 Okay, this section marks an area that a lot of students often dislike, so I'm telling you that just to manage expectations. Some students think this is a little bit rough. We are going to talk about a topic called specific heat. It's a relatively new concept. The way that I introduce it is to ask a couple of thought experiment questions. So here's the thought experiment. Imagine that I have one gram of water. So that's not much. That's about a paperclips worth of water. So think of this as a tiny little container with one gram of water. And the temperature of the water is currently 20 degrees Celsius. So that's about room temperature, a little less than room temperature. And I want to heat up my paperclips worth of water, and I want to heat it up by one degree Celsius. In other words, I want to warm it up, let's say, and the way that I can warm it up is I have access to a bunch of candles. All of the candles are identical to each other. I get them from some factory in California, let's say. And it turns out that I need to burn three candles. Here are my three candles burning completely. To raise the temperature of my one paperclips worth of water by one degree Celsius. So let's just pretend that that's true. Here's the first question. Imagine that I have two grams of water, so twice as much two paperclips worth, so this is two grams. The temperature of the water is initially at 20 degrees Celsius, just like it was over here. And I want to raise the temperature of my water to 21 degrees Celsius as well. The question is, how many candles do I need to burn to do this? You can pause for a moment if you want, and unpause at this point. And if you said that it's going to take six candles, you're correct. The reason it takes six candles is because there's twice as much water. You're raising the temperature by the same amount, so it stands to reason that you're going to need twice as many candles to do the job. This is not specific heat yet, we'll be getting there, but this is to get you in the right frame of mind. Here's the second thought experiment. Imagine that I have a piece of metal, so this is a piece of metal. It also weighs one gram. It's also at 20 degrees Celsius, and I also want to raise the temperature of my little block of metal by one degree Celsius. I want to change it to 21 degrees. This time it turns out that I can use the same candles, but what I'm going to tell you in the setup is that I only need to burn one candle to raise the temperature of my one gram of metal by one degree Celsius. If I have two grams of my metal, in other words, two paper clips worth of my metal, and the metal is at 20 degrees Celsius, and I want to raise the temperature again to 21 degrees Celsius, how many candles am I going to need to burn to do this? Again you can pause. The answer, hopefully, is what you said, is two candles, because we have twice as much of the metal, and it stands to reason that it's going to take twice as many candles to do the job, and so there are a couple of punch lines here. If you understood what we just went over. The first punch line is this one that I'm circling off to the right. Some materials absorb heat easily, and others not so easily. So in my examples off to the left, you can ask yourself which one, which material was easier to heat up? Was the water easier to heat up, or was the metal easier to heat up? Again you can pause and try and answer this question. The answer is going to be the metal, in this case, was easier to heat than the water. Because it only took one candle's worth of energy or heat to raise the temperature of the metal by one degree. It took three times as much energy or heat to raise the temperature of the water. So this is a relatively big punch line. You might not have guessed it, but different materials absorb heat and get warmed up or cooled down more easily than other materials. So I'm going to ask this question again. This is supposed to be one gram of water. How much heat or energy? I'm going to use the terms heat and energy interchangeably. You can think of them as being roughly the same thing. How much energy does it take to raise the temperature of one gram of water by one degree Celsius? Well in my cartoon example I was saying it's three candles, whatever a candle's worth of energy is. It turns out that this is the amount of energy it takes to raise the temperature of one gram of water by one degree has a very formal definition. It's called a calorie. I know that people think about calories all the time, but it actually has a very formal definition in chemistry and the definition is basically here. It's the amount of energy it takes to raise the temperature of one gram of water by one degree Celsius and it turns out to be one calorie. So if that's true, how much energy does it take to raise the temperature of one gram of my metal by one degree Celsius? You can pause and think this through or work it out on a sheet of paper if you're unpausing. Basically what we are saying in my pretend example is that one calorie equals three candles. So if one calorie is equal to three candles, then one candle is equal to one third of a calorie and that's sort of shown here. So apparently in my pretend example it takes a third of a calorie to raise the temperature of one gram of my pretend metal by one degree Celsius. So we have now introduced, even though I didn't announce it, we've introduced the topic of specific heat. Specific heat is going to be defined on the next page. It's the amount of energy or heat that it takes to raise the temperature of one gram of something by one degree Celsius. And it's going to be, it will be different amounts of energy or different somethings. So as an example, and this is where sometimes students get confused because you're going to have to get this definition under your belt. For example, the specific heat of water is what I told you on the previous slide. It's one calorie per, that's what this line means, degree Celsius per gram of water. So, in other words, there are no numbers written in the denominator here, but there are numbers that are implied to be there. Typically when there's no number written, the number one is implied, so I'm going to put the ones in, one gram as well. So what this equation is saying with the ones written in is that if I have one gram of water, that's the formula for water, and I want to raise the temperature by one degree Celsius, it's going to cost me one calorie. So whenever a specific heat value is written, it will be some number, I'm just going to call that x, let's say calories, and it'll usually be written per one degree Celsius per one gram. Sometimes people will leave out the ones, but they're implied to be there. And so it's this x, whatever number goes in this place, in the place of x will change for different materials. For water, the x happens to be a one, it costs you one calorie of energy to raise the temperature of one gram of water by one degree Celsius. But for other materials, it's going to be a different number, it won't always be a one. As an example, aluminum, you're all familiar with aluminum hopefully, it has a different specific heat. It has a specific heat of 0.215 calories per one degree Celsius per one gram. Again, what this means in English is if I have one gram of aluminum, pretend I have a little piece of aluminum foil, and I want to heat up my piece of aluminum foil by one degree, it's going to cost me 0.215 calories. It doesn't cost me one like it did with water, it costs me less. 0.215 is obviously less than one calorie. So what you can say is the aluminum is actually easier to heat up than the water is because it costs less energy. And all of these numbers, the 0.215 calories per degree Celsius per gram, and the one calorie per degree Celsius per gram for water, these little fractions here, they are specific heats for different materials. This here is the specific heat for aluminum, this is the specific heat for water. You don't need to memorize these numbers, I would give them to you on a quiz or a test, and honestly they're a complete waste of time to memorize. But you should know what they mean. They are basically numerical ways of explaining to people how easy or how difficult it is to warm up or cool down a substance. Another thing that I'd like to point out is we just said a minute ago that, let's see, we just said a minute ago that the aluminum, aluminum is easier to heat up than an equal amount of water, than water. The reason that we can see numerically is because this number takes less energy to heat up the aluminum by a certain amount than it does to heat up the water by a certain amount. So aluminum is easier to heat up than water. This is true for most metals that I can think of, maybe all metals. Metals are typically much easier to heat up than water. An example of this is if you've ever boiled water in a metal pan on your stove, it takes a long time before you can burn yourself by sticking your hand in the water. In other words, as you start to heat up water in a metal pan, you can stick your hand in the water for a pretty long time before the water gets really hot. And that's because it's relatively difficult to heat up the water. However, if you touch the metal on the pan pretty soon after you turn the heat on, there's a good chance that you're going to get burned. The reason is that metals are easier to heat up than water. There's specific heats. This is a specific heat. There's specific heats are lower than the specific heat of water, which is just a fancy way of saying they're easier to heat up. And so one of the reasons why it's easier to get burned when you're heating up a metal pan of water by touching the metal is because it's easier to heat up the metal. It'll get warmed up faster than the water. There are other reasons, too, but that's the point I'm trying to make is with specific heat here. All right. So here's a specific heat problem. This is a typical type of problem that you might see on a quiz or a test. There are many variations of this, but this is an example of what you might see. How much energy or heat does it take to raise the temperature of 12 grams of aluminum from 11 degrees Celsius to 21 degrees Celsius? And I tell you that the specific heat of aluminum is written right here. So I don't want you to commit this to memory. I would tell you what the specific heat of the material is, otherwise you couldn't do the problem. So what you're basically doing is you're saying, look, one gram of aluminum, if I want to raise it by one degree Celsius, it's going to cost me this many calories. But what you have to realize is that in my problem I don't have one gram of aluminum. I have 12 grams of aluminum. So what I am going to do, the way that I solve this problem, how much heat is I make an equal fraction off to the right of my specific heat fraction. So this is my specific heat fraction. And off to the right I'm going to make a new fraction. But instead of one gram, I have 12 grams. So I'm going to put 12 grams in the same spot. I'm not raising this. I don't want to raise the temperature by one degree Celsius. I want to raise it by how much? I want to raise it by 10 degrees Celsius because I'm going from 11 to 21. So instead of writing a one degree Celsius in my new fraction, I'm going to write 10 degrees Celsius. And the question wants to know how much heat or energy. So the units could be in calories like they are in the numerator here. So in the place of 0.215 calories, I don't know how many calories I'm going to need. So I could put a question mark there to be a little bit more mathy. I'm going to put an X there. So it's going to take me, I don't know how many calories or X calories. And then because these fractions are going to be equal to each other, I can cross multiply and solve for X. So what I can do is I can do this part of the cross multiplication. All of those numbers get multiplied together. 0.215 calories times 10 degrees Celsius times 12 grams is equal to the other part of the cross. All of these numbers multiplied together, which is 1 degree Celsius times 1 gram times X calories. And we want to solve for X. We want to just find out what X is. So we have to get rid of this 1 degree Celsius and we have to get rid of this 1 gram. The way that you can do that is you can divide both sides by 1 degree Celsius. So I'm dividing the left side also by 1 degree Celsius. So here the 1 degree Celsius is canceled. On the left side, the degree Celsius is canceled. I also want to get rid of the 1 gram. I'm not sure if you can hear that but someone has a re-arranging chairs in the room above me. So that's driving me a little crazy. If I want to get rid of the 1 gram I also divide the right side by 1 gram and I have to do the same thing to the other side to keep them equal. So I divide the left side by 1 gram as well. On the right side, the 1 grams cancel. On the left side, the grams cancel. And so X calories is equal to 0.215 calories times 10 times 12. I'm not sure what that is. Turns out to be about 25 0.8 calories is equal to X calories. So what we have done is we basically said look it takes this many calories to heat up 1 gram of aluminum by 1 degree Celsius. But I don't have 1 gram. I have 12 times as much. And I'm not raising it just by 1 degree Celsius. I'm raising it by 10 times as much of a temperature by 10 degrees. And if that's true, if I have 12 times as much aluminum and 10 times as much of a temperature raise, then how many calories is it going to take? We do this calculation and it turns out that it's going to cost us 25.8 calories worth of heat or energy. Don't worry about significant digits and rounding. Just try to understand the concept here. Here's a different calculation. It's almost identical to the one on the previous slide except we're dealing with water now. So how much energy does it take to raise the temperature of 12 grams of water from 11 degrees Celsius to 21 degrees Celsius? And again, don't memorize this because I give it to you in the problem. I say, hey look, if you have 1 gram of water and you want to raise the temperature by 1 degree Celsius it's going to cost you 1 calorie. And that's called the specific heat of water. But we don't have 1 gram of water. We have 12 times as much. So I'm going to make an equal fraction over here and instead of 1 gram in the denominator I'm going to write 12 grams. We're not raising the temperature by 1 degree Celsius. We're raising it by 10 times as much. We're raising it by 10 degrees Celsius. So I'm going to write a 10 degree Celsius in the denominator and up here I'm going to write x calories because we don't know how much energy it's going to take. But we do the same calculation as before. We can cross multiply to solve for x. We can say 1 calorie times 10 degrees Celsius times 12 grams is equal to 1 degree Celsius. So we're multiplying the other part now. Times 1 gram times x calories. I should point out that the abbreviation for calorie is cal. So I'm probably going to start using that in a little bit. But, okay, here we have an equation to things equal to each other. We want to figure out what x is to get rid of the 1 degree Celsius and the 1 gram. We're going to divide both sides by 1 degree Celsius and also by 1 gram. And if we do that, the 1 degree Celsius is cancelled out on the right. The degree Celsius cancels out on the left. The 1 gram cancels out on the right. The grams cancel out on the left. The only thing we have left over on the right side is x calories. And the only thing we have left on the left is calories and some multiplications. 1 times 10 times 12 calories. This turns out to be 120 calories is equal to x. So it's going to cost us 120 calories worth of energy to raise the temperature of 12 grams of water by 10 degrees Celsius. And let's compare that to the previous energy. The previous energy for aluminum the amount of energy it took to heat up an equal amount of aluminum by 10 degrees was 25.8 calories. So again, you can see that if I have equal weights of aluminum and water, and I want to raise them by the same temperature, it's going to be easier to raise the temperature of the aluminum because it only costs me 25.8 calories. It's going to be more difficult to raise the temperature of the water. It's going to cost me 120 calories. So this is the introduction to specific heat. Part of why this is sometimes difficult for students is to just sort of get the definition or what specific heat means under your belt and then you're also going to have to do some kind of calculation like this. So that's it for specific heat. Thank you very much.