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Cooling a Chunk of Iron

Patrick Ford
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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