Intro to Calorimetry - Online Tutor, Practice Problems & Exam Prep

Calorimetry involves calculating the final equilibrium temperature when two materials at different temperatures mix. The heat gained by the colder material equals the heat lost by the hotter one, expressed as $Q$_A=-Q_B. For multiple materials, the general equation is $T$_final=m_Ac_AT_A+m_Bc_BT_B+m_Ac_A+m_Bc_B. Understanding these principles is crucial for solving thermal equilibrium problems effectively.

Solving Calorimetry Problems

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_BcΔT_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_finals 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.

Mixing Hot & Cold Water

Hey, everybody. So welcome back. Hopefully, you got a chance to try this one on your own. So we're going to be mixing 2 cups of water together. Let's get started. I'm going to draw this out first. We have 1 cup of water that's 0.5 kilograms at 15 degrees Celsius. So I've got a cup of water like this. We know it's 15 degrees Celsius and it's going to be 0.5 kilograms. Now we're going to add this to another cup of water except this water is going to be boiling. Right? Right. So we've got boiling water at 100 degrees Celsius. So we've got some amount of water. We don't know how much it is. All we know is that it's at 100 degrees Celsius, so it's really really close to boiling, and then we're going to add them together so that the final mixture is exactly 80 degrees Celsius. When you combine these two things in a larger container, you're going to end up with this amount of water here, something larger, and this final mixture here is going to be 80 degrees Celsius. Now, this is a pretty classic calorimetry type question because we're going to be combining 2 or more materials in a container and they're both going to reach some equilibrium temperature. Alright. So we're going to go ahead and stick to our steps here. The first thing we're going to do is write Q_A = −Q_B. Basically, what's going on here is I've got the heat that the 15 Celsius water gains is going to be equal to the heat that the 100 degrees Celsius water loses until they finally reach something that's somewhere in between. So the first step here is writing q_a = −q_b. Now, really what's going on here is I want to work my way towards how much boiling water I need to add so that this mixture is 80 degrees Celsius. So if you think about what's happening here is this is cup a and this is cup b. The mass of this water here is 0.5 kilograms, and what I'm looking for here is m_b, the mass of the boiling water. That's really what I'm looking towards. Alright. So let's move on with the second step now, which is we're going to replace the Q's with mc∆T. Right? That's our Q = m c∆T equation m∆T. So this is going to be m_a × c for water, Right? We're using c for water that's 4186 just in case you forget it. And then times delta a, delta t for a. Now this is equal to negative m_b × c for water × ∆T_b. So all I have to do here is move on to the last step, which is solving for my target variable and that's this m_b in this case. So let's go ahead and start plugging in some numbers. One thing I can do is I can actually just cross off the specific heats for water on both sides of the equation, and I can only do this because it's the same substance. I'm dealing with water and water, so it kind of just goes away from the equation. That's the first thing I can do, and then I'm just going to start plugging in some numbers. Right? So this is going to be 0.5. What about the change in temperature for the water? Well, that's always a final minus initial. So this is going to be the T final for both of them, whereas this is going to be the T_A initial for a and this is going to be T_B initial for b. So really this is going to be my final temperature of 80 minus my initial temperature of 15. Now I also don't have to convert it to Kelvins because when you're working with delta T's, remember it could be Celsius or Kelvin. So that's the equation. On the right side, what we have is negative m_b, and then for delta T_B, what we have is final minus initial. So this is going to be 80. That's still the final for both of them minus the initial of 100. So all I can do now is just go ahead and simplify, and let's see. What I've got here is when you combine this, this is going to be 0.5 times, and this is going to be 65 here. And then when you divide this to the other side here, what you're going to get is you're going to get negative 20 over like this. Except what you also have to do is realize, that there's also another negative sign that's out here. So you can also move this out over here. Basically, what happens is the negative signs are going to cancel out since you end up with something positive. Now it makes sense because we're solving for mass, so we shouldn't get a negative number. So we just should get negative, m_b and this is going to equal 1.625, and this is going to be kilograms. So this is how much boiling water we need. It kind of makes sense that we have this number here that was kind of bigger than this number because the final temperature ends up being much closer to the 100 degrees Celsius water and that's just because there's more of it. Anyway, so that's the answer guys. Let me know if you have any questions.

Equilibrium Temperature Equation in Calorimetry Problems

Hey, guys. So in previous videos, we saw how to use the calorimetry steps to go ahead and solve for the target variable, which was the equilibrium temperature of a mixture of materials. We're going to have to do this a lot in calorimetry problems. So what I'm going to do in this video is show you one last time how to navigate through these steps here, and what we're going to come up with is a general equation for the equilibrium temperature. And this is actually going to work for any number of materials. So it's going to work for 2 or 3 or 4 or even more. Most of the time, we're just going to be working with 2 or 3 materials. So that's really all there is to it. I'm going to go ahead and show you, using this example real quick. So, we have this mass of water, this quantity of water here at 10 degrees Celsius, and then we are going to add a block of aluminum that's at a hotter temperature. So what I always like to do is I call the colder one the A, so this is going to be the mass of A, which is the colder one. This is 0.4 kilograms and the temperature for that is going to be 10 degrees Celsius. The block of aluminum on the other hand, that mass is 0.2 and the temperature of that aluminum is 80 degrees Celsius. So we're going to throw these things in an insulated container and in the first part of the problem we want to derive an expression for the final equilibrium temperature. So we actually just want to go ahead and use a bunch of letters rather than numbers and we're going to see why it's really useful here. So that's what we're going to do in part A. So all we have to do here is actually just go ahead and stick to the steps. We're going to start out with our calorimetry equation Q_A=-Q_B and then we're just going to work out and sort of get towards that final equilibrium temperature. So we have Q_A=-Q_B. Remember, we're just going to stick with letters here. It's going to be a little bit annoying. There's a lot of algebra involved, but it's actually not very complicated. Let me show you. So all we have to do now in the second step is replace the Qs with MCΔTs. Right? Those are our Q expressions. So this is going to be m_ac_aΔt_a and this is going to equal -m_bc_bΔt_b. Now I'm just going to use the c_a, that's for water, and the c_b, that's going to be for the aluminum, and I just have those specific heats right here. So now what I want to do is I want to calculate the final equilibrium temperature. So remember those that's going to be locked up inside of the ΔT terms over here. So all we have to do is expand those. So this is going to be m_ac_a×, and then remember ΔT is just final minus initial. For both of these objects, their final temperature is going to be that equilibrium temperature, so it's going to be T_final minus T_a. So then we're going to have -m_bc_b, and then again this is going to be the same thing. It's going to be T_final minus T_b. Alright? So now what we have to do is basically just isolate this T_final here. That is the equilibrium temperature. We just need to figure out what that expression is and then move everything else off to the other side. It's a lot of algebra, but it's not very complicated. So let me go ahead and walk you through it. So if we want this t_final here that's on both sides of this equation here, we're actually going to have to factor everything out. So you're basically going to have to distribute this and sort of expand this. So this is going to be m_a c_a times t_a minus m_ac_a times t_a. Alright. Oh, sorry. This is going to be t_final. That's t_final. Okay. And then over here, we're going to get -m_bc_bt_final, and then we're going to have a plus because this negative here distributes this minus sign. So this is going to be plus m_b c_b t_b. Alright. So now all we have to do, right, we just expanded everything else. We're just going to have to group together these t_final terms. So what I'm going to do is I'm going to bring this one over to the other side, so it becomes positive, and I'm going to do the same thing with this term except I'm going to move it to the right. So we're basically just trading those two things. So wha

Pouring Hot Water in an Aluminum Cup

Hey, guys. So hopefully, you got a chance to work this one out on your own. Let's check this problem out here. So we have some amount of water, 150 grams of water at 35 Celsius. You're going to pour it into a cup. This cup is 65 grams with 11 degrees Celsius. Now, what happens is we're going to mix these two things together, and it's going to reach some final equilibrium temperature. We know this is a classic calorimetry problem. You're mixing two things together until they reach some equilibrium temperature. So if I can kind of draw this out here, what happens is I have some amount of water. This is 35 degrees Celsius, and we have 150 grams of water. You're going to add this to some cup. The cup itself, the material of the cup, is at 11 degrees Celsius, and this is going to be 65 grams. Right? So this is the water, and this is the aluminum. You mix them together, and basically, what you'd end up with is just a cup filled with water and the final temperature is what you're looking for. Alright? So we know this is a calorimetry type problem. All we have to do is start off with these steps. We're just going to write our q=-q equation. However, since we already know what target variable we're looking for, the final temperature, we can actually just skip the rest of the steps. We're actually going to go right to this equation over here. That's the equilibrium temperature equation. So what happens here is we have the QAl=-qwater. The aluminum gains some heat, the water loses some heat, and the final temperature is going to be somewhere in the middle between the two initial temperatures. It's going to be somewhere between 11 and 35. So just go ahead and shortcut all of that. I'm just going to start plugging stuff into the equation because I've given it to you. Right? So what happens here is this final temperature is going to be according to this format. You're basically going to go do mcTmcT divided by m, c + m c. So this is going to be m Al c Al T m water c water T of the aluminum, plus the m of the water, c of the water, and then the T of the water. So you're going to do that, and then you're going to divide this by the m of the aluminum, c of the aluminum, without the Ts, plus m of the water, c of the water. Alright. So it's pretty straightforward. It's kind of just plugging and chugging a bunch of stuff in. So the final temperature here is going to be Well, let's see. This is, the mass of the aluminum. Now, this is given as 65 grams, but it's really important that you convert this. So this is equal to 0.065 kilograms. So that's what we have to plug in, 0.065. Then the specific heat for the aluminum is 910. Notice how this is pretty small in comparison to water, and that makes sense. Aluminum is a metal, so it doesn't require as much energy to change its temperature. It changes temperatures much easier than water does. That's 910, and the initial temperature here is going to be the 11 degrees Celsius. It actually doesn't really matter in this case whether you use Kelvin or Celsius because remember that this Tfinal here actually comes from a ΔT, so it's actually okay if you use Celsius or Kelvin, you'll still get the right answer. Alright. So we're going to add this to the mass of the water. Now just as we convert this we'll have to convert this to 0.150 kilograms. That's water. So this is going to be 0.150. The specific heat for water is 4186, and the initial temperature of the water is 35 degrees Celsius. Right? So it's just kind of a bunch of plugging and chugging. So this is the mass of the aluminum, 0.065 times, 910 plus 0.150 times 4186. So just really just do this kind of carefully. You can do the top part first and then the bottom part, but basically, what you'll end up here with when you'll plug this in is you'll end up with a temperature of 32.9 degrees Celsius. Notice how what we said was true. It was somewhere in between 11 and 35, and it kind of makes sense that this temperature is much closer to the water because we have more of the water, 150 versus 65 grams, and also the specific heat of water is greater than aluminum. So the number should have been way closer to 35 than it was to 11. Anyway, so that's it for this one, guys. Let me know if you have any questions.