you could now consider any process where we started out with certain pressure P1 and T1 does not have to be standard okay and we now go through the process of the reaction to get completed and you know you now reach a different state P2 and T2 which has got nothing to do with the standard all right if there is a case then we note that we are not using the standard heat of formation here okay we are saying whatever is the heat of formation of all the species that are involved in the reaction taking care that if you if ni double prime were 0 that means it is not really a product okay I am sorry if ni double prime where it is correct is a if ni double prime where 0 then it is not a product if ni single prime where 0 then it is not a reactant so that will take care of that so we essentially can sum over all the heats of formation not the standard heat of formation but heats of formation at certain pressure and temperature that is for the products minus the heats of formation for the reactants so effectively this is like for the products and this is for the reactants I have changed notation a little bit here for you and I am going to explain this further on we are now beginning to use small hedge okay because particularly because this is this is associated with per unit mole and you multiply by the number of moles okay so in a in a in a in some sense you can now call this as a molar specific enthalpy that means you are talking about it as a per unit mole okay of course we have been talking about having this as per mole as well okay but this is the standard heat that we are looking at this is not the standard heat of formation yet I do not know if this makes sense right away if I now write it like this I am actually trying to do something a little bit circuitous in the sense if I have a reaction that is going on let us suppose that I have methane plus oxygen gives carbon dioxide and water let us suppose that that is the reaction that we are looking at now we are looking at here the formation heat of formation of carbon dioxide and heat of formation of water here we are looking at heat of formation of methane and heat of formation of oxygen okay. Now we know that the standard heat of formation of oxygen is 0 and the heat of the formation reaction for methane for example would be like carbon at solid form plus hydrogen gives you to hydrogen gives CH4 that is a reaction that we have to go through for the heats of the formation reactions for carbon dioxide and water or like C plus O2 gives CO2 and H2 plus half O2 gives H2 and so on those are not the reactions that we are finally looking at right. So we are finally looking at CH4 plus O2 gives CO2 plus H2O so why did we actually have to get into formation enthalpies so essentially what means is we are beginning a hypothetical process in our minds where you are writing this and I want you to sort of get to understand that which is let us suppose that you started out with some reactants in this case in this example it is hydrogen and oxygen or the other example that I mentioned as methane and oxygen what it means is and then they are at some pressure and temperature okay. So what we are really basically saying when you are including the heats of formation is we are now going through unbonding their bonds okay so CH4 we now want to actually impart some heat to the system to form C and H2 which are the formation constituents there is a reference elements O2 is a reference element already okay as far as see and then we now have the reference elements we now regroup the reference elements so we now have C H2 and O2 these are the reference elements with which you could now form your carbon dioxide and water. So it is sort of like you got the reactants you now unpack or just debond them and then get back the reference elements that means you have exchanged some heat in this process keep that as an account okay and then get the reference elements to form the products that you are looking for alright and you will now get some heat out so minus the heat that is supplied is the net heat that you are going to get so you are going through a hypothetical reaction in your mind through a reverse formation reaction for the reactants to form the reference elements and then a formation a set of formation reactions for the products to get some heat transaction in both these steps okay and then you are looking at the net heat transfer to get you this keep Qp this is as it is beginning to get you into some hypothetical reactions that are not happening at all okay okay. The next step for us is so here our hf small hfmi as I said was specific molar enthalpy enthalpy at any P and T in this case it is P1 and T1 in this case it is P2 and T2 okay so here we can now write we now write hfmi at it at a certain temperature and note that I am basically writing it only as a function of temperature why because we are assuming ideal gases okay and because we are assuming ideal gas which is thermally perfect we expect that the enthalpy the heat is now shown to be an enthalpy and the enthalpy should be only a function of temperature okay we can now write this as delta hf0 298 of the species plus a hfmi at T- hf0 298 mi if you do not like 298 you can replace this by T ref any reference temperature that you want but you should look for standard tables that give you these things at that temperature that you are looking at it is sort of like in this step it is not exactly algebraically adding and subtracting things you would find that this is showing up in these two places so it is sort of like am I adding something and subtracting something in some sense yes but there is a physical meaning at those associated with this what you are then saying is this is the standard heat of formation right and then this is what is called as the sensible enthalpy let me write it in capital letters sensible right that is what is called as a sensible enthalpy so what is going on why are we talking about something called a sensible enthalpy is that because that is the one that makes sense whereas the other one is nonsense any ideas yes yeah yeah yeah so so algebraically this is not going to make sense so do not worry about it okay so what you are essentially doing is we are splitting this into two parts one is what is called as the standard heat of formation that we have been talking about all the time okay and then the next one is what we call as a sensible enthalpy so I am going to explain to you what these two mean so here we are writing it in this way primarily from a notation point of view okay so you do not go here and start cancelling things from left hand side aha here yeah no then it can be this and this this and this you could you could you could say that you could say that okay you could say that because as I told you this H is not necessarily I mean this is supposed to be like a change okay that is why we have the delta all right but it is changed from 0 okay so this is like enthalpy if heat of formation of H minus heat of formation if you can apply this this equation this equation to the formation react formation reaction itself what will happen is this is this is the heat of formation of H minus heat of formation of H2 which is 0 right therefore the change is nothing but the value itself because you are basically taking a value and subtracting 0 from there okay so you can write it this is a good question because it also firms up the notation that we are using so the change as well as the value are pretty much the same okay because if you take the value and subtract 0 from this for the reference elements heat of formation you get the change so if this is essentially the same okay but what is more important for us from now on is not how we split this it is about what they mean okay so you are now looking at this particular thing as one term called the sensible enthalpy the other one is called the standard heat of formation which we are now familiar with the question is why well why what is meant by the sensible enthalpy right so what is sensible about this enthalpy further why further well this any increase in enthalpy that the sensible enthalpy shows is solely because of a temperature rise okay all other words when you now talk about enthalpy of a species we can now say this species now has a basic enthalpy associated with it at standard conditions okay you can take the standard at whatever level you want okay your P and standard pressure and temperature can be whatever it is and then you say the standard heat of formation at this pressure and temperature is this this arbitrary the assignment of the pressure the standard pressure and temperature at which the standard heat of formation is measured that choice is somewhat arbitrary but at that conditions of those conditions you now have a standard heat of formation that refers to the energy that is contained in the bonds by itself for simply for it to simply remain in that particular molecular combination of atoms at that particular standard temperature and pressure it needs some energy okay that is what this is any further change in the enthalpy of the species from there on is purely by a change in temperature okay so that means you now stick in a thermometer or a thermocouple or anything and then you measure the temperature you should be able to tell what the sensible enthalpy is so in other words you this will say this is a part of the enthalpy that can be sensed by a thermometer or a thermocouple so since it is sensible by a temperature measuring instrument we call this the sensible enthalpy so many times when you pretty much recklessly write that the enthalpy is directly proportional to temperature we are always talking about the sensible enthalpy we are not talking about the heat of formation but the heat of formation is hidden okay so when you are talking about non-reacting fluid mechanics where you are not bothered about chemical reactions and so on you are simply saying you know H is ? H is equal to Cp ? T that is because the change in enthalpy is solely coming out of only the change in temperature and therefore you are essentially interested only in the sensible enthalpy and you do not even bother about the formation enthalpy because the formation enthalpy is there and the species is now going to change you are not looking at any chemical reactions that will change the species but here we are burdened with the situation where the species is going to change it is going to eat it is going to get the if it is a reactant it is going to get deeply that if it is a product is going to get produced right therefore we have to now begin to pay attention to the energy of the species itself because it just exists at a standard temperature and pressure and that is by virtue of the bonds and we arbitrarily assign the bonds that are present in like these diaromic molecules like H2 and so on to have 0 energy that is like a datum relative to which a atomic species that does not have any bonds actually accrues some energy for it to be remaining in atomic state because it is not naturally existing therefore some energy must have been given to get it get it to that state you see so that is so even a lack of bond in that sense actually has some energy associated with it because of the reference element that we take to be what it is so this is a very very important concept here that we are looking at where this could now be cataloged in these tables and then we primarily look at only the change in temperature for the rest of the enthalpy change for any species right so we we can we can express the sensible enthalpy sensible enthalpy as Hfmi of T-Hf not let us say T-Rafmi right equal to integral T-Raf to T Cpmi of Tdt that comes from the definition of Cp so Cp is a partial derivative of enthalpy with respect to temperature and therefore if you want to get the enthalpy back then so again you now think about this like a constant of integration if you are not trying to integrate Cp with respect to temperature for a calorically perfect gas we can write this further as Hfmi of T-Hf not Trefmi as Cpmi T-Tref now I want you to note that for the first time we are recognizing that each and every species may have different Cp than the other okay so Cp is actually species specific keep this in mind it is going to now haunt you through the rest of your lives if you are doing combustion for the rest of your lives okay usually we are saying Tref is 298.15 K so usually Tref 298.15 K if you want two decimal places accuracy on the temperature there so therefore heat of a reaction this is any reaction that we are now looking at let us for example let us say methane oxidation to carbon dioxide and water you can get Cp equals delta H equals H2-H1 equal to sigma I equals 1 to n ni double prime open your curly brackets here and then write delta Hf not Trefmi plus Hf T2-Trefmi minus Hf not Tref right take that back a little bit so you could write combinedly for this for each species Mi minus sigma I equals 1 to n ni single prime delta Hf not Trefmi plus Hf T1 minus Hf not Tref for species Mi right so you can see that T2 and T1 are showing up explicitly when we had these things parenthetically there okay but what about P1 and P2 they do not show up you are looking at ideal gases where the enthalpies are only functions of temperature therefore the sensible enthalpy depends on the temperature all right so you can explicitly see T2 and T1 show up here okay but what about pressures do we have to bother about them or do you do not we have to bother about them keep bothering about whether we want to bother about them or not okay so we will just proceed to what we want to do which is look at the adiabatic flame temperature we are now ready to look at the adiabatic flame temperature so what do we what do we understand by adiabatic flame temperature why is it called adiabatic we still do not know why it is called flame temperature or do we so in the first class we said the flame is nothing but a reaction zone and we are basically looking at any reaction which is happening to start with that P1 and T1 and ending at P2 and T2 and we know how to write an expression for the heat released in this reaction given the heats of formation the standard heats of formation of the species and the sensible enthalpies of the species which are supposed to be tabulated so that the tables will actually come contain a left hand column of temperature starting from let us say 298.16 and then keep on going up to fairly high temperatures maybe 4000 Kelvin 2000 3000 Kelvin or something like that okay and then for every row of temperature you will now get the sensible enthalpy for that species right so you have the sensible enthalpy as a function of temperature or if you do not want to deal with sensible enthalpy you can deal with CP's and the way you want to deal with CP as a function of temperature you can fit in polynomials and then do the integration or if you want to assume a calorically perfect gas you can now consider a CP that is constant over the range of temperature that is in that we are interested in right so you can do all these things so if you are given the ni double primes and the ni single primes okay and the heat of formation the standard heat of formation and the sensible enthalpies you could you could in principle find out the Qp but what see situation here what are we looking for what are we trying to find out we are trying to find out the adiabatic flame temperature okay that means we do not know the temperature at which the products will be when the reactants have burned that you get the products okay and we want to actually find out what will be the temperature at the time by the products when they have reacted from the reactants without losing heat to the surroundings at all therefore they get the maximum temperature possible because if you lose some heat to the surroundings the products will attain less temperature okay so not losing heat to the surroundings is what we mean by adiabatic so if you now have a adiabatic situation then you get the maximum temperature that is in principle possible for the products to attain from the reactants yeah so what would we want to do in this equation what do we what do we know and what do we do know yeah previously given T2 and T1 and P1 and P2 and N1 double prime and N2 double prime and you get the sensible enthalpy from the tables and the heats of formation from the tables you could find Qp but here what is given and what is not given and I double prime P2 if you assume constant pressure P2 is known and T2 is unknown you do not know what else is given sorry that's fine that that's been given even before so we are a fine T2 is a question right so previously we were given T2 and we could find Qp here what do we do you have to find T2 you have to find T2 so what do you what about Qp Qp is taken a 0 so starting point for us is to take Qp is equal to 0 so adiabatic simply means Qp equal to 0 right that means H2 is equal to H1 right and when you say H2 equal to H1 the implicit assumption is constant pressure okay so if somebody does not mention constant pressure and talks about adiabatic flame temperature it is implicitly assumed that you are basically in hearing about the constant pressure process okay so this refers to this unless you want to explicitly talk about constant volume adiabatic flame temperature then you have to say so okay because without that we are simply going to be equating H2 to H1 if you want to look at constant volume adiabatic flame temperature what do we want to see we want to say Qv is equal to 0 that means we have to be equating e2 to e1 alright here we are looking at Qp is equal to 0 therefore H2 is equal to H1 implicitly assuming constant pressure so adiabatic flame temperature assumes a constant enthalpy and constant pressure a constant enthalpy given by H2 is equal to H1 and the definitions of heats essentially having constant pressure in built in them so what this means is sigma i equals 1 to n ni double prime open up the equally brackets and then say delta Hf0 t ref plus Hf0 at t add okay and Hf0 t ref all for species mi equal to sigma i equals 1 to n delta Hf the right hand side remains the same as before t1 Hf I am sorry yeah okay that is fine t ref mi okay now for a calorically perfect gas of course we can assume Cp as a constant and then again explicitly so now what happens is t add is showing up as what we want it is actually a function of the sensible enthalpy and so what you have to look for is you now know you can find out what the right hand side is okay the right hand side is unknown you know and n y ni single prime and you know your t1 you can calculate these so what you have to look for is if you were if you knew your ni double prime okay ni double prime refers to the composition of the products given the composition of the reactants and we will talk about that soon okay if you are given an i double prime you have to find out so you know you now have to actually look at one particular temperature for which you could you could now look at the tables for the different species okay at that temperature and find out what this is and then see if it matches with the RHS and then the temperature that you chose for looking at the tables for each and every species in the sensible enthalpy is right if not you start with a different temperature so so again you can do this a bit iteratively okay or if you now assume a Cp as a function of temperature you will now get polynomial in temperature and you have to solve the solve this polynomial so you will now get a temperature explicitly showing up as a polynomial which has to be a solve for to get to get this temperature if you want to get rid of all these hassles and hopefully that is what you will face in things like exams because you can do this very quickly you assume a calorically perfect guess which means we can simply say Cp mi of t add- t reference and here you are going to have Cp mi t1- t difference so t t ribatic is actually showing up linearly as opposed to a polynomial okay and then you can just rearrange and get your temperature all right okay so that that is easy for a calorically perfect guess but not otherwise so given ni single prime and ni double prime one can find one can find t t ribatic from the above equation right okay. So sensible enthalpy is also tabulated as I said tabulated in Jan of tables as a function of temperature these are now available on the web for typical species that you would be interested in and so on okay now I want to talk about two things here okay and these are the two most important things as far as this is concerned so we now say this is equal to here because Qp is equal to 0 previously we had a negative sign and then the whole thing would be equal to Qp but now we have set Qp is equal to 0 split this apart so this refers to products this refers to reactants yeah so we have to keep in mind we are thinking about combustion and combustion is about burning fields okay and we want to identify but then the reaction that we took was something very algebraically general okay something like sigma i equals 1 to N ni single prime script mi gives sigma i equals 1 to N N i double prime script mi what was the fuel there how is it getting oxidized how do you know which is the fuel how do you know which is the oxidizer so we have this question okay so we wanted to be able to tell a fuel when you see it how do you do this what are we looking for when we burn what happens when when you burn a fuel it releases heat so how do you feel it it is pretty hot okay so we are basically looking for a fairly high value of theory abaric huh what are the typical values of theory abaric any idea or figure about theory abaric we do not even know whether you are having heat loss to the surroundings are not in the flames that you normally encounter okay but can you guess whatever is the temperature that in that you that you that you have in the flames that you encounter let us just take an example of the flame that you encounter example let us let us pick up pick a flame that you encounter any anything I do not even know what the premix flame is this is a second class or third class now I want to talk about something that I encounter candle flame so you now burn a candle okay what do you think is the temperature there but the peak temperature anywhere inside the candle flame ballpark figure are we looking at a few degrees Celsius or tens of degrees of Celsius hundreds of degrees of Celsius thousands of degrees of Celsius or 10,000 I am going on logarithmic scale what are we talking about right a few degrees Celsius is just this barely more than ice obviously it is not it is not that cold okay a few tens of degrees is big basically just barely more than body temperature or around body temperature but when I touch it it is pretty hot so of course I mean I cannot touch anything more than about 70 80 degrees I am going to get one okay the human body is so fragile you cannot dip your finger into boiling water which is 100 degrees C so we are not even talking about getting touching the flame that means I guess the temperature is now grading greater than 100 degree C right so is it hundreds or even worse thousands thousands so you light up a candle and you can get a temporary you do not even have to light up a candle if you are to light up a candle you have to strike a mat stick you strike a mat stick okay and you get a little fire there that flame is actually going to have a temperature that is of the order of 2000 Kelvin 2000 to 3000 it is so hot okay but if you are referring to plasma physics and so on it is only only so hot depends on what kind of person you are okay but we are talking about temperatures that are of the order of 2000 2000 C 2000 K does not matter when you are in 2000 whether to see okay okay is a matter of the same order okay so that is what we are looking at if you want to get that higher temperature here that means the sensible enthalpy has to be high okay if your sensible enthalpy has to be high the heat of formation here has to be low that is the reason why we said is stoichiometric mixture is the one where the products will have the highest negative heats of formation so now we are getting back to why we were talking about that okay so we want to have products that have the highest negative values here so that this will be high to compensate and get you the total and then we have to look at the total the total will be given by the right hand side okay so the right hand side is already a given you take a particular mixture of certain composition of methane and oxygen let us say okay the right hand side is fixed at a particular pressure and temperature you take everything okay temperature pressure composition these the species everything this right hand side is fixed then how do I know that I am actually working with the fuel and not some garbage that means how do I know that methane is actually going to work like a fuel that means I want to have a high total to begin with that this gives you the right hand side should be a fairly high total and keep in mind T1 is not going to be very high for you you do not want to actually heat up methane to 2000 Kelvin and then burn to get a carbon dioxide and water at 2000 Kelvin does not make sense so what is our methane at room temperature okay so room temperature is like okay 298 Kelvin let us say okay if you are not talking about a fairly low sensible enthalpy how is it possible for us to now have a very high total of the sensible enthalpy and the heat of formation together only if you have very high heat of formation for the reactant okay now if you have methane and oxygen oxygen is not going to contribute to your standard heat of formation at all because it is a reference element and then therefore we take it as 0 okay so the only way you are going to make sure that your T adiabatic is high is if you work with a particular species which gave you a very high standard heat of formation so you now look at a species and find that it has a high high standard heat of formation you know it is a fuel you see that is the way the catch you get that and then you can burn and you can get very high temperatures and that is the equation that tells you how to do this right so this pretty much contains like a lot of information about combustion which is very very interesting okay the second question that I have is I can give you ni single prime all right do I know ni double prime what is that a piece if I took a certain composition of reactants how do I know what would be the composition of the products is it straightforward see you Thursday