Hey, guys. So in previous videos, we talked about the specific heat and how that heat dealt with temperature changes. In this video, we're going to talk about latent heats and how those deal with phase changes in a material. Now a lot of problems are going to combine specific heats and latent heat, so I'm also going to show you a step-by-step process to navigate these kinds of problems. Let's go ahead and check it out.

Let's first talk about the phase of a material. So if you've ever taken another science course like chemistry, a phase of material is really just like the state of matter. We use 3 commonly in physics: solids, liquids, and gases. And these phases usually just depend on the temperature of the material. So for example, the compound H_{2}O, which we all commonly know as water. Water is in liquid form, but if you cool it down, if you put water in the freezer, then it becomes ice. It's the same thing; it's just in a solid form and if you heat water up enough, it becomes steam. Still the same compound, it's just in a gaseous form.

Now remember, when a material absorbs or loses heat (plus or minus Q), it changes temperature like heating or cooling water, but it can also change phase. But the most important thing is it can only do one or the other and not both. To show you what I mean by this, I'm going to show you a graph in which we're plotting the temperature versus heat for water. The idea here is we're going to have an ice cube that starts at 0 degrees Kelvin, and we're going to see how the temperature responds as we inject more heat into the system. Let's go ahead and check this out.

So imagine we have this ice cube that's here at 0 and you stick a flame underneath it and you start to heat it up. Well, basically what happens is you're going to start raising the temperature of that ice. The sort of graph for this is going to look like a straight line and the equation that this is going to obey is Q=mcΔt. This is just the temperature change equation. So we're going to use the c for ice times delta t. So the idea here is that we're only just dealing with one phase of matter, just ice. And so in these temperature change sort of phases, you're just going to use Q=mcΔt, and what happens here is that the temperature is going to be changing, but the phase remains constant. Still ice, but now your temperature is rising. And then eventually you're going to get to this one point right here, which is the melting point of water, 273 Kelvin or 0 degrees Celsius. And what happens here is actually something different because now what happens is the temperature is not going to rise anymore. Instead, what you're going to notice is that now the ice, instead of raising its temperature, starts to transform into water.

So what's going on here is you've actually hit this part of the diagram, which is a phase change. The idea here is that now the temperature actually remains constant, but now the phase is going to be changing. All the additional heat that you're putting into the ice is not going into raising the temperature. It's actually going to break the chemical bonds of ice so that this thing can transform into water. So that's the difference between these two. In one, the temperature is changing and the phase is constant, and the other, it's the opposite. The temperature is constant and the phase is changing. The equation that we use for these phase change parts of these diagrams is Q=mL. So that's the equation that we use for these things here. This L here is called the latent heat. Some books will also call it the heat of transformation. This is a constant that's very similar to the specific heat. It just depends on the material per kilogram, but it also depends on the phase change. So it depends on what material you're talking about like water or mercury or whatever, but it also depends on which two states you're traveling between. So we actually have a couple of different names. If we're going from a solid to a liquid like we were going here, ice to water, we call this the latent heat of fusion, so this is L_{f}.

Alright. So if you keep now injecting more heat, basically what happens is that at this point, all of the ice is transformed into water and now you're just going to have another diagonal section like this. So now the only phase that you have here is water and you're just going to have another Q=mcΔt equation. This is going to be m c for water times delta t and then what happens is you're going to hit this other temperature, 100 degrees Celsius or 373 Kelvin. That's the boiling point of water. Now you have another phase change. Here, what happens is that all the water stops heating up and now it just becomes steam. And so now you have another Q=mL, except now this L here when you're changing from liquid to gas is called the latent heat of vaporization. We call this L_{v}. Alright? So this is another sort of phase change portion. And then finally what happens is that once all the stuff has turned into steam, now you can actually continue increasing this thing, basically on to infinity here. So this is steam and you're going to have one final Q=mcΔt equation. You use c for steam. Alright? So that's how these sort of diagrams work. Now a lot of problems are actually going to combine specific and latent heats in your problems. You might have one or the other. You might actually even have a combination of both. So I'm going to show you what actually, the sort of step-by-step process for solving these kinds of problems. So let's take a look at our first one here.

We're supposed to calculate the heat required for the following heating processes assuming we have 400 grams of water. That's 0.4 kilograms. In the first part here, we're just going to heat this water up from 80 to 100 degrees. So let's go ahead and take a look at the steps. The first thing we're going to do is always draw out these T versus Q graphs. These are really going to help us visualize exactly what it is that we're trying to do and how much heat is going to be required. So we're trying to go from 80 to 100 degrees Celsius. So basically, for water, we're going from 0 to 100, so 80 would be somewhere around here. It doesn't actually have to be to scale. This is kind of just to help us visualize what's going on. So really what happens is we're going from here on the diagram and we're trying to get it to 100, but we're actually not trying to boil the water. So all we're trying to do here is we're trying to go from here to here. So the first thing we're going to do is draw the T versus Q graph, identify your initial and final temperatures. The second thing you're going to do is you're going to draw the path from the initial to the final, which is what we did here. So here is our T initial and here is our T final. The next thing you're going to do is you're going to write out your Q total equation. So the Q total here, the total heat is really just going to be the heat required to raise the temperature from 80 to 100. Remember, the equation that we're going to use for these diagonal sections here is going to be Q=mcΔt. So we're really just going to use a Q equals m c delta t equation. So that's really all there is to it. So we use Q equals delta t, and this is really just going to be m c delta t like this. So we've actually got all of our constants like this. We've got the mass, we've got the specific heat and the temperature change. So you're just going to go ahead and plug and chug. So the Q total is just going to be the mass, which is 0.4 times 4186 times the delta T, which is 20. If you go and plug this in, you're going to get 33,500 joules, and that's the answer to the first part. Let's move on to the second one.

Now what we're going to do is we're actually going to boil exactly half of the water that's already starting from 100 degrees Celsius. Let's go through the steps. We've got our diagram. Here is our initial temperature, but now what we're trying to do is not trying to raise the temperature, we're trying to boil it. That's going to be a phase change. So what happens here is we're trying to boil this water here. We know that this, between these two points here, we're going to have basically a phase change. So this, this would be, the total heat required to do, to boil all of this would be this entire line segment right here. This would be Q=mL assuming that we had all 400 grams of water to boil. But remember, we're only going to boil half of the water. So what happens is we're basically going to go from here and not all the way to here. We're actually to stop exactly halfway in between. So really this is the part that we're kind of working with here and we're actually going to ignore all of this other stuff. So the idea here is that exactly half, what that means is that we're only going to boil half of the mass of the water, which is going to be 0.2 kilograms. Alright. So we've drawn the path we're traveling to. We're going to use Q=mL. So that's going to be our Q total equation. So Q equals, this is going to be Q for the phase change, and this is just going to be the m L, and we're going to use the latent heat of what we're going from, 100 degrees Celsius. That's going to be from liquid to gas, water to steam. So we're going to use the latent heat of vaporization. Alright? So this Q total here is just going to be, the mass that we have, which is 0.2 times the latent heat, which is going to be 2.256 x 10^{6}, and I just actually just pulled that from this table over here. That's where I got that value from. So if you go ahead and work this out, what you're going to get is that Q is equal to, I get a value of 451,000 joules. So that's it. Alright. Let's do the last one very quickly.

So the last one is going to be we're going to completely boil all of the water starting from 80 degrees Celsius. So I'm starting from 80 over here and that's going to be over here, but now I want to completely boil it. Remember that point is going to be over here. So if you take a look, there are actually two things I need to do. The first line segment I need to travel on is this one over here, And we know that on this equation, we're in this segment here, we're going to use Q=mcΔt. Then we're going to actually travel on this line segment in which we're going to boil all of the water, that's going to be Q=mL. So in this case, we actually have heats that are for both of them. It costs some energy to raise it to 100 degrees first and then it costs more energy to boil all of it, so that it turns to steam. Alright? So that means that the Q total equation is really just going to be the Q from the delta t, plus the Q of the phase change. So I'm going to call this the water to steam. So really what we're going to do is we're just going to combine our equations. Q equals m c delta t plus m times L_{v}. Right? We have the masses, we have all that stuff. So we're just going to go ahead and plug this in. So this is going to be 0.4 times the specific heat 4186 times the temperature, which is going to be 20, then we're going to add it to the m L, which is going to be 0.4. Here we're using all of the mass, all 400 grams of it and then times the latent heat of vaporization. So 2.256 x 10^{6}. You go ahead and plug all this stuff into your calculator and your Q total, the total amount of heat is going to be 9.36 x 10^{5} joules, and that's the answer, guys. So hopefully, this makes sense. This is a very important type of problem for you to get really familiar with. So let's go ahead and take a look at some practice problems.