Start typing, then use the up and down arrows to select an option from the list.

Table of contents

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.

Related Videos

Related Practice

03:37

04:47

14:04

07:50

04:29

06:56

07:44

08:13

03:40

06:20

03:56

© 1996–2023 Pearson All rights reserved.