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# Specific heat, heat of fusion and vaporization example

## Video transcript

A couple of videos ago, we learned that if we started with solid water or ice at a reasonably low temperature-- maybe this temperature right here is minus 10 degrees Celsius. And we can deal with Celsius when we're dealing with these phase changes, because we really just care about the difference in temperature, and not necessarily the absolute temperature. So one degree in Celsius is the same thing is one degree in Kelvin so the differences are the same whether you're dealing with Celsius or Kelvin. So we're starting with minus 10 degree Celsius ice or solid water. And we learned that as you heat it up, as you add heat energy to the water, the temperature goes up. The molecules, at least while they're in that ice lattice network, they just start vibrating. And their average kinetic energy goes up until we get to zero degrees. Which is the melting point of water. And at zero degrees, we already learned something interesting happens. The added heat in the system does not increase the temperature of the ice anymore. At least over that little period right here. What's happening is that heat energy is being used to kind of break the lattice structure. To add potential energy to the ice. Or essentially melt it. So for it here, right here, we're ice. Right at this point, we're zero degree ice. And then as we add more and more heat we get to zero degree water. So at zero degrees, you can either have water or ice. And if you have water, to turn it into ice, you have to take heat out of it. And if you have ice, and you want to turn it into water, you have to put heat into it. And then the heat is used again to warm up the water at some rate. And then at 100 degrees, which is the boiling point of water, right here, a similar phase change happens. Where the increased heat is not used to increase the temperature of the water, it's used to put potential energy into the system. So the water molecules are forced away from each other. The same way that if I'm forced away from the planet Earth, I have potential energy because I can fall back to the earth. Similarly, they have the potential energy of falling back to each other. But this energy, right here is the energy necessary to vaporize the water. Right here you have 100 degree water, 100 degree liquid. And here you have 100 degree vapor, water vapor. And then as you add more and more heat, once again it increases the temperature. But you, Sal, I already learned this a few videos ago, I have the intuition. But I want to deal with real numbers. I want to know exactly how much heat is required for these different things to happen. And for that, we can get these numbers. And these are specific to the different states of water. If you looked up any other element or molecule, you would have different values for these numbers we're going to be dealing with right now. But this first number right here is the heat of fusion. And this is the amount of heat that's required to fuse 100 degree water into 100 degree ice. Or the amount of energy you have to take out of the water. So this distance right here, or along this axis, is 333 joules. If you're going in the the leftward direction, you have to take that much out of the system to turn into ice. If you're going in the rightward direction, you have to add that much to turn into water. So heat of fusion. It's called the heat of fusion because when you fuse something together you make it solid. So it could also be considered the heat of melting. Just two different words for the same, thing depending on what direction you going. The important thing is the number, 333. Similarly, you have the heat of vaporization. 2257 joules per gram. That's this distance along this axis right here. So if you had one gram of 100 degree liquid water and you wanted to turn it into one gram of 100 degree liquid vapor. And in all of this, we're assuming that nothing silly is happening to the pressure, that we're under constant pressure. You would have to put 2257 joules into the system. If you had 100 degree vapor and you wanted to condense it, you would have to take that much energy out of the system. OK, fine, you know how much energy is required for the phase changes. But what about these parts right here? How much energy is required to warm up a gram of ice by one degree Celsius or Kelvin? And for that we look at the specific heat. It takes 2 joules of energy to warm up 1 gram 1 degree Kelvin. When water is in the solid state. When it's in the liquid state, it takes about double that. It takes about 4 joules per gram to raise it 1 degree Kelvin. And when you're in the vapor state, it's actually more similar to the solid state. So given what we know now, we can actually figure out how much energy it would take to go from minus 10 degree ice to 110 degree vapor. Let's work this out. So the first thing we're going to be doing is, we're going to be going from minus 10 degree ice to zero degree ice. So we're going to go 10 degrees. We have to figure out how much heat does it take to warm up ice by 10 degrees. So the heat is going to be equal to the change in temperature. So actually let me write the specific heat first. So 2.05 joules per gram Kelvin. Oh, and I should tell you, we can't different values for the amount of ice we're warming up to vapor, so let's say we're dealing with 200 grams. So it'll be the specific heat times the number of grams we're warming up of ice times the change in temperature that we're trying to get. So, times 10 degrees Kelvin. Le'ts just say it's a 10 degrees Kelvin change, it doesn't matter if we're using Kelvin or Celsius. I could have written a Celsius here. Let's put Celsius right there. So what is that equal to? Get the calculator. Clear it out. 2.05 times 200 times times 10 is equal to 4,100 joules. Let me do this in a different color. This is 4,100 joules. Fair enough. Now, so what we've done is just this part right here. This distance right here is 4,100 joules. Now we have to turn that zero degree ice into zero degree water. And for that we use the heat of fusion. We're adding that amount of heat. It's 333.5 joules per gram. So that's equal to 335.55 joules per gram times, we have 200 grams of ice that we're trying to melt. So that is what? So let me get the calculator out. So I have 335.55 times 200 is equal to 67,110. Let me do it in that color so can sum them up at the end. 67,110 joules to melt the water. Or take it from ice to water. Now, we have to go from, and this is the big one. We have to go from zero degree water to 100 degree water. Or liquid water. This is in the liquid state. So now we take the specific heat of water which was 4.178. For some reason I'm thinking it's 4.176 but it doesn't matter. So it's 4.178. I might off a little bit on that number, but that digit is not that significant. Joules per gram Celsius times 200 grams times 100 degrees Celsius. And notice, the Celsius and the Celsius cancel out. The grams and the grams cancel out. So we are left with joules, which is what we want. We want to know how much heat or how much energy we're adding to the system. Let me get the calculator out. So this stage is going to be 4.178 times 200 times 100 is equal to 83,560 joules. Does that look about right? Let's see. 4 times 200 is 800, 800 times 100; yeah, that's about right. Now, we're dealing with 100 degree water vapor, and we have to turn that 100 degree water vapor to 110 degree vapor. So we use the specific heat of vapor. 1.89 joules per gram Kelvin. Multiplied by the amount of vapor we're dealing with, 200 grams. That obviously doesn't change. We're not adding or taking away mass from the system. Times the temperature change, times 10. So what is that? Let me get the calculator out again. So we're dealing with a 10 degree change. 10 degree Celsius times 200 times 1.89 is equal to 3,780. And I just realized, I made a horrible mistake. It's not an irrevocable mistake, otherwise I would rerecord the video. But I just figured out the amount of energy to take it from zero degree water to 100 degree water, which is this energy right here. And now I just calculated how much energy to go from 100 degree vapor to 110 degree vapor. Which is this right here, that distance right here. I forgot to figure out how much energy to turn that 100 degree water into 100 degree vapor. So that's key. So I really should have done that up here before I calculated the vapor. But I'll do it down here. So to do 100 degree water to 100 degree vapor. That's this step right here, this is the phase change. I multiply the heat of vaporization, which is 2,257 joules per gram times 200 grams. And this is equal to 451,400. I'll do it in that blue color. 451,400 joules. So this piece right here is 451,000 for our sample of 200 grams. This piece right here was 83,000 joules. This piece right here was 3,780 joules. So to know the total amount of energy, the total amount of heat that we had to put in the system to go from minus 10 degree ice all the way to 110 degree vapor, we just add up all of the energies we had to do in all of these steps. Let's see. And I'll do them in order this time. So to go from minus 10 degree ice to zero degree ice. Of course we have 200 grams of it. It was 4,100. Plus the 67,000. I'm just doing this off the screen. So plus 67,110. Plus 83,000. That's to go from zero degree water to 100 degree water. Plus 83,560. So we're at 154,000 right now just to get to 100 degree water. And then we need to turn that 100 degree water into 100 degree vapor. So you add the 451,000. So, plus 451,400 is equal to 606. And then finally, we're at 100 degree vapor, and we want to convert that to 110 degree vapor. So it's another 3,700 joules. So plus 3,780 is equal to 609,950 joules. So this whole thing when we're dealing with 200 grams, when you're going from minus 10 to 110. And remember, this is for 200 grams. It took us 609,950 joules. Or 609 kilojoules to do this. So that by itself is an interesting thing. You might think, wow, this is a huge number. But actually, it turns out joules isn't a lot of work. Kilojoules starts to become a little bit interesting. You might realize, 200 grams of ice, that's about half a pound of ice. So to take half a pound of ice and warm it up on your stove would take 609 kilojoules of heat to do that. So that's something that you probably could do on your stove at home.