Hey, guys. So in earlier videos, we saw the equation for heat and temperature change. We used a q equals mcat equation, but that was only just for one object. In this video, I'm going to show you how to solve calorimetry problems. These calorimetry problems basically involve 2 or more materials that are at 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 calorimetry problems and they 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. In our problem, we actually have 2 quantities of water. We have 1 kilogram of water at 20 degrees and the other one we're going to add is at 90 degrees. I've got a diagram here 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 at different temperatures, then there's going to be some heat transfer. Heat's caused by changes in temperature. So therefore, heat flows from hotter to colder. There's some q that gets transferred. And the idea here is that the colder water gains some heat. So this plus q here means that the temperature is going to increase. The hotter water is going to lose some heat and therefore its temperature is going to go down. These things are going to mix together and basically the 20 degrees is going to increase, the 90 degrees 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 2 qs. The idea behind calorimetry 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 2 q's are equal to each other. That's the whole point of these calorimetry problems. So the equation that we use for that is that Q_{A}=−Q_{B}. The heat that's gained by 1 is exactly equal to the heat that's lost by the other. So that's how to deal with these kinds of problems.

Let's go ahead and check out our example here, we're going to 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 going to be 1 kilogram and the temperature is going to be 20 degrees Celsius. The second mass we're going to add is 5 kilograms of water and the temperature is going to be at 90 degrees. When they mix together, they're going to 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 Q_{A}=−Q_{B}. That's the start of basically all of our calorimetry problems. So we're going to start off with this equation. So the second step is we know that these heats are going to produce temperature changes using these Q=mcΔT equations. So therefore, we're just going to replace both of these qs with mcats.

So this Q_{A}, really just becomes m_{A}
cΔT_{A} and this negative Q_{B} becomes negative m_{B}cΔT_{B}. 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 where, in order to do that, we're going to expand out what these temperatures, these delta Ts 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 going to cancel. On both sides, we can just cancel them out. We also know that this m_{A} here is just going to be 1, so it actually doesn't really do anything to our problem.

So what happens here is we're going to expand out these delta Ts. So remember, delta T is always just final minus initial. So delta T here is going to be T_{final}−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 going to have negative m_{B}, right, the c goes away, and this is going to be T_{final}−T_{B} initial. The T_{final} is the same for both of the objects because remember, they're going to mix until they reach some equilibrium temperature. So the T_{final}s are the same for both of them. So all we really have to do here is just go ahead and solve for this TF.

So what I'm going to do here is now I'm going to go ahead and start plugging in some numbers. So I've got my T_{final} minus the initial temperature for this TA here, remember, this is TA initial and this is TB initial. I've got 20 over here. Then I've got the mass, which is 5 kilograms, and we're going to have to distribute this m_{B} into this expression over here. So what we end up getting here is minus 5 T_{final}, and then we distribute this negative sign over here. This could be on the plus sign. This is going to be 5 T, B final. So this is actually going to be 5 times the 90 degrees. So all we have to do now is go ahead and solve for this T_{final}. That's our only target variable. We're going to move this over to the other side, and then we're going to move this negative twenty to the other side. We end up getting here is 6 T_{final} equals, this ends up being, 450, and then we have the plus the 20, so this ends up being 470 over here. So now finally, our T_{final} here is just going to be 470 divided by 6, And if you go ahead and work this out, what you're going to get is 78.3 degrees Celsius. So just as we expected, we got a number that was between 2090.

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 1 kilogram and a 50 kilogram and they're at 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 kilograms 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 2 different masses of materials, but one is colder and one is hotter, the hotter one, that has more mass is basically going to skew the final temperature towards that side. So this T_{final}, this equilibrium temperature is going to be closer to 100. That's why we got something that was much closer to 90 than it was to 20 and it's because we had 5 kilograms of hot water. So that's how to deal with these kinds of problems here. Let me know if you have any questions.