 In this segment, what we'll do will conclude the example problem that we're dealing with where we have octane gas is being combusted with excess air and we're told that there is 60% relative humidity and 30% excess air. So we've gone through and we've come up with our final reaction equation. We did that by doing frist stoichiometric balance. Then we added in the excess air and then we counted for the moisture, the 60% relative humidity coming up with an expression for the number of moles of water vapor that is being carried through. So what we're now going to do, we're now going to apply the first law and this will be first law for a steady flow system with chemical reactions or combustion in this particular case. So let's write out our first law. So we have our products and our reactants. Now the first thing that we can do, we can neglect work because it's a steady flow system. There are no moving boundaries or anything like that. So work disappears. The next thing we have to do when you're solving these problems, you kind of got to be careful. There are different places you can make mistakes. But what we're going to do, we're going to look up the enthalpies from tables in the back of your book. And in doing this, we need to look up the values for the different temperatures that we have within our system. And we were told that things come in at 25 degrees C and they leave at 600 Kelvin. So to begin within this table, what I'm going to do is write out the substance that we're dealing with. The next thing that we will have is the heat or enthalpy of formation. Then we will have enthalpy at the base state 298 Kelvin, enthalpy at the temperature of the products, which was 600 Kelvin. And we will pull those out of the tables in the back of the book. So let's begin. What we have is octane gas or C8H18. And that is in the gaseous state. And when we look up the tables in the back of the book, what I'll do is I'll just write the numbers out and then we'll use them in the next part. Now for this one, we don't need to do anything because this is a reactant coming in. It's not a product. So it's coming in at 25 degrees C and we can neglect the 600 Kelvin term. But the other things that we have, oxygen, nitrogen, we have water vapor in the gaseous states, that's H2O, then we have CO2. So I'm just now going to write in the values. So those are all of the different values that we'll be using to solve this. What we're going to do is take these values and we will plug them up into the first law that we have coupled with knowing information about the reaction. Remember, this is our final reaction equation. So we know the number of kilomoles for all of the different substances that we have within our reaction. There is our table. So what we will now do is proceed with what I call bookkeeping. And it's just basically straightforward applying the first law and plugging in all the different values. So I will write that out now. So that's the bookkeeping that we have. And if you look back at our first law, what we're doing, we're starting with the products and then we go on to the reactants. So the products in this equation are here, here, here, and here. And then when we're subtracting, those are the reactants. And you'll notice the reactants were 25 degrees C. So in certain cases where we have no enthalpy of formation, for example, when it's 16.25, if we go back to our equation, our reaction 16.25 was it's either oxygen or nitrogen. And for there, what we have is zero for being the enthalpy of formation because either oxygen or nitrogen, the value would be zero on the table. As well, it's at 25 degrees C. So the 86-82 is just cancelling with itself. But I carried it out just to be rigorous and let you know how all these numbers turn out. But when we do this and we evaluate Q, what we get for the heat transfer, we get that for the number of kilojoules per kilomole. Now, they wanted us to evaluate this per unit mass. So in order to do that, we need to calculate the molar mass of C8H18 or octane. And so that's the final answer that we obtain. And you can see it's negative implying that the heat is leaving our system, which we would expect it to be for an exothermic reaction like combustion of octane gas. So that concludes the example problem of applying the first law to a system with reaction. You can see it's kind of long and laborious. A lot of places to make little errors. You got to be careful. But what I would recommend is going through the procedure that I did by coming up, first of all, starting with the stoichiometric balance, getting your reaction equation, and then writing out that table with all the values and then proceeding to apply the first law. And by doing that, you minimize your possibility of making small errors or mistakes while you're solving the problem. So that concludes this example problem. In the last lecture that we will look at, I will be covering the topic of the adiabatic flame temperature. So thank you for your time and we'll move on to adiabatic flame temperature next time. Thank you. Bye bye.