23. The Second Law of Thermodynamics

# Entropy and the Second Law of Thermodynamics

1

concept

## Intro to Entropy

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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 gonna 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 some things 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 gonna 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 systems 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 solid. All of the atoms are sort of locked together like this and 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 that 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 in 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 per kelvin and that's something you can just get from this equation here. Alright, so let's get started with our first example, we're gonna come back to this in just a second here. In our first example we're gonna 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 there's only one temperature, it is gonna be the 27 but we actually have to convert it to Kelvin because this tea 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 gonna 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 going to be delta. S. We're just going to 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 for face changes. We had this old equation here that Q equals M. L. We use this a lot in kalorama tree. So you're just gonna replace this Q. With M. L. And we're going to 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 then 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 this just 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.

2

Problem

3 moles of an ideal gas are compressed isothermally at 20°C. During this compression, 1850 J of work is done on the gas. What is the change of entropy of the gas?

A

–24.9 J/K

B

6.31 J/K

C

–6.31 J/K

D

–9158 J/K

3

Problem

You have a block of ice at 0°C. Heat is added to the ice, causing an increase in entropy of 120J/K. How much ice melts into water in this process?

A

0.098 kg

B

0 kg

C

1.32×10

^{-6}kgD

1.47×10

^{5}kg4

example

## Entropy Increase When Braking

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everybody. So let's take a look at our problem. So we have a car that's driving along the road, it's gonna suddenly slam on the brakes. And we want to calculate what's the change in entropy of the air. So in other words, what we want to calculate here is delta S. And I want to say delta S. For the air. But before we get started, I want to sort of draw out what's going on here. Right? So we've got this little road, I'm gonna draw this little car like this, don't make fun of my car drawings like that's and what happens is it's chugging along initially at some initial speed, which is 20 m per second. Then what happens is over some time it's gonna slam to a stop. So eventually what happens is that the final is going to be zero. Alright, so what's going on here? If you want to calculate the change in the entropy, we're gonna have to use the change in entropy equation now. Remember we can only use this equation Q over t as long as the temperature remains constant. But it does because we're told in this problem that when you slam on the brakes, the air which is like so massive and large, Remains at a constant temperature of 20°C. So what happens is if you add a little bit of heat to it, the air is obviously so massive that it doesn't really change temperature. So we have delta S for the air is the heat that's added to the air divided by that temperature, right? So we have what the temperature is. So the question here is, well, how much heat is actually added to the air? We don't have an equation for this, right? We can't use like Q equals M. Cat or something like that because we don't have the mass of the air. We have the specific heat. So, what's really going on? But what happens here is that we have to realize that this initial velocity for the car means that this car has some kinetic energy. So, in other words, this kinetic energy, I'm gonna I'm gonna say K initial is equal to one half M. V. Naught squared. Actually, I'll call this K. Not, so I'll call this K zero. And eventually what happens is that when you slow to a stop, right? You slam on the brakes eventually when you get to this point here, you're K. Final is equal to zero. So in other words, you had some kinetic energy and then you lost it because you slammed on the brakes. So then where did all of this energy go? Remember that energy can only change forms. It can't be created or destroyed. It doesn't go to potential energy because it's still staying at the same height. So basically what happens here is that all of the K. That gets transferred, basically goes into friction. It goes into the friction of the brakes and that gets dissipated as heat. Remember friction. When you rub something, it's really just heat transfer. So what happens here is that the delta K. The change in the kinetic energy really just becomes Q. It becomes the cue that gets given off to the air. Alright, so that's what's going on. So basically what we can do here is we can say well the Q air really comes from the change in the kinetic energy. Now. Really, what happens is this change is really just the one half Mv not squared because the K final is zero so we don't have to do final minus initial. So this really just becomes 1/2 and this is going to be M V not square. We have the mass of the car and the initial speed so that we can go ahead and do that. So we're gonna have one half of the mass which is 1500 times the initial velocity which is 20 squared. So now you divide that by the initial temperature. Now the temperature is 20 degrees Celsius, but we have to convert that to kelvin. So this just becomes 293 Kelvin and that's what we plug into our denominator here. So if you work this out, what you're gonna get is that the change in the entropy for the air is equal to 1.02 times 10 to the third. And this is going to be jewels per Kelvin. Alright, so that's how much entropy gets, sort of dissipated. That's how much energy you've now dissipated. The energy is a little bit more random and therefore the energy of the universe increases. Alright, so that's it for this one. Guys, let me know if you have any questions.

5

concept

## Calculating Entropy Changes for Systems of Objects

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Hey everyone. So in the last couple of videos we saw how to calculate the change in entropy, but that was just for one object and a lot of problems, you'll have to calculate the entropy change for a system of multiple objects, it could be two or more or in some cases the change in entropy of the universe. For example the problem we're going to work out in this video together has a hot reservoir that's connected to a cold reservoir, it could be something like a heat engine or something like that. And what we wanna do is calculate the net change in entropy of the system. So we're gonna do is work out this problem together because I want to show you a step by step process to solve these problems. So you get the right answer every single time. So let's just go ahead and get to it. The very first step that we want to do is actually just draw a diagram. So again we have a hot reservoir connected to a cold reservoir. I'm just gonna draw a sketch out real quick, we're told that this hot reservoir has a temperature of 200 Celsius, but we want to convert that to kelvin and that's just going to be 473 kelvin. And we know it's connected, we don't know exactly how to a cold reservoir. So it's connected to this cold reservoir here and this cold reservoir has a temperature of 100 Celsius, Which is 373 Kelvin. Now again we don't know how they're connected? All we know is that 400 joules of heat gets transferred from the hot to the cold? Remember heat flows from hot to cold like this? So we have Q equals 400. Alright, so that's the first step here. We have drawn the diagram. The next thing we want to do is we want to write our delta S. Equation. But because this is a system, we're going to write delta S total. So this is gonna be delta S for the total system here. Now, the only equation that we've seen for delta S is this one down here, that's Q over T. And remember this only works for ice. A thermal processes. However, if we look at the problem it is ice a thermal because the reservoirs are so big that the heat exchange has no effect on their temperature. So we absolutely can use this equation. The problem is is that this equation only worked when we had one object like the hot reservoir or the cold reservoir. What we want to do is complete the change in entropy of the system. So we have to consider both of those objects. So in general, what happens is whenever you're writing the delta as total equation, that's just gonna be delta as of one delta S. S. Two, delta S of three. Basically you just add up whatever objects in your system are exchanging heats and changing n trapeze and that will be the total change in entropy. So we're gonna come back to this last part in just a second here. So basically what happens is this hot reservoir is losing heat and so that's going to change the entropy. I'm gonna call this delta S. H. R. And I'm gonna this cold reservoir is absorbing heat. And so it's going to change entropy as well. I'm gonna call this delta scr So the total is just gonna be delta S. H. R. Plus delta S. C. R. So now what I can do is just replace these equations in the next step. So I have the delta S total is going to equal, well this is going to be Q over T. Plus and this is gonna be Q over T. However these things are a different temperature so we don't want to confuse them. So I'm gonna call this one th and this is one is going to be T. C. We have those values over here. So now we just start replacing all these numbers with the values that we know what's the heat that's added or what's the heat that is flowing out of the hot reservoir? It's 400 jewels. So because of that, what happens is he is being removed and we add a negative sign. So it's going to be negative 400 over 473. Now what happens with the cold reservoir, well the hot reservoir is transferring 400 to it. So if the hot reservoir loses 400 the cold reservoir gains 400. So this is going to be positive 400 over 3 73. Right? So those are your numbers. Basically what you can end up with here is that delta? S total is equal to when you plug this in. You're gonna get 0.85 plus. This is gonna be 1.07. And then when you add the whole thing up, you're gonna get zero points 22 jewels per kelvin. And that is the answer. All right. So what happens is that some part of our system here lost entropy or decrease in entropy? Another part of our system increase in entropy and the overall sort of effect Was that the overall change in system of entropy in the system was positive 0.22 jewels per kelvin. And in fact this is always going to happen. This is actually one of the statements of the second law of thermodynamics called the entropy statement. Remember the second law is just a bunch of statements. This one is called the entropy statement. And what it says is that it's impossible for any process to cause an overall decrease in the entropy of a system or in some cases the universe. And that's actually the most important part of that sentence there, the entropy of the universe because what we saw here is that one part of our system did in fact decrease in entropy. So it's that's perfectly fine if any part of your system decreases in entropy, all that happens is that another parts like the cold reservoir in our problem must increase in entropy at least the same amount if not more. Alright, so that's how to solve these kinds of problems guys. So basically when you go back to this delta as total equation here, the sort of last piece is that when you add up all the changes in entropy for a system, it will always be greater than or equal to zero. Another way of saying this is that the entropy of the universe is always increasing. The best you could possibly hope for is that the delta is a change in entropy for a system is zero. That's the best case scenario. All right. So really that's all there is to it guys. The last point I want to make is that sometimes for this reason entropy is called time's Arrow. Which is basically just a useful way to explain why processes happen the way they do, why they happened irreversibly. For instance, like heat transfers. Like we just saw in our problem here. So heat transfer happens only from hot to cold and it never flows the other way around. And that's because this system here increases with this process increases in entropy. You can't get that you can't get the energy back. The energy is more spread out. Another example is friction. When you rub your hands together, you're generating heat, you're spreading out more energy. That's a process that increases the entropy of the universe. So it never will happen backwards if you put your hand on a coffee cup or something and it's hot. That heat will flow from coffee to your hands and never the other way around. That's just another example. Alright guys, So basically time flows or moves in the direction of increasing entropy. So that's it for this one guys, let me know if you have any questions.

6

Problem

A non-Carnot heat engine operates between a hot reservoir at 610K and a cold reservoir at 320K. In a cycle, it takes in 6400 J of heat and does 2200 J of work. What is the total change in entropy of the universe over the cycle?

A

2.6 J/K

B

9.5 J/K

C

16.3 J/K

D

–2.6 J/K

7

example

## Entropy of Carnot Engine

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Hey everybody. So let's check out our example problem. We have a Carnot heat engine that's taking in some heat from the hot reservoir, expelling it to a cold. And we want to calculate the total change in entropy of the universe. Now we already know this is gonna be a entropy problem that involves multiple objects because the hot and the cold reservoirs are both exchanging heats at different temperatures. So let's go ahead and draw our quick little diagram here. That's my hot, that's my engine, that's my cold. We've got the temperature of the hot is 500 the temperature of the cold is 3 50. All this stuff is in kelvin and so we have the heat that's taken in from the hot reservoir is 2000 jewels. So that's my cue. H. right? That's Q. H. equals 2000 jewels. So then what happens is expel some heat to the cold reservoir but we actually don't know what that Q. C. Is. So the next thing I wanna do is just set up our delta S total equation. The total change in entropy for the universe is going to be adding up the two n. Trapeze entropy changes for the hot and cold reservoirs. In other words, DELTA S. H. Plus delta S. C. Alright, so what happens here is that this delta S. H. Because we can assume that this happens at constant temperature is going to be qh over th but remember we have to add a negative sign because the hot reservoir loses some heat to the engine. Whereas this delta S. C. Here is gonna be plus que si over T. C. And that's because this cold reservoir is absorbing heat from the engine. Right? So one's positive, one's negative. Okay, So basically this question figure this out, right? So, we've got delta or so you have negative QH which is going to be negative Divided by the 500. Now, we have the T. C. Here, we have this 3 50 but we don't actually have what the Q. C. Is. So, we're gonna have to go figure that out now. The way that we did this in another problem was that we actually had the work done, and so if we used the work equation, right? This w equation, we calculate Q. C. But we actually don't have that here. So, in order to figure out what this Q. C. Here is, I'm gonna need another equation. So, let's go ahead and do that. So, which equation can we use for the Carnot cycle? We actually have a couple of them. If we need the efficiency we would use this equation, but we don't have the efficiency. We can't use this equation here. So, the only one that that I can use that sort of special to Carnot engine is going to be this one over here. This is the ratio of the heats. Does he go to the ratio of the temperatures? Remember, we can only use this for a carnot cycle. So, let's set this up here. So, we've got Q. C over Q. H. Is equal to T. C over th All right. So, if you look at this, I have three out of four variables. Right? I've got all the heats and the temperatures. So I can calculate this real quick here. This Q. C. Is really just gonna be well, T. C over th is 3 50 over 500. So, we're going to multiply that times the qh which was 2000. If you go ahead and work this out, what you're gonna get here is you're gonna get um Let's see, I get 1400. So, this is 1400 jewels. That's what we plug into this equation now. Alright, so, basically this just becomes negative. 2000 over 500 plus 1400 over 3 50. Now, when you work this out, what you're gonna get is that this delta s total here is actually equal to zero. So, essentially happens is these two things will cancel each other out. So, what's going on here is that because carnot heat engines are ideal and they are perfectly reversible, then, essentially what that means is that the change in entropy for a card lifecycle is always equal to zero. So, in other words, carnot engines are perfect and perfectly reversible, then, basically what happens is that the best outcome happens, you have no total change in entropy of the universe. All right. So, that's sort of like the best case scenario that you could hope for. All right, guys. So that's it for this one.

Additional resources for Entropy and the Second Law of Thermodynamics

PRACTICE PROBLEMS AND ACTIVITIES (5)

- CALC Two moles of an ideal gas occupy a volume V. The gas expands isothermally and reversibly to a volume 3V. ...
- CALC You make tea with 0.250 kg of 85.0°C water and let it cool to room temperature 120.0°C2. (a) Calculate th...
- A 15.0-kg block of ice at 0.0°C melts to liquid water at 0.0°C inside a large room at 20.0°C. Treat the ice an...
- CALC You decide to take a nice hot bath but discover that your thoughtless roommate has used up most of the ho...
- A sophomore with nothing better to do adds heat to 0.350 kg of ice at 0.0°C until it is all melted. (a) What i...