Solving Calorimetry Problems

by Patrick Ford
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Hey guys. So in earlier videos we saw the equation for heat and temperature change. We used a Q equals M cat equation, but that was only just for one object in this video, I'm gonna show you how to solve kalorama tree problems. These kalorama tree problems basically involve two or more materials that are different temperatures that are mixing together in some kind of container and they're going to do so until they reach the same final temperature. Remember we call that thermal equilibrium. Now there are lots of different variations to these kinds of telemetry problems and there can be kind of tricky. So I want to show you a step by step process for how to get the right answer. Let's take a look here. So in our problem, we actually have two quantities of water, We have one kg of water at 20° and the other one we're going to add is at 90°. I've got a diagram here, it's kind of visualized to kind of visualize what's going on here. So we've got one mass of water that's at 20 degrees and the other one is at 90. Now we know from a previous video that if you mix together materials that are different temperatures and there's gonna be some heat transfer heat caused by changes in temperature. So therefore heat flows from hotter to colder, there's some cue that gets transferred and the idea here is that the colder water gains some heat. So this plus que here means that the temperature is going to increase the hotter water is gonna lose some heats and therefore its temperature is going to go down. So these things are gonna mix together and basically the 20° is going to increase, the 90° is going to decrease until they finally reach some sort of balance some thermal equilibrium temperature. That's going to be somewhere in the middle here. At that point there's no more heat transfer. So how do we actually solve these kinds of problems here? Well, there's a relationship between this Q. These two cues. The idea behind kalorama tree problems is that if the container, the cup that they're in is thermally isolated, which means that it doesn't exchange the heat with the outside world, doesn't gain or lose anything to the outside, then that means that all of the thermal energy inside this cup has to remain conserved. So what that means here is that the heat that is lost by one material, like the hotter water is exactly equal to the heat that is gained by the other material. These two cues are equal to each other. That's the whole point to these kalorama tree problems. So the equation that we use for that is that Q. A. Is equal to the negative of QB. The heat that's gained by one is exactly equal to the heat that's lost by the other. That's how to deal with these kinds of problems. Let's go ahead and check out our example here, we're gonna come back to this in just a second. So we've got these two quantities of water. We've said that the first mass here is gonna be one kg and the temperature is going to be 20°C. The second mass we're gonna add is five kg of water and the temperature is going to be at 90 degrees when they mix together. They're gonna reach some equilibrium temperature. And that's what we want to find here. That's going to be somewhere between 20 and 90. So how do we actually solve these problems? Well, the first step to solving these problems is writing out QA equals negative QB. That's the start of basically all of our calorie mystery problems. So we're gonna start off with this equation Q. A. Is equal to the negative of Q. B. So the second step is we know that these heats are gonna produce temperature changes. Using these Q equals M. Cat equations. So therefore we're just gonna replace both of these cues with them cats. So that's the 1st and 2nd steps here. So this QA he really just becomes M eight times the sea for water times delta T. A. And this negative Q. B becomes negative M. B. C. For water. And then tell to TB. Now the last step is we just have to go ahead and solve for the target variable. So what we're looking for here is we're looking for the final equilibrium temperature here and that's why in order to do that, we're gonna have to expand out what these temperatures these delta teas are. But first we can actually go ahead and simplify this equation a little bit. We know we're dealing with water on both sides of the equation, so therefore the specific heats for both of them are gonna cancel on both sides, we can just cancel them out. We also know that this Emma here is just gonna be one. So it actually doesn't really gonna do anything to our problem. So what happens here is that we're gonna expand out these delta tease. So remember delta T. Is always just final minus initial. So delta T. Here is gonna be T final minus T. A. Initial. And this is really what our target variable is, what is that final equilibrium temperature. Now we expand out for this one, we're gonna have negative MB right, the C. Goes away and this is gonna be t final minus T B initial. The t final is the same for both of the objects because remember they're gonna mix until they reach some equilibrium temperatures, the T finals are the same for both of them. So all we really have to do here is just go ahead and solve with this T. F. So what I'm gonna do here is I'm gonna go ahead and start plugging in some numbers. So I've got my T F minus the initial temperature for this. Ta here, remember this is T initial and this is T B initial is I've got 20 over here and then I've got the mass, which is five kg and we're gonna have to distribute this MB into this expression over here. So we end up getting here is -5, so we end up getting here is minus five TF and then we distribute this negative sign over here, this is gonna be on a plus sign, this is going to be five T B finals, so this is actually gonna be five times the 90 degrees. So all we have to do now is go ahead and go ahead and solve for this T final, that's our only target variable. So we're gonna move this over to the other side and then we're gonna move this negative 20 to the other side. We end up getting here is 60 final equals this ends up being 450 then we have the plus the 20 so this ends up being 470 over here. So now, finally R. T final here is just gonna be 4 70 divided by six. And if you go ahead and work this out, you're gonna get is 78.3°C. So just as we expected, we got a number that was between 20 and 90. So, one analogy I always like to make with these thermal equilibrium temperature problems, is that they're kind of very similar to a previous concept that we've seen before called the center of mass. The idea behind the center of mass is that if you have these two objects, like, let's say, one kg there 50 kg and there are these two different positions. The center of mass gets skewed towards the heavier one. So the center of mass here, if you have one object that's 50 kg at X equals 100 is going to be close to 100 it's going to be somewhere over here because there's so much mass that's concentrated on that side. It's the same idea here with this thermal equilibrium temperature here. If we have two different masses of materials, but one is colder and one is hotter. The hotter one, that has more mass is basically gonna it's gonna skew the final temperature towards that side. So this t final this equilibrium temperature is going to be closer to the one that is heavier and also has the hotter temperature. So that's the final is gonna be closer to 100. That's why we got something that was much closer to 90 than it was 2 20. And it's because we had five kg of hot water. So that's how to deal with these kinds of problems here. Let me know if you have any questions