23. The Second Law of Thermodynamics

Entropy Equations for Special Processes

# Entropy & Calorimetry

Patrick Ford

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everybody. So in this problem here, we have a mixing of two different temperatures of water. So, we have 100 and 95 kg at 30 Celsius and five kg of boiling. And then the first part, we're gonna calculate the final temperature of the mixture of water. So, remember, this is gonna be like a classic kalorama tree 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 started with an equation, I just want to 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 1 95 and this temperature here, this tea is gonna be 30 Celsius. I'm gonna convert this to Calvin 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 five kg at initial temperature of boiling. Remember boiling point for water is just th equals 100. So, in other words, it's gonna be 100 which is going to 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 going to be C and T C. All right, so, you mix these two waters together and then ultimately what you end up with here, As you just end up with a total amount of water of 200 kg, right, The five plus the 195. And it said in a new equilibrium temperature, we've got these two waters that mix different temperatures, there's going to be an exchange of heat from the hot to the cold and they're going to end up with 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 kalorama tree, but we've basically come up with an equation for the equilibrium temperature. It's this really long one that it's like M C. T. S, M C. T. S and M. C. S and M. C. S. You're 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 MCC for water and T. C. We're gonna use C. For water for everything. Plus this is going to be M H. C. For water th and then you divide this by M. C. C. W. Plus M H C. W. So if you plug this in, where you're gonna get is 1 95 41 times the initial temperature. So this is going to be the 303. Alright, this equation you can use Celsius or kelvin. I'm just gonna use kelvin. So this is gonna be plus five times 41 Pull Times 3, 73. And then you're going to divide by uh 1 41 86 plus five and then 41 86. Alright, so it's kind of tedious but when you plug all that stuff and 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 3 73. So when you mix them together the final temperature should be pretty close to the initial temperature of the colder water. Right? So we've 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 entropy is 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, or delta, delta is total equation and it's gonna 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. In other words, the delta S total is gonna be delta S. H plus delta S. C. So now we 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 thing we should check one, the first equation should check is if you can even use Q over T. Is it a nice a thermal process? And the thing is, it's not remember, it's not a nicer thermal process because both of these waters here start off after 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, it's 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 M C L N A 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 going to be NH cw 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 M. C. L. N. Terms together you plug everything in and when you add them together that should be your delta S total. So when you plug everything which you should get is you should get five times 86 times the Ln of T. Final. Which is the 304.8 divided by the 73 Plus. Or sorry? Yeah. Plus this is going to be 195 41 times Ln Of 3 04.8 divided by 303. And then when you work this out you should get here, is that this number here just becomes negative. 4222. And this number here works out to 4835. So when you add them together, your delta s total Should be jules 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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