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Calorimetry Problems with Temperature AND Phase Changes

Patrick Ford
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Hey guys. So up until now we've seen calorie mystery problems that have only involved temperature changes. But you're gonna need to know how to solve calorie mystery problems where materials are not only just changing temperature but also phase. So what I'm gonna show you this video is we're going to combine what we know about kalorama tree. We're gonna use a lot of the same sequence of steps with just a couple of new ones and we're gonna combine it with what we know about latent heats and face changes. So we're gonna be using these heat and temperature diagrams to kind of visualize what's going on inside of your calorie mystery problems. Let's go ahead and check this out here. We're gonna jump straight into our example. So, I have an insulated cup that combines two different materials. I've got some mass of water that's at 15 and then I'm gonna put some ice in it to cool it down And the ice is at negative 20°. And I want to combine these things that two things happen. I want exactly half of the ice to melt. And I also want the final temperature of that mixture to be zero degrees Celsius. All right, So, we know we're gonna have to use kalorama tree equations. This Q is negative QB. But the very first thing I like to do, what I would like to call step zero is actually draw the temperature versus Q diagram and figure out what the initial temperatures are. So that you can sort of work backwards and find out what that final temperature is going to be. So let's go ahead and check this out. The water is at 15°C, which is going to be here in the water part of the Diagram. So this is 15. The ice on the other hand is somewhere over here below the freezing point. So this is gonna be negative 20°C. Now, both of these things, right, are gonna exchange heats and they're gonna meet somewhere in the middle of what we want is exactly half the ice to melt. And we also want the temperature to be zero. So what that means is that the temperature is zero over here on this sort of dotted line, but it's gonna be anywhere along this horizontal line here, remember the temperature is the same throughout this whole process. It's just that the phase is changing from ice to water. So what's going on here is if you kind of visualize as you input more heat at this point, this is the freezing point melting starts, so we have melting starts here and then what happens is you input more heat, more ice gets converted to water. And then basically what happens over here is here is where you have complete melting So this is complete melting over here. So we want is we want this final mixture to exactly be half ice melted. So what that means is that we're actually gonna be looking here at this halfway point, we know the temperature that's final here is gonna be 0°C because we're along this line and we want is exactly half the ice to melt. So we're gonna go exactly halfway across this horizontal line. So that's really it. So let's go ahead and get to now our first step which is going to be writing out the Kalorama tree equation. So we got that Q. A. Is equal to negative QB. And we have here is we're gonna have one Q per change what's going on here is if you think about it, we actually have two processes that are going on for the ice. The ice in order for it to get from negative 20 degrees Celsius all the way to melted. Has to do two things. The first thing I see has to do is actually get up to the freezing 20.0 degrees. I'm gonna call this QA one And the next part is that it actually has to melt somewhat or you know partially so that half the ice melts. So it's gonna be QA to the water on the other hand only actually has to do one thing. The water initially starts at 15° and you're gonna have sort of half mixture of ice and water. The hot water only has to lose some temperature. This is QB. It only has to go down to 0°. So these things actually don't necessarily meet. You don't actually have to draw the path for hotter objects cooling down to this point. It doesn't work that way. So remember that he always flows from hot to cold. So what happens is only the colder objects are going to change both temperature and phase. Hotter objects will never cool down and also change phase. So it's kind of like one direction on this because heat always flows from hot to cold. Alright. So what this means here is that we actually have to expand this equation here, because we have two cues for a So those two processes going on. So remember only colder materials may have more than one Q. So the idea here is that we're gonna expand this to QA one plus Q A two is equal to negative Q B. So now that brings us to the second step, we're gonna replace all our cues with either M cats or delta mls we're gonna talk about what this Delta means in just a second here. So, remember we have two processes QA one and QA to see the idea here, is that this diagonal portion. Remember in these heat versus temperature graphs, you're gonna use Q equals M cats, and then whatever you have horizontal pieces, you're gonna use the Q equals ml the phase change equations. Alright, And the heat on the other hand, for B is just gonna be a Q equals M cat. Right? So it's gonna decrease temperature. Alright, so let's go ahead and plug in some numbers, we know that M Four A. That's the ice is actually gonna be unknown. We're trying to figure out is the amount of ice and kilograms that we have to Put into this problem and notice for all these things to work out. Right? So, the massive ice is actually what I'm trying to find here. So, this is my target variable. The temperature of the ice is going to be negative 20 and so what? That's going to be negative 20. All right. And then the mass of the water is going to be 0.25. And the temperature for the water is going to be uh 15 degrees Celsius. Alright. So, we're gonna go ahead and replace these equations here, QA one remember this QA one just becomes an M. Cat. This is M. A. Times C. A. Times DELTA T. And then this QA to just becomes a delta M. A. Times L. And this just becomes negative M. B. Times C. B. Times DELTA T. B. So, I just want to point out some things. The sea ice is gonna be our C. A. The C. B. Is gonna be for the the water specifically for water? Now, one question is, are these two things going to be the same? These two masses? And the answer is no. So what's happening here is remember that the face? So that the temperature change for this first part of the ice, it actually has to warm up all the way to 0°C. But then what happens is that not all of the ice melts into water, only half of it does. So the idea here is that if not all the material changes phases, then we actually have to sort of separate these two variables. We're gonna use delta M. For the partial mass that we use inside of our latent heat equation because only some of the material might melt or change phase. And then we're gonna use regular M. For the total mass. And that's what we're gonna use inside of our Q equals M. Cat equations. Alright, so there's sort of two different masses were told here is that we want half the ice to melt. So that means that at this point this delta M. A. Is actually just equal to one half A. So I'm just basically going to replace that variable with one half M. A. Whatever that Emma actually ends up being, we don't know what that is, right, that's our target variable. So let's go ahead and rearrange this. This is gonna be Emma. And I'm just gonna start plugging some stuff in this is 2100. This is uh the temperature change for a. So this is gonna be zero, that's the final minus the initial temperature, which is negative. 20. Be careful with your signs then. Plus. And this is gonna be delta M. Which are gonna replace with one half M. A. And then the latent heats. Remember we're changing from ice to solid. So we have two different latent heats. We have the fusion and vaporization. Remember the fusion is the one that you use when you have solids to liquids like ice to water. So we're gonna use the latent heat of fusion. And that's gonna be this guy over here. That's 3.34 times 10 to the fifth. And then finally on the right side we're going to have is negative 0.25. Uh And this is going to be the C. Four water which is 41 86. And then this final temperature of minus initial is gonna be zero minus 15. Lots of plugging in. So all we really have to do is actually just go ahead and solve for this M. A. And basically we're almost done. So I'm just gonna go ahead and start plugging in some numbers. This is going to be M. A. Times 42,000. It's 42,000 plus. And this is going to be once you group this together, I have one half that can multiply into this 3.34 times blah blah blah. And I'm gonna get 1.67 times 10 to the fifth. And then we're just gonna have on this side we're gonna have this negative is actually gonna cancel that with this negative over here because you're gonna have 0 -15 and what you're gonna end up with is um you're gonna end up with uh this is going to be 15 98. Alright So let's see basically what we end up with over here is we're gonna end up with M. A. Times 209. Um When you when you add these two things together, you're actually gonna get let's see we rearrange this, this is gonna be two oh 9000 M. A. Equals 15 6 98. So if you go ahead and work this out you're gonna get is that M. A. That's massive ice that you need is 15 6 98 divided by 20. 9000. And it's going to be 0.0 75 kg of ice. That's how much ice you need in order for these two things to happen here. Alright guys, so that's it this one let me know if you have any questions and I'll see you in the next one.
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