by Patrick Ford

everyone. So in this video I'm going to introduce a concept called entropy. I'm super excited for this video because entropy is one of those tricky sort of hard to understand conceptual things, but hopefully by the end of this you're gonna have a great solid understanding of what entropy is all about. And I'm also going to show you the equation that you need to solve problems. So let's go ahead and get started here. What is entropy all about? Well, almost certainly the definition you're going to get is that it's a measure of something called disorder but I never really liked this definition because disorder is kind of vague. What does it mean to be disordered? How do you measure or calculate something's disorder nous. So I think a better definition is that it's a measure of an object or a systems randomness. So it kind of has to do with statistics and probability. Let me show you an example here. Imagine I had these two jars of different colored balls and I was going to mix them together in a big jar and shake the whole thing up now before I've mixed anything, the two sort of colors are separated in their own containers and this system doesn't have a whole lot of randomness but when you combine them and you shake the whole thing up, you're gonna end up with something that could look like this in which the ball sort of mix around to get more randomized and if you keep shaking this, you're just gonna end up with more and more random configurations, the balls will never separate back into their own colors again. So what happens here is because there's less randomness in the before case before I mix them up, there's lower entropy. But once I mix everything up, there's more randomness and therefore higher entropy. Alright, so what does this have to do with thermodynamics? Well, you can imagine that instead of mixing these containers of balls together, I'm mixing jars of different gas particles together. So that's sort of the connection there. So in thermodynamics specifically entropy measures how randomly a system's energy is spread out. So let me show you another example of this. Imagine I had two jars but now once full of ice and one is full of water. And I ask you which one is more disordered. Now you might be tempted to say it's the ice. Notice how the ice cubes are sort of in their random positions, whereas the water is sort of relatively flat. It's not really doing anything. But actually this would be the wrong answer. The ice is actually the one that has less randomness. And so what happens here is you have to look down at the atomic level to see how the energy is spread out. Remember that for ice which is a solid, All of the atoms are sort of locked together like this in their positions and they can't move around. Which means that the energy can't really spread out enough, whereas the water which is a liquid. Remember these liquid particles, they're actually able to freely move around because the bonds have broken. So because these things are able to move around freely and therefore more randomly, the energy is more spread out. So the whole idea here is that the water has more randomness, even though it looks like it is less disordered and therefore higher entropy, whereas the ice has lower entropy. So in general because things with higher temperatures have more energy, the energy can get spread around a little bit more. They have higher randomness and therefore higher entropy. Alright, so hopefully that's a great conceptual understanding of what entropy is. Now, let's go ahead and talk about the equation. So in all problems you're you're going to calculate the change in entropy or in most problems rather than the entropy itself. The symbol we use for the change in entropy is delta. S. Number, delta means change and S is for entropy. The equation is actually really straightforward. It's actually just Q over T where Q is just the heat. That's a variable we've seen over and over again. And t. Is the temperature which must be in kelvin Now, it's really important about this equation, is that it only works for ice, a thermal processes. This only works when there is no change in temperature. One way to remember this is that this equation doesn't have a delta T. It just has one t so you don't need multiple tees usually in a problem, you'll just be given one and that's what you plug into this problem here. Now just again, the units also just gonna be in jewels, jewels, but kelvin and that's something you can just get from this equation here. Alright, so let's get get started with our first example, we're gonna come back to this in just a second here. In our first example, we're going to add 2400 joules of energy to a body of water that's so large that its temperature remains constant. And that's really important because remember our entropy equation only works for constant temperature and we want to calculate the change in entropy and that's just delta S. So we want to calculate delta S. And in order to do that, we're just going to use our new equation that's going to be Q divided by T. So we need to figure out how much heat gets transferred divided by the temperature. Now we're told in this problem that we're gonna add 2400 joules of heat, that's gonna be our Q. It's 2400. Now we just have to plug in the temperature, do we plug in 27 or something else? Remember that? Only one temperature, it is gonna be this 27 but we actually have to convert it to kelvin because this thi has to be in kelvin, So we're gonna have to add to 73 to this. So we're gonna do 27 plus two, and this just becomes 2400 divided by 300. When you work this out here, you're gonna have eight jewels per kelvin and that is the answer. So one way you can kind of think about this is that you added energy to a system a body of water. The temperature didn't change. But what happens is you basically spread out a little bit more energy throughout that body of water. So the entropy did increase by eight jewels per Calvin. All right, let's move on to our second example here, we're going to calculate the change in entropy when two kg of water freezes completely to ice. Now, what does that mean? We've seen these kinds of problems before where we have a substance that's changing phases. We have water that's freezing to ice and we want to calculate the change into entropy that's just gonna be delta. S. And we're just gonna use Q. Over T. Now, what happens here is we have to figure out how much heat is transferred into this is transferred when you have water freezing to ice. What happens is, well, for face changes, we had this old equation here that Q equals M. L. We used this a lot in kalorama tree. So you're just gonna replace this queue with M. L. And we're gonna use the appropriate latent heats. Remember there was two of them for water? We had the latent heat of fusion and the latent heat of vaporization when something turned to esteem. Now we're actually gonna use this L. F. Because that's when water is changing to ice and vice versa. So we're gonna use M. L. F. And they're really divided by the temperature. Now, what is this temperature here? Well it's the temperature at which water freezes to ice. So remember that the delta T. For a phase change is equal to zero. And that's really important here because that means that we can use this equation here. Q over T. If delta T. Wasn't zero, we can use this equation. So there's no change in temperature when you have a phase change and this temperature for freezing is just zero Celsius, which is just 2 73 kelvin. So what happens here is district becomes T freeze And now we can just plug everything in. So we're gonna have the mass which is to the latent heat of fusion which is 3.34 times 10 to the fifth. And now you're gonna divide this by 273. Now, the one thing that we forgot though is a minus sign, remember that we have water that's turning and changing to ice. So we actually have to remove heat in order to do that. So this queue here actually has a negative sign. So this is gonna be negative M. L. F. And we're gonna have a little negative sign here. Now when you work this out here, let me just move this down. What you're gonna end up with is that delta S. Is equal to negative 2447 jewels per kelvin. So we can see here is that in some cases the change entropy will be positive and in other cases it will be negative and the sign of delta S is always gonna be the same as Q. And basically the rule is that if you add heat which means that Q is a positive number, then the entropy increases. And that's what we saw for the first example if you remove heat, then Q. Is a negative and then the change in entropy or the entropy decreases. Alright, so that's it for this one guys, let me know if you have any questions.

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