1

concept

## Calorimetry Problems with Temperature AND Phase Changes

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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.

2

example

## Cooling a Chunk of Iron

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Hey everybody, so welcome back. I've got a great problem. It's gonna pull together a lot of great stuff from kalorama tree. So I want you to imagine that you're a metal worker, you've got a really hot 1.6 kg chunk of iron that's at Celsius, it's piping hot not to cool this thing down very quickly. You're gonna drip some water over it at 20 degrees Celsius. A lot of metal workers do this because water has a really, really high specific heat so it can absorb a lot of energy from that really hot iron and cool down very quickly. So you do this and the iron cools down to 1 30 then all of the water that you dripped, boils away once it makes contact with the iron. So basically we want to figure out this problem is how much water did we drip over the iron? So that's going to be M. W. Over here, right, that's the mass of water. So the first thing I'd like to do in these problems is set up the diagrams for both of these substances. That's what I want to do here. So, I've got the water and the iron right here, remember what this looks like for water? We have the ice portion, the phase change the water portion, the phase change, and then the steam portion, right? And what happens is for iron, iron has a really high melting point of 1500 way hotter than anything. We're gonna work within this problem. So really just only has one sort of part here, the temperature change, it does eventually melt, but just way after anything we'll deal with this problem. So let's get started, we have to figure out the initial and final temperatures. So in the iron graph, I'm just gonna plot 600, it's gonna be somewhere over here. Does't have to be to scale this is gonna be my initial point for the iron. The iron cools down to 1 30 so it goes down the graph and ends up somewhere over here. So the iron really only has one change, it's just doing a temperature change, it's moving down the graph. The water on the other hand is going to start off at 20 degrees Celsius. It's gonna start off somewhere here in the water part of the diagram remember this esteem and this is ice. Now what happens is the water absorbs all the heat from the iron and it starts to warm up. So we're gonna start off here, this is my initial and we're gonna go all the way up until we reach the boiling point because we're told that all of the water starts to boil away, which means that once you get up to the phase change, you're gonna move all the way across because all of the water boils away and you end up over here on the diagram. Alright. So what happens here is that the water has two steps? This first diagonal pieces gonna be Q. W. One, the second one is gonna be Q. W. Two. It's two different steps. So we're gonna need two different equations. Alright, so now that we've done this, it's gonna make the rest of the problem much simpler because the next step here is we're gonna write are calibrated tree equation Q equals negative QB. And we're gonna use one Q per change. So what happens is on the left side, the thing that gains the heat is gonna be the water. So this is gonna be Q. W. And then on the right side we're gonna have negative queue for the iron. Now the w. Here remember it has to change is we have Q. W. One Q. W. Two. So I'm gonna do Q. W. One plus Q. W. Two equals negative queue for the iron. Now the iron only has one change because it's just only doing one uh temperature change like this, Right? So there's only one term. And the next step is we just have to replace our cues with either M cats or delta mls. Remember the rule is M. Cats for the diagonal parts, Delta mls for the flat parts. So this Q. W. One here is a diagonal part. So this is going to be M C delta T. But this is gonna be for water. So this is going to be M. W. C. W. And then delta T. For the W. Then for the flat part over here, we're gonna have delta M. L. And this is the boiling point. So we're gonna use the vapor of latent heat of vaporization. So this is gonna be delta M. For LV. But we use delta when we have when we have like a partial melting or boiling or something like that. Now on the right side we're only going to have A M. C delta T. For this piece is going to be negative M. Uh this is going to be M I C I delta T. I like this. So really what happens is we're gonna have a negative M. I C I delta T. I. Alright, so we want to do really is figure out what's this water, this mass of the water over here. Now, I should also mention that, remember you use the delta M. S. Whenever you go, only sort of part way across the phase change. If you do the entire thing like we have in this problem here, then what happens is this M. And this delta M. Are the same. So all the water heats up and then all the water boils away. So that means that delta M. Is equal to M. So what this means here is we can actually sort of combine these two terms. So we have M. W. Then we have C. W delta T. W plus Mw Lv. So this is gonna make our equation a little bit simpler, like this. Alright, so now what I can do is these are my target variables. So what I can do here is just using algebra trick, right? I've got these two are the sort of greatest common factors of these terms. So I want to combine them. If I add them together, I can do something like this, I could do M. W. And this is going to be C W delta T W plus lv, write if you distribute back, you should get back to this expression over here. So this is gonna be equal to negative M I C I delta T. I. Now all I have to do, the last step here is just to move this to the other side and I can just start plugging in some numbers, but we're basically almost done. The last thing we have to do is just solve for our target variable. Okay, so that means R M. W is gonna be, I got negative now, what's the mass of the iron? So this is gonna be 1.6. Now, the specific heat for the iron is actually given right here as 470s, this is 470. Now, what's the temperature change? Well, basically, it's just final minus initial, This is my final and this is my initial right? The 1:30 in the 600. So basically what I can do is say 1 30 minus divided by And then I just have the specific heat for water, which is 41-86 times the delta T for the water. Now, what happens is the water goes from 20 and then it finally ends up at 100. So this is kind of one of those rare kalorama tree problems where the substances don't necessarily have to end at the same temperature. So really what this is gonna this is gonna be DELTA T. Is just gonna be 100 minus 20. Then we have to do is add the L. V. So this is the latent heat of vaporization 2.256 times 10 to the sixth power. Now just be really careful when you plug this out in your calculator. But this is basically just all the numbers. Just you can just do the first part and then the top part and then the second part. But what you'll end up with here and you'll end up with 0. 36 kg of water. So it's actually not a whole a lot amount of water and that's kind of what we said, right, water has a really high specific heat. So it actually doesn't take a whole lot of it to cool something down very rapidly anyway, so that's about this one guys, let me know if you have any questions

Additional resources for Calorimetry with Temperature and Phase Changes

PRACTICE PROBLEMS AND ACTIVITIES (6)

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- An insulated beaker with negligible mass contains 0.250 kg of water at 75.0°C. How many kilograms of ice at -2...
- A laboratory technician drops a 0.0850-kg sample of unknown solid material, at 100.0°C, into a calorimeter. Th...
- An asteroid with a diameter of 10 km and a mass of 2.60*10^15 kg impacts the earth at a speed of 32.0 km/s, la...
- An ice-cube tray of negligible mass contains 0.290 kg of water at 18.0°C. How much heat must be removed to coo...