Hey, everybody. So in this problem here, we have a mixing of 2 different temperatures of water. We have 195 kilograms at 30 Celsius and 5 kilograms of boiling. In the first part, we're going to calculate the final temperature of the mixture of water. So remember, this is going to be like a classic calorimetry problem. In the second part, we're going to calculate the total change in entropy. So let's go ahead and get started here. Before I get started with an equation, I just want to go ahead and draw a diagram. We've got this hot tub like this, and I filled it with some massive water. This is going to be 195 and this temperature here, this T, is going to be 30 Celsius. I'm going to convert this to Kelvin real quick, which is going to be 303. Then what I'm going to do is I'm going to add some hot water to it. This hot water here is going to be 5 kilograms at an initial temperature of boiling. Remember, the boiling point for water is just at 100. So in other words, it's going to be 100, which is going to be 373 Kelvin.

So what I'm going to do is here I'm going to call the hot water MH and TH. The colder water is going to be MC and TC. Alright? 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 at different temperatures. There's going to be an exchange of heat from the hot to the cold, and they're going to end up at some equilibrium temperature. And that's actually the first thing that I want to calculate here. In part a, I want to 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 go ahead and plug and chug this. This is kind of a tedious part of this problem, but basically, this is just going to be the MC for water and TC. We're going to use C for water for everything. Plus, this is going to be MH, C for water, TH, and then you divide this by MC_w+MH_w. So if you plug this in, what you're going to get is 195 * 4186 * the initial temperature. So this is going to be the 303. Alright. This equation you can use Celsius or Kelvin. I'm just going to use Kelvin. So this is going to be plus 5 * 4186, times 373. And then you're going to divide by, 195 * 4186 plus 5 * 4186. Alright. So it's kind of 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 going to 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 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? The hotter water gives off some heat to the colder water. So in other words, there's a ΔS total here. That's the second step, which is right our ΔS total equation. And it's going to be made up of two 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 two waters that are exchanging. So in other words, the ΔS total is going to be ΔSh + ΔSc.

So now what we have to do here is to solve for this variable here. We're going to have to figure out which equation we use for ΔS. So remember that we've actually come up with a couple of different equations for different processes. So the first one we 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 going to 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 going to use. So it's going to be MCLn of T_finalover T_initial. That's that, and that's actually going to be the case for both of these terms here. So the first one is going to be MH* CW * Ln of T_final, that's our equilibrium temperature, over T_initial, which is just the temperature of the hot water, plus the MC * CW * Ln of T_final 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 ΔS total.

So when you plug everything in, what you should get is you should get 5 * 4186 * the Ln of T_final, which is the 304.8 divided by the 373. Plus or sorry. Yeah. Plus this is going to be 195 * 4186 * 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 ΔS total should be 613 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.