Male Speaker: Okay, so its not the most exciting lab to do in the world but you know it has one of the most important and impressive types of calculations to be able to do and here is the thing about the ice cube lab that's pretty much standard lab that we do everywhere as chemistry teachers. We put the ice cube into water and you think at yourself, it is pretty straight forward kind of a thing, we have an ice cube undergoing a phase change that's nH and we have the water undergoing a temperature change that's mc delta t, so there is the formula and lets say the question was to calculate the molar heat of melting of ice, well that would be isolate the term big H and be able to solve the question but there is actually problem with that. There is something else here in this system that has to be taken into account in the calculation, what is now I'm going to give you a hint. There is something else other than the ice that's gaining heat here. Something else other than the ice is gaining heat. Now here is where most students always say, they say oh! It's the air, it's the cup, it is the thermometer, it is the retort stand that is sitting on and think about it though. If everything here starts off at room temperature and the ice is put into the water, then the water is getting colder, then the water is getting, since the water is getting colder its absorbing heat now from the cup, from the retort stand, from the air, from the thermometer. See this is what it is. It's the water coming off the ice. You understand that that as the ice is melting it's turning into liquid water. The liquid water itself is gaining heat in this investigation, so its not just the ice cube that's gaining heat, its the water coming off the ice, it's undergoing a temperature change and you have to include that into the formula, so here is what looks like. So to tackle this calculation, heat loss equals heat gain. There is water and there is ice but remember there is also that water that's coming off the ice that we have to take into account, its undergoing a temperature change, so there is another heat gaining here and its the ice water or the water coming off the ice which is called ice water, so the temperature change of the calorimeter water is going to be mc delta T. The change in ice is nH because it's undergoing a phase change but the water coming off the ice is undergoing a temperature change, both on the heat gaining side, they are both gaining heat from calorimeter water. So now we plug the numbers in and this is what we get well first of all, we got to rewrite the formula and then plug it in and that's very important, rearrange the formula first and then you go for a plugging in all the numbers. So if you want to isolate big H, we have to subtract ice water from each side, so we get the mc delta T of the calorimeter water minus the mc delta T of the ice water and that's all divided by n, which is the number of moles of the ice and that will get you that molar heat of melting of the ice. So when we plug the numbers in, now look what I did, I just kept it in grams and joules, so our answer is going to be numbers of molar heat, we are looking for big H, so it's going to be per mole, joules per mole in this case. So 100 grams of calorimeter water times 4.19, which is that specific heat capacity of water there is the temperature change of five degrees written now properly then the mass of the ice water is the mass of the ice. Its 15 grams times 4.19, here that was ice, no the taste of water, water, water liquid, liquid so that's 4.19, that's the liquid, right that's melted off of the ice. The temperature change is from the final temperature or the final temperature isn't 25 is it, it's 20 because this goes from 25 down to 20, so this goes up from 0 to 20, which is the final temperature, so when you do this calculation here getting joules, here getting joules subtracting from each other and then dividing by the moles of the ice, which is 15 grams divide