 Alright in this video we will do a couple more heat capacity calculations but they'll be slightly longer and more complex ones. Alright so let's look at this first one. You've got 95.1 grams of water in a beaker and you measure its temperature to be 42 degrees Celsius. You take a lump of aluminium mass equals 43.5 grams and you heat it to exactly 100 degrees Celsius and you're told that water has a heat capacity of 4.19 joules per gram Kelvin. You then drop the metal, the hot metal, into the water, the 95.1 grams of water and you stir it well. The temperature of the water rises to 47.2 degrees Celsius and you have to calculate the heat capacity of the aluminium. Alright so always the way to start with these problems is to write down what you know. So we've got water and we've got aluminium in this problem, let's write down what we know about the water. Alright so we know that the mass of the water is 95.1 grams. We know that the heat capacity of the water is 4.19 joules per gram Kelvin. We know that the initial temperature of the water is 42.0 degrees Celsius and we know the final temperature of the water once the aluminium has been put into it and we've stirred it until all the heat has been transferred. The final temperature is 47.2 degrees Celsius. Okay from those last two values we can work out delta T, the change in temperature. That's just going to be 47.2 minus 42.0 which is 5.2 degrees Celsius. Okay so that's what we know about the water. Next let's write down what we know about the aluminium. We know that the mass of the aluminium is 43.5 grams. We know the initial temperature of the aluminium is 100 degrees Celsius and I've told you that it's exactly 100 degrees Celsius so from that wording you can assume that we've got three significant figures here. It's right on 100. I haven't told you it's 100.0 so we don't have four sig figs but we're going to take it that we've got three there. Its final temperature is 47.2 and it's worth here perhaps remembering that when two things come in contact and they're at a different temperature the hot one will cool down and the cold one will warm up until they're both at exactly the same temperature. So when you put the hot metal into the cool water the hot metal cools down, the cool water warms up and they both end up being at 47.2 degrees Celsius. Okay so what can we do with this? We're aiming to get the heat capacity of the aluminium. So at the moment with the aluminium we've got the mass, we've got delta T but we don't have Q the heat transferred and neither do we have the heat capacity so we can't do anything with that at the moment. However with the water we have mass, we have heat capacity and we have delta T so we can calculate the amount of heat that has been transferred to the water and the thing to remember is that the energy that's lost by the aluminium as it cools down is equal to the energy that's gained by the water as it warms up assuming that you haven't lost any heat in the meantime. Okay so let's calculate the heat that was lost by water. We've got Q equals mass times heat capacity times delta T. You can get that from rearranging the form of the equation I gave you in the last video which was C equals Q on M delta T and then we just plug the values in. We've got the mass and the heat capacity and the change in temperature and when you calculate that you will get 2,072 joules, never forget your units. So we know that the water in warming up has gained 2,072 joules. Now that's the amount of energy that must have been lost by the aluminium. So Q over here must equal 2,072 joules. So since we now know the mass of the aluminium delta T, we should work that out, delta T for the aluminium equals 100 minus 47.2 which is 52.8 degrees Celsius. So we now know the mass and delta T and Q and that means that we can calculate the heat capacity for aluminium. So the heat capacity for aluminium is Q, the heat that was lost, divided by the mass which is 43.5 grams times the temperature change which is 52.8. And when you plug that in you will get 0.91 joules per gram. Now you can write it as joules per gram degrees Celsius or you can write it as joules per gram Kelvin. We'll discuss this more in class just to be sure you're clear on this but because we're talking about a change in temperature rather than an absolute temperature it doesn't matter what units it is in. I should say it doesn't matter whether it's degrees Celsius or Kelvin. It would matter if it was in Fahrenheit but we don't use Fahrenheit. Okay now just one check. We need to check sig figs. I've given this value 0.91 to 2 sig figs. If we go back and look at all the values that were used in the calculation, the mass of the water was 3, heat capacity was 3. Delta T turned out to be 2 although the original temperatures were to 3 sig figs. When you subtract one from the other remember that the rule is lowest number of decimal places. So that left us with a delta T value of 2 decimal, sorry 2 significant figures. So that means that our final one here is going to be 2 as well. We had 3 for the mass of the aluminium, 3 for the initial temperature and final temperature, 3 for delta T for aluminium. So 2 sig figs is what we're after. And if you check that out on the web you'll find that that is in fact the heat capacity of aluminium.