23. The Second Law of Thermodynamics
Entropy Equations for Special Processes
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Hey, everybody. So in this problem here, we have a mixing of 2 different temperatures of water. So we have a 195 kilograms at 30 Celsius and 5 kilograms of boiling. And in the first part, we're gonna calculate the final temperature of the mixture of water. So remember, this is gonna be like a classic calorimetry problem. And in the second part, we're gonna calculate the total change in the entropy. So let's go ahead and get started here. Before I get started with an equation, I just wanna go ahead and draw a diagram. So we've got this hot tub like this and I filled it with some massive water. So this is gonna be 195 and this temperature here, this T, is gonna be 30 Celsius. I'm gonna convert this to Kelvin real quick which is gonna be 303. Then what I'm gonna do is I'm gonna add some hot water to it. This hot water here is gonna be 5 kilograms at an initial temperature of boiling. Remember boiling point for water is just t h equals a 100. So in other words it's gonna be a 100 which is gonna be 373 Kelvin. So what I'm gonna do is here I'm gonna call the hot water MH and TH. The colder water is gonna be c or and t c. Alright? So you mix these two waters together, and then ultimately what you end up with here is you just end up with a total amount of water of 200 kilograms. Right? The 5 plus the 195. And it's at a new equilibrium temperature. We've got these two waters that mix different temperatures. There's gonna be an exchange of heat from the hot to the cold, and they're gonna end up at some equilibrium temperature. And that's actually the first thing that I wanna calculate here. In part a, I wanna calculate the equilibrium temperature. Now we've done this in an earlier video on calorimetry, but we've basically come up with an equation for the equilibrium temperature. It's this really long one that it's like MCTs, MCTs, and MCs, and MCs. You basically just gonna go ahead and plug and chug this. This is kind of a tedious part of this problem, but basically this is just gonna be, the m cc for water and tc. We're gonna use c for water for everything. Plus, this is gonna be m h, c for water, t h, and then you divide this by mccw+mhcw. So if you plug this in, what you're gonna get is 195 4186 times the initial temperature. So this is gonna be the 303. Alright. This equation you can use Celsius or Kelvin. I'm just gonna use Kelvin. So this is gonna be plus 5 times 4186, pull times 373. And then you're going to divide by, 195, 4186 plus 5 and then 4186. Alright. So it's kinda tedious, but when you plug all that stuff in, just make sure you enter it carefully in your calculator with parentheses and all that stuff. What you're gonna get here is 304.8 Kelvin. Now if you look at this number, this should make perfect sense to you. Remember, this colder water here is at 303. There's much less of the hotter water which is at 373. So when you mix them together, the final temperature should be pretty close to the initial temperature of the the colder water. Right? So we got 304.8, and that makes total sense. Alright. So now let's move on to the second part here. What is the change in entropy? So in this problem here, what we have is we have multiple objects that are exchanging heats and therefore changing entropies. Right? So the hotter water gives off some heat to the colder water. So in other words, there's a delta s total here. That's the second step which is right our delta s total equation. And it's gonna be made up of 2 things. The water doesn't exchange heat with the hot tub or the air or anything like that. The only thing that's sort of happening in this problem is that we just have the 2 waters that are exchanging. So in other words, the delta s total is gonna be delta s h plus delta s c. So now what we'd have to do here is to solve for this variable here. We're gonna have to figure out which equation we use for delta s. So remember that we've actually come up with a couple of different equations for different processes. So the first one we should check 1 the first equation you should check is if you can even use q over t. Is it an isothermal process? And the thing is it's not. Remember, it's not an isothermal process because both of these waters here start out at their initial temperatures and then the final one is gonna be different. So this is not a constant temperature process. So, is it a phase change? Well, we don't have water changing to ice or gas or anything like that. It's not a phase change. What about the temperature change? And actually this is the equation we're gonna use. So it's gonna be mcln of t final over t initial. That's that, and that's actually gonna be the case for both of these terms here. So the first one is gonna be m h c w and then times ln of t final, that's our equilibrium temperature, over t initial, which is just the temperature of the hot water, plus the m c c w times the ln of final temperature over the temperature of the cold water. So you just add those MCLN terms together, you plug everything in, and when you add them together that should be your delta s total. So when you plug everything in, what you should get is you should get 5 times 4186 times the ln of t final, which is the 304.8 divided by the 373. Plus or sorry. Yeah. Plus this is gonna be 195 4186 times ln of 304.8 divided by 303. And then when you work this out, what you should get here is that this number here just becomes negative 4,222. And this number here works out to 4,835. So when you add them together your delta S total should be 609 joules per Kelvin. Alright. So this is basically just confirming the second law of thermodynamics. We had these two processes and basically one of them resulted in a decrease in entropy, but there was another part of the problem that increased more entropy. So the overall change here in entropy of the universe is positive, and that's exactly what we should expect. Alright, folks. So that's it for this one.
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