 Okay, so let's try solving some problems using our equation. Heat capacity equals energy over mass times the temperature change. So let's take this first problem. We're going to calculate the heat capacity of water given that it takes 37,674 joules to raise the temperature of 0.250 kilograms of water by 36 degrees Celsius. So the first thing to do in any problem like this is to write down what you know. So we know the energy, that's Q. We also know the mass of the water. That's given to us in kilograms. Now it's perfectly possible to calculate heat capacity in terms of kilograms rather than grams as long as you show that in the units of the value of the heat capacity. But most of the values that we'll be dealing with have it in grams. So I'm going to change this to grams. So 0.250 kilograms, but comes 250 grams. I'm putting a dot after the 250 to indicate that that's three significant figures, not two. Alright, we also know that the change in temperature is 36 degrees Celsius. Note that a change in temperature, it doesn't matter whether it's degrees Celsius or degrees Kelvin. The size of a degree Celsius is the same as the size of a degree Kelvin. So when you take the difference between two values, the two units end up being exactly the same. Alright, so now it's simply a matter of plugging it into the equation. We've got that the heat capacity equals the energy over the mass times the change in temperature. And if you plug that into your calculator, you will get 4.186 and the units, our energy was Joules, mass was grams and temperature is degrees Celsius. So our units would be Joules, gram, degrees Celsius. Now there's something wrong with this. Spot what it is. It's significant figures. If we go back to our original values, 37,674 Joules has five significant figures. Our mass has three significant figures, but our temperature has only two. That means that our final value must have two as well. So we're going to round that to 4.2. Okay, let's do the second one. We've got the lead. We're given the heat capacity of the lead. So obviously we're not going to be calculating heat capacity. We're going to be calculating something else. And then we're asked how much energy, that's Q, how much energy in Joules is needed to raise the temperature of 62 grams of lead from 20 to 65 degrees Celsius, which is where it's just too hot to touch. So let's write down what we know. We have the heat capacity of lead. Now that's another way of writing the units. It means exactly the same thing. Joules per gram per Kelvin. That second way of writing it there is probably more correct. We also know the mass of the lead, 62 grams. And we can work out the temperature change. We've got delta T. Our final temperature is 65.0. Our initial temperature was 20.0. So our change in temperature is 45.0 degrees Celsius. Don't lose that 0.0. That's a significant figure. Okay, now we're calculating energy. That's Q. So we're going to rearrange the equation. I'm going to show you how it ends up being rearranged. But I'd like to make sure that you can rearrange it for yourself. So once you've rearranged it, you should end up with Q equals Cp times the mass times delta T. And if we plug that in, we get 0.13 times 62 times 45.0. And if you put that into your calculator, you should end up with 362.7 Joules. We check our sig figs again. We find that our lowest number of sig figs is both the heat capacity and the mass at 2. And so our final answer rounds off to 360 Joules. Final one. If you haven't been doing this already, stop the video now and try and solve this one for yourself before you watch the answer. This is in lieu of a task at the end of the video. When you've done it, come back, watch the solution and see how it went. Okay, so once again, we're going to write down what we know. We're given the heat capacity of olive oil. 1.97 Joules per gram Kelvin. We're told that a hot plate is used to put a certain amount of Joules. That means energy. So that's going to be our Q. 2.56 times 10 to the third Joules. Remember that Joules is just the unit. It's not the name of the thing. The name of the thing we're talking about here is energy. And it is measured in Joules. So a hot plate is used to put 2.56 times 10 to the 3 Joules of energy into 30 grams of olive oil. So we've got that mass as well, 3.0 grams. We're told that the initial temperature, we'll call that T i, is 25 degrees Celsius and we want to know what is its final temperature. So to work out the final temperature, we're going to need to know the change in temperature. So that's what we're figuring out. So we're going to need to rearrange that equation again. Once we'll try it for yourself, we'll make sure you have tried it for yourself, you should end up with delta T equals Q over M times the heat capacity. When we plug that in, Q is 256 times 10 to the mass because it's actually 30 grams. And our heat capacity is 1.97. When you evaluate that, you will get our 43.3 degrees Celsius to 3 significant figures. So we now know our change in temperature. Our initial temperature was 25. So our final temperature is going to be the initial temperature plus a change in temperature and by plus 1.3, which will be 68 degrees Celsius. Recall your significant figure rules for when you're adding, when you're adding or subtracting the thing that you're looking for is the number with the lowest number of decimal places and that's what tells you how to write your answer. Okay, in the next video I'll go through a couple of slightly more complicated examples using the heat capacity equation, but this shows you how to use it for the most basic calculations.