A laboratory technician drops a 0.0850-kg sample of unknown solid material, at 100.0°C, into a calorimeter. The calorimeter can, initially at 19.0°C, is made of 0.150 kg of copper and contains 0.200 kg of water. The final temperature of the calorimeter can and contents is 26.1°C. Compute the specific heat of the sample

Video transcript

Hey everyone in this problem, we're told that the temperature of water in a styrofoam cup increases from degrees Celsius to 33 degrees Celsius. When a 10 g cylinder of unknown metal at 430 degrees Celsius is dropped in the water. The massive water is 250 g. The mass of the styrofoam cup is 8.1 g. And its specific heat capacity is 1131 joules per kilogram Calvin. And we're going to assume the cup is a perfect insulator and were asked to calculate the specific heat capacity of the unknown metal. All right. So, we have some water and a cup. We're gonna drop a ball into that water or a cylinder, I guess of metal into that water. It's going to change the temperature in that cup. And we're gonna try to figure out the heat capacity of that unknown matter. All right. So, when we're talking about relationships, we have information about temperature change. We have information about mass. We have information about specific heat capacities. Okay, We're going to think about Q. The quantity of heat. Now, let's recall because we have energy conserved. The quantity of heat Q. For the system is going to be equal to zero. Well, what is the quantity of heat for the system consist of? Well, the quantity of heat for the system Q. System, it's gonna consist of the heat related to the water. The heat related to the cup itself and the heat related to the metal and this is important. Not to forget, the cup, that can be something that's easy to forget. But don't forget the heat associated with that cup. Okay. Alright. Let's start with the water. What's going to happen to the water while we're told that the water warms from 30°C to 33°C. Okay so the water is going to warm But it's not gonna undergo a phase change from 30°C to 33°C. It's going to remain water. And so when we're looking at the quantity of Q for the water, we're only gonna have a change in temperature in the heat associated with a change in temperature, it's going to be equal to M. C. Delta T. And in this case we have the mass M. Of water. See the specific heat capacity of water and delta T. The change in temperature of the water. We know these values the massive water we're told is 250g, 250 g. We want to put this into kg. So we divide by 1000 and we get 0.25 kg. Hey, the specific heat capacity for water, this is something you can look up in a table in your textbook or that your professor provided. It's gonna be 4190 jewels per kilogram Calvin. Okay and then the change in temperature. Well this is going to be the final temperature minus the initial temperature. Okay, the final temperature is 33°C. We're gonna add 273.15 to convert this into Calvin. Okay, we have per kilogram kelvin here in our specific heat capacity. So we want to make sure we have the same unit for temperature when we're talking about delta T. And then we're going to subtract the initial temperature which was 30°C. and again add 273.15 Catherine. When we work this out, we're going to get 3142.5 K in our unit. Well, the unit of kilogram is going to cancel. The unit of Calvin will cancel we're left with just a unit of jewel, which is a unit we want for heat. So our units do work out there. Alright. So we figured out what's going on with the water. Now, let's do the same for the cup. Now, the cup we're told is a perfect thermal insulator. Okay, And what that means is it's going to be in thermal equilibrium with the water inside of it. Okay. And so it's gonna warm the exact same way that the water does. Okay. It's gonna warm by the same amount. Okay, so again, we have a change in temperature and when we have a change in temperature, the heat Q. It's going to be equal to m in this case m the cup. See see the specific heat capacity of the cup and delta T. Let's give ourselves some room to work this out. Now, the mass of the cup, we're told that the cup is 8.1 g. Okay, we want to divide that by 1000 to put this into kilograms. So we get 0. kg. Okay, we have this specific heat capacity for the cup which is styrofoam, which we're told is 1131 with a unit jewel per kilogram Calvin. And then we have a change in temperature and again, because it's in thermal equilibrium with the water because it's a perfect insulator, we're gonna have the same change in temperature that we did for the water. So we're going to have 33 degrees Celsius plus 273.15 to convert to Calvin minus the initial temperature 30 degrees Celsius. And again plus 273.15 to convert to Calvin. And when we work this out, this is going to give us a Q value for the cup. 27. jules. Alright, so we've got water. We got the cup. Now we're onto the last thing we need to consider which is the metal mm No, The metal is starting at a very high temperature. They were told that it has a temperature of 430°C and it's gonna have the same final temperature as everything else. So the metal is going to cool it's gonna go through again, a change in temperature. It is not going to go through a phase change though. And so Q. For the medal, it's going to be the same. That's the other equations we've used. Okay, it's going to be the mass M. In terms of specific heat capacity, C times DELTA T. The change in temperature. Okay. But in this case we get the mass M of metal, we get see metal, then we get delta T. And I'm just gonna put em here because the change in temperature is different here, we start at a different temperature. Okay, So delta T. Is going to be a little bit different. Alright, what we'll notice is that in this equation we have C metal and recall that's what we're looking for. We want to find the specific heat capacity for this metal, this unknown metal that we're putting into our cup. Um And so it's gonna be there. Alright, so we are told that the metal is g. Again, we want to put this into kilograms. So we're going to divide by 1000. We're gonna have a mass of 0.1 kg. We have C. Metal which we're trying to find And now the change in temperature of the metal. Now the final temperature is gonna be the same as the final temperature for the cup in the water. 33°C. We had 273.15 to convert this to Calvin. And then we're gonna subtract the initial temperature which we're told is 430°C. And again we add 273.15 to convert to Calvin. And when we multiply this we're gonna have negative 3.97. Okay. And what is our unit? Well we have kilogram times Calvin. So we get kilogram Calvin time C. Metal. Alright so we've got this equation with C. Metal. We have values for Q. W. And Q. Cup. Let's go back up remember where we started? We started by knowing that the heat Q. For the system was equal to zero. And that that was made up of the water. The cup and the metal. Okay so let's use that. We know the Q. System which is equal to Q. The quantity of heat for the water plus quantity of heat for the cup. This quantity of he for the middle Is equal to zero. Hey Let's go ahead and plug in our values for. For so for water we have 3,142. jewels For the couple of .4833 jewels For the metal we have negative 3. kg Calvin Times. See metal. Okay and this is all equal to zero and again we are trying to find. See metal. Alright. So if we rearrange we get see metal Is going to be equal to 3,142.5 jewels plus 27.4833 jewels is gonna give us 3,169. jewels. And we divide by 3. kg Calvin. And this is going to give us a specific heat capacity for the medal of 798.48 in our unit is jules per kilogram Calvin. Exactly what we would want for a specific heat capacity case or units check out which is a good um kind of check to make sure that we did use the right numbers in the right place. Okay. All right. So if we go back up to our answer choices, we see that we have option D. Okay, specific heat capacity C. Medal for that unknown metal is 798.5 jewels per kilogram Calvin. Thanks everyone for watching. I hope this video helped see you in the next one