Okay, let's look at this problem. We've got some calcium oxide that we're placing in water. And this reaction is taking place, tells us how much calcium oxide we're putting into a certain amount of water. And we see that the temperature is rising. As soon as we see that, we know that this is an exothermic reaction. And we can keep in mind that the delta H when we're all said and done should be negative in sign. 'Kay? So what are we trying to determine? We're trying to determine the delta H, so we're trying to figure this value out for the reaction as balanced. Calorimetry has a starting point of the q of the system, is equal to negative the q surroundings. We're going to keep in mind that the system is the reaction that we are interested in. And my surroundings is made up of two things. The calorimeter, so I want to get the q of the calorimeter, and the water. Now, they tell me up here, kind of messy, but you get q, calorimeter. They tell me to assume that this that the heat capacity of the calorimeter is zero. So it's negligible. It's really not going to be zero, it will absorb some heat, but it's such a small amount of heat that we won't have to worry about it. So we know that if we broke this a little further out, it would be the heat capacity of the calorimeter times the change in temperature, plus the mass of the water, or the solution I should say, because it's really not separated out, specific heat of the solution in the change in temperature of the solution. Now, this is the one that we're assuming is equal to zero. So if that is true, then we have. We can get rid of this whole term and we would have the heat of the reaction is equal to minus the mass of the solution, the specific heat, that's a solution. And the change in temperature, which is the final minus the initial. So let's plug in what we know. Well, the solution is not just the water. It consists of two things, the water and the dissolved reactant, which turns into products actually when it's all said and done, but we could add those two things together. So the mass would be 100.876. Now that's carrying extra significant figures. I really only know it to this place when you're adding, but I might as well carry them all. The units for that will be grams. It says to assume that it's the same as that of pure water for the specific heat. And we can look that up but I'm telling you it's 4.184 joules per gram degrees Celsius. And then final, which is 28.07 minus the initial 25.00. Okay, so this is going to give me the q of the reaction. And the value for this is a negative 1296. And that will have units, everything will cancel except, for the joules. And this is for the quantity that you reacted. Okay, this is for 0.876. But, we want to know what for one mole. So this negative 1296, is the energy 40.876 grams of calcium oxide? I don't know, want to know it for grams of calcium oxide, I want to know it for a mole of calcium oxide. So what I will need is a molar mass of calcium oxide. That is 56.079 grams in a mole. And so when we divide these values out, I would get a negative 8.29 times 10 the fifth joules. And that would be joules per mole of calcium oxide. Let's convert that to kilojoules. That would be 800 in 29 kilojoules per mole. Now as we look at the balanced equation we see it is for one mole of this, so we have now determined that the Delta H of this reaction. The Delta H of the reaction, as it determined by the calorimeter is a negative 829 kilojules. Now let me just fix something down here, I lost a minus sign, so let's make sure that minus sign is right there. Okay? So, typical calorimetry. You start with the reaction is equaling to the heat absorbed or given off by the surroundings. And, so that's why the minus sign, whatever one does the other one does the opposite in sign. We expanded out, we plug in the information that we know. And this is always, when we determine in a calorimeter for the quantities you reacted, if you're just trying to determine it for the overall balanced reaction, you have this final process to go from the joules for the little amount to how much is for a whole mole.
