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

2

example

Mixing Hot & Cold Water

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4m

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Hey everybody. So welcome back. Hopefully got a chance to try this one on your own. So we're gonna be mixing two cups of water together. So let's get started, I'm gonna draw this out first. So we have one cup of water that's 0.5 kg at 15 degrees Celsius. So I've got a cup of water like this, You know it's 15°C and it's going to be 0.5 kg. Now we're gonna add this to another cup of water except this water is gonna be boiling, right? So we've got another, we've got boiling water at 100°C. So we've got some amount of water, we don't know how much it is. All we know is that it's about 100°C. So it's really really really close to boiling And they were going to add them together so that the final mixture is exactly 80°C. So when you combine these two things in a larger container, you're gonna end up with this amount of water here, something larger. And this final mixture here is gonna be 80°C. Now this is a pretty classic kalorama tree type question because we're gonna be combining two or more materials in a container and they're both going to reach some equilibrium temperature. Alright, so we're gonna go ahead and stick to our steps here. The first thing we're gonna do is write Q. A. Equals negative QB. Basically. What's going on here is I've got the heat that the 15C 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 that's the first step here is writing Q. A. Is equal to negative 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°C. So if you think about what's happening here is this is cup a and this is cup be the mass of this water here is 0.5 kg. And what I'm looking for here is m be the mass of the boiling water. That's really what I'm looking towards. Alright, so let's go move on with the second step. Now, which is we're going to replace the cues with them Cats. Right? That's our cue equals M. Cat equation. EMC delta T. So this is going to be M. A, times C. For water, right? And we're using C. For water, that's 41 86. Just in case you forget it. And then times delta A adult T for a. Now this is equal to negative MB, times C. For water, times delta T for B. So really all I have to do here is just move on to the last step, which is salt for my target variable. And that's this M. B. In this case. So let's go ahead and start plugging in some numbers. So what I've got here. One thing I can do is I can actually just cross off the specific heats for water for 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. So that's the first thing I can do and then I'm just gonna start plugging in some numbers. Right? So this is gonna be 0.5. What about the change in the temperature for the water? Well that's always a final minus initial. So this is gonna be the T final for both of them, whereas this is gonna be the T. A. initial for uh for a. And this is gonna be TB initial for B. So really this is gonna be my final temperature of 80 minus my initial temperature of 15. And I also don't have to convert it to kelvin's because when you're working with Delta Tes remember it could be Celsius or kelvin. So that's the equation on the right side, What we have is negative MB. 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 - the initial of 100. So All I can do now is just go ahead and simplify. Um and let's see what I've got here is when you combine this is going to be 0.5 times this is going to be 65 here And then when you divide this to the other side here, what you're gonna get is you're gonna get negative 20 over like this, except what you also have to do is realize um 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. And that 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 gonna be kilograms. So this is how much boiling wheel water we need. Um 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 much much closer To the 100°C 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

3

concept

Equilibrium Temperature Equation in Calorimetry Problems

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6m

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Hey guys. So in previous videos, we saw how to use the kalorama tree steps to go ahead and solve the target variable, which was the equilibrium temperature of a mixture of materials. We're gonna have to do this a lot in kalorama tree problems. So what I'm gonna do in this video, I show you one last time, how to navigate through these steps here and we're going to come up with is a general equation for the equilibrium temperature. And this is actually gonna work for any number of materials. So it's gonna work for two or three or four or even more most of the time. We're just gonna be working with two or three materials. So that's really all there is to it. I'm gonna go ahead and show you using this example real quick. So we have this massive water, this quantity of water here at 10 degrees Celsius. And then we have, we're 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. A so this is gonna be the massive A, which is the colder one. This is 0.4 kg. 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 gonna 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 gonna see what it's really useful here. So that's we're gonna do in part A. So all we have to do here is actually just go ahead and six of the steps, we're gonna start out with our calorie mystery equation, Q. Is negative QB. And then we're just gonna work out and sort of get towards that final equilibrium temperature. So we have Q A -4 equals negative QB. Remember we're just gonna stick with letters here, it's gonna 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 cues with them cats, right? Those are Q expressions. So this is gonna be M. A. C. A. Times delta T. A. And this is gonna equal negative M. B. C. B. Times delta T. B. Now, just I'm just gonna use the C. A. That's for water and the C. B. That's gonna be for the aluminum and I just have those specific heats right here. So now what I wanna do is I want to calculate the final equilibrium temperature is as I remember those that's gonna be locked up inside of the delta T terms over here. So all we have to do is expand those. So this is going to be M. A. Times C. Eight times. and then remember delta T. Is just final minus initial for both of these objects, their final temperature is gonna be the equilibrium temperature. So it's gonna be T final minus T. A. So then we're gonna have negative M. B. Times C. B. And then again this is gonna be the same thing that's gonna be T final minus T. B. Alright, so now 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 on 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 gonna have to factor everything out. So we're basically gonna have to distribute this and sort of expand this. So this is gonna be M A C A times T A minus M A C A times T A. All right, so I'm sorry, this is gonna be the final that's T. Final. Okay? And then over here we're gonna get negative M B C B T. Final and then we're gonna have a plus because this negative here distributes this minus sign. So it's gonna be plus M B C B T B. All right. So now all we have to do, right, we just expanded everything else. We're just gonna have to group together these T final terms. So what I'm gonna do is I'm going to bring this one over to the other side. So it becomes positive and I'm gonna do the same thing with this term, except I'm gonna move it to the right. So we're basically just trading those two things. So what I end up with is a See 80 final plus MB CBT final equals. And then on the right side I have M A C A T A plus M B C B T B. All right. So now all we have to do here is we just have to sort of isolate this T final notice how they both are involved in those two terms. So we can actually pull those out as a common factor that T final. And what you end up with is a parentheses M A C A plus M B C B. And that's T final equals. And then Emma, I'm actually just gonna go ahead and see if I can copy paste this just to save a little bit of time. Cool, cool. So we can just get this over on the right side. Alright, so now the last thing I have to do is really just divide by this whole entire expression here and then I'll get to the final by itself. So this T final here is just gonna equal and again I have just I'm just gonna copy paste this over here and then we're gonna divide this by and then we're gonna divide this by M A C A plus, M B C B. And that's it. We're done. We got to the finals equals something else. So this is actually the answer to part A. That's the expression which just involves a bunch of numbers here um for the final equilibrium temperature. Now, this actually works just for the two materials that we have involved here, the water and the aluminum. But we can actually sort of generalize this equation to work for any number of materials, whether you have two or three or four, however many you have here, the general equation is going to be, it's going to be M A C A times T A plus M B C B T B plus and then so on and so forth. If you had another third material, you would just keep doing em C times T and then, you know, so on and so forth. Right? And then on the bottom here you're just gonna get M A C A plus MB CB and then plus. And then again, you would just keep on going however many materials that you have. Alright, so this is our equation or this expression for just two materials, let's go ahead and now do part B in part B. Here. We want to actually basically plug in all the values into our expression to find out what that equilibrium temperature is. So we're basically done. All we have to do is just plug and chug. Alright, so what I get here is 0.4. And remember the C. A. Here is gonna be the specific heat for the water. So this is gonna be 41 86. We're pretty familiar with that one times the initial temperature which is 10. And then for the aluminum it's gonna be 0.2. This CB here is gonna be the specifically for aluminum which we're told is 900 then the initial temperature is 80 degrees. So you're gonna multiply all that stuff. Just remember, do it carefully. You got really just two terms here that you can plug into your calculator And on the bottom you're just going to get 0.4 uh times 41 plus this is gonna be 0.2 times. And then this is gonna be the 900. Alright, so it's kind of tedious. That's palatable. But it's actually very straightforward if you go out and plug all of this in, what you're gonna get is a final equilibrium temperature of 16.7 degrees Celsius. Alright, so that's it for this one, let me know if you have any questions

4

example

Pouring Hot Water in an Aluminum Cup

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4m

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Hey guys, So hopefully 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 g of water at 35 Celsius. You're gonna pour it into a cup as cup is 65 g with 11 degrees Celsius. Now, what happens is we're gonna mix these two things together and it's going to reach some final equilibrium temperature. So we know this is a classic kalorama tree problem. You're mixing two things together until they reach some equilibrium temperature. So I can kind of draw this out here. What happens to have some of that out of water? This is 35 degrees Celsius and we have 150 g 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 g. Right? So this is the water and this is the aluminum. You mix them together. And basically what you end up with is just a cup that's filled with water. And the final temperature is what you're looking for. All right, so we know this is a calorie mystery type problem, all we have to do is start off with these steps, we're just gonna write our Q equals negative Q. Equation. However, since we already know what type what what target variable we're looking for the final temperature. We can actually just skip the rest of the steps and we're actually gonna go right to this equation over here. That's the equilibrium temperature equation. So what happens here is we have the Q. A. L. Equals the negative Q. Of the water, right? The aluminum gain some heat, the water loses some heat. And the final temperature is gonna be somewhere in the middle between the two initial temperatures. It's gonna be somewhere between 11 and 35. So just go ahead and shortcut all of that. I'm just gonna 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 gonna go do M C. T. M C. T divided by M. C. Plus EMC. So this is going to be m aluminum, See aluminum and then the temperature of the aluminum plus the m of the water. Sea of the water and then the t of the water. So you're gonna do that and then you're gonna divide this by the M. Of the aluminum C. Of the aluminum without the teas plus m of the water. Sea 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 this is the mass of the aluminum. Now this is given a 65 g but it's really important that you convert this. So this is equal to 0.065 kg. So that's what we have to plug in, 0.065. And 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 temperature is much easier than water does. That's 9 10 and the initial temperature here is going to be the 11°C. It actually doesn't really matter in this case whether you use kelvin or Celsius, because remember that this T final here actually comes from a delta T. So it's actually okay if you use Celsius or kelvin, you'll still get the right answer. Alright, we're gonna add this to the mass of the water. Now, just as we converted this will have to convert this to 0.15, I'm sorry, 0.15 kg. That's water. So this is going to be 0.15. The specific heat for water is 41 86 And the initial temperature of the water is 35°C. Alright, so just kind of a bunch of plugging and chugging. So this is the mass of the aluminum, 0.065 times plus 0.15 times 86. So just really just do this kind of carefully. You can do the top part first, then the bottom part, but basically what you'll end up here with you and you 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 g. And also the specific heat of water is greater than aluminum. So the number should have been way closer to 35 than it was 2 11. Anyway, so that's it for this one. Guys, let me know if you have any questions.

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