 Okay, let's try another problem. Alright in this one you have 90 mls of water at 58.0 degrees Celsius and your lab partner also has a beaker of water and she measures its temperature and finds that it's 75.0 degrees Celsius. Now she adds some of her water to yours but she forgets to measure how much, she doesn't measure its volume or its mass. So you stir the mixed water thoroughly and you find that the final temperature of that mixed water is 69.0 degrees Celsius. So the question is how much water did your lab partner add? And to solve this we're going to assume that the density of water is 1 gram per mill. So for every mill of water it's 1 gram of water. Okay, so the way to approach it again is to write down what you know. So let's write down what we know about your water. We know the volume of your water is 90 mls so we're going to say that its mass is 90.0 grams. We know that its initial temperature is 58.0 degrees Celsius. We know that its final temperature, once it's been mixed with the other water its final temperature is 69. And from those two things we can work out delta T, 69 minus 58 equals 11 degrees Celsius. Don't forget that point zero, it's an important significant figure. The other thing you know about water is, and I haven't specified it here but you should be beginning to remember it by now, is that the heat capacity of water is 4.19 joules per gram Kelvin. Okay, let's write down also what we know about your partner's water. We don't know the mass of the partner's water, that's what we're working out. But we do know its initial temperature, which was 75 degrees Celsius. And we do know its final temperature, which is also 69. So we can work out delta T, 75 minus 69 is 6 degrees Celsius. And again because it's also water we do know its heat capacity, so 4.19 joules per gram Kelvin. Okay, for your partner's water we don't know the mass and we also don't know the energy transfer so we can't do anything with that. However for your water we know mass, we know delta T and we know heat capacity so we can calculate Q, the heat transferred. So we plug those values in, the mass is 90, the heat capacity is 4.19 and delta T is 11. And if you calculate that out you'll find that that is 4,148 joules. So that means we know that the heat lost by your partner's water was also 4,148 joules. Alright, and now with Q, C and delta T we can work out the mass of your partner's water that was added. So we rearrange our equation to make mass the subject and that's going to give us mass equals Q over C delta T, which equals 4,148 over the heat capacity times delta T for her water was 6.0. So plug that in and the mass that you get out is 165 grams. Let's check the sig figs situation. We've got 3 here and 3 here and 3 here. If your partner's water, we've got 2 for her temperature once you've done the subtraction and 3 here so 2 is going to be our final number of sig figs. So we have to round our 165 up to 170 grams. So she added 170 grams of water at 75 degrees Celsius to your 90 grams of water at 58 degrees Celsius and the consequence was that you ended up with 260 grams of water all at 69 degrees Celsius.