Advanced Calorimetry: Equilibrium Temperature with Phase Changes

Hey guys, so up until now, when we've seen calorie mystery problems that included both temperature and phase changes, we were told what the equilibrium temperature was gonna be. But in some cases you're actually gonna have to calculate that, you're gonna have to calculate what that final equilibrium temperature is. Unfortunately, that makes these problems a little bit harder. And these problems are kind of rare, but they are trickier. So in this video, I'm gonna show you a different sequence of steps than the ones that we've seen so far to solve these kinds of problems. Let's go ahead and check out our example here, we've got these two substances, we've got copper at 1 50 we've got mixing with water at 30 degrees. So we're gonna close them, mix them in a container and we want to calculate with the final temperature of this mixture is So remember before we actually write out our kalorama tree equations, the first thing we wanna do do is draw our diagrams and indicate what the initial and final temperatures are. That's the sort of zeroth step. So I've got these two different substances, copper and water. Copper is always a solid up until thousands of degrees Celsius. This one's gonna start here at 150 degrees, That's our hotter one. Doesn't have to do this scale, it doesn't matter, right. The water on the other hand, is going to be at 30 degrees Celsius and we've got our sort of familiar diagram here. So here's what's going on, the copper is gonna lose some heat and therefore lose some temperature, it's only gonna go down in temperature, it can experience a phase change, but the water, on the other hand, can actually do a couple of different things. So we have to do is indicate where the initial or the final temperature is, but we actually don't know where that is. So the problem with this is that there's three possibilities here. The water could have absorbed enough heat from the copper so that it goes up in temperature only. So imagine this basically just goes higher here on this diagram and we only experience a temperature change. There could be even more heat though from the copper, such that the water actually starts to boil so it starts to transition into steam. And then if there's enough heat there's enough heat to actually turn this thing completely into steam that the temperature actually could continue rising. The problem is we actually don't know upfront which one of these possibilities could actually be true. So what happens here is we have three possibilities. Right, so we have water, you have em water, the specific heat for water turning all the way to steam, or sorry, turning all the way to 100 degrees, then we've got some of that water which could melt or sorry, boil into steam and then we've got some of this steam that can actually increase in temperature as well. So these are the three different terms that we could possibly have inside of our telemetry equation on the right side. We're only gonna have the copper, right? M. C C for copper and the delta T for copper. So, the problem, what we're basically going to do here before I get into the sort of messy steps is in order to figure out how many of these terms actually exist inside of our equations. We're gonna have to do a little bit of trial and error. Basically, in these problems, you're gonna have to figure out the number of M cat or mL equations inside of your kalorama tree equations and we do that by calculating some stuff by making some assumptions. We're gonna try some things and if it doesn't work, we're gonna have to go back here and change some things and then just try again. It's a lot of trial and error. All right, So let's go ahead and get to the steps now before we actually So, instead of writing r kalorama tree equation, we're actually just gonna jump straight into step two, which is calculating the final temperature and we're gonna do is we're gonna assume that there's no phase change. Basically we're working with this assumption right here that there's only just been some increase in temperature and no phase change. So basically what we're gonna do here is we're gonna calculate the final temperature by using a familiar equation that we've seen before with for the equilibrium temperature. Alright. So, I'm just gonna jump straight into that this T final here, it's gonna be I'm gonna have the M. For water sea for water and then T. Initial for water plus M. Four copper C. For copper and then delta and then the sorry T. Initial for copper. And I'm just gonna divide this by and this is going to be m water seawater plus M copper. See copper, I'm just gonna start plugging in some numbers. So the M water we're told is 0.1 kg. The specific heat for water is 41 86. We've seen that so far and the initial temperature is gonna be the 30 the copper is going to be two times 3 90. That's the specific heat which I've indicated right here and the initial temperature is 1 50. So now we're just gonna divide this by 0.1 times 86 plus two times 3 90. If you go and work this out, what you're gonna get is a final temperature Of 108°C. So the question is, is this actually correct? And the answer is actually no, this can't be right because in the second step here we're going to basically come up with two results. If your final temperature is less than the temperature of the phase change, which means your assumption was correct, then you can just go ahead and solve. But what we got was actually something that was greater than the temperature change. Uh temperature. The temperature of the phase change for water. So this one oh eight Celsius is higher than 100 degrees, that's the boiling point for water. So what happens here is we actually can't go all the way to 100 degrees Celsius because before that happens some of the water would actually start to turn into steam. Therefore our assumption was actually wrong. There wasn't just the temperature change, there was also a phase change. So we have to do here is this is not our correct answer. And we're actually gonna have to go back to our telemetry equation and we're gonna have to add on this second term. This is basically what the term that we're gonna add on, is this A. Q. Equals M. L. So let's go ahead and do that. Now we're gonna do now in our third step is now with this new QA and QB equation, we're going to compare these two values. Basically what I want to do is figure out how much heat is required to change from water all the way to steam and then how much heat is required for the copper to go down to that 100 degrees Celsius. And we're going to assume that the final temperature is going to be somewhere along this line here. We're going to assume that this final temperature is somewhere at 100°C. So we have this Q. A. How does this compare with QB and we're only just gonna work with absolute values for right now just so everything is positive. So let's go ahead and calculate Q. A. Q. A. Is going to be with our new equation QA one plus Q. A. Two. So that's gonna be the mass for water, sea for water, delta. T. For water plus. And now we're gonna have delta M. And then we're gonna have the latent heat of vaporization. So this is going to equal uh let's see, we have 0.1 times 41 86. And then the final temperature is gonna be 100 minus the initial temperature of 30. So plus and then we're gonna have all the mass. We're actually just gonna assume that all of the mass vaporizes. So we're still gonna use the same 0.1. Except now we're going to use this the latent heat of vaporization 2.256 times to the six. Remember that? That's what happens when you have liquids turn into gasses. Alright, so what you end up getting here is that Q. A. Is equal to I get 29,300 for the first term plus 2.256 times 10 to the fifth for the second term which end up with is a grand total of 254,000 900. So that's some jewels. So let's see how that works out with the QB with QB. Remember we're only gonna have one process for BB is just gonna go down in temperature. Right? So we're gonna have this is going to be M. B. Are actually Sir em for the copper C. For the copper delta T. For the copper. So this is gonna be the two times 3 90. and then the final temperature we're assuming is a 100 minus the initial temperature is 1 which end up getting here for Q. B. Is negative 39,000. So what we have to do is compare these two numbers here and hopefully you guys realize that this number is much much bigger than this one. So what happens here is we've ended up with the fact that QA is greater than Q. B. Basically what this means is that the amount of heat that's required for for water to go all the way up to 100 degrees Celsius. And also completely boil is too much, is more than the the heat that copper can actually give off when it's going down to 100 degrees Celsius. So what this means here is that if QA is bigger than QB, then that means that only some of the water has turned into steam, Which means we actually know what our final temperature is now. So because we got this result, we know that this temperature is going to equal 100°C. What this result means is that we're actually somewhere along this line here. So basically we figured out that number three is not a possibility because we're gonna be somewhere along this line, some of the water has turned into steam and therefore the equilibrium temperature is at 100°C. Alright, so we know that's actually the first part, so that's the first part of our problem. The second part now is figuring out how much if any water turns into steam. So to do that, basically we're gonna go back to our equation here. This is gonna be part B we're really looking for is delta M. That's the partial mass. Remember that turns from water to steam. So to solve this, we're actually gonna go back and use this equation here now that we know how many terms we have. So I'm just gonna go ahead and set this up. Our equation is going to be M water, seawater, delta T. Water plus delta M. Times latent heat of vaporization equals negative M C C C. And then times delta T. For the copper. The good thing is we've already calculated a lot of these values. So I'm just gonna go ahead and plug this and remember this value here towards 29,300 is when we calculated this and that's just gonna be the same number. So it's gonna be 29,300. What we're looking for here is this delta M. So this is gonna be plus delta M. Times latent heat of vaporization equals. Now we have to be a little bit careful here. We calculated QB. Which was negative 39,000, But now we actually have another negative sign in front of this. So this is just gonna be negative negative 39,000. Just be very careful with the signs And this turns into a positive, right? So basically the last step is that Delta M is going to be uh to negative 29,300. I'm just gonna move this over to the other side, and then we're gonna do plus 39,000, divided by 2.256 times 10 to the sixth. Which you end up with is you get to end up with a mass a partial mass of 0.43 kg. That's how much ice has now become water. It's not all of the 0.1 kg. It's just only a small amount of it. All right, so that's how you solve these kinds of problems, guys. Hopefully this made sense. Let me know if you have any questions