A copper calorimeter can with mass 0.100 kg contains 0.160 kg of ...

Question

A copper calorimeter can with mass 0.100 kg contains 0.160 kg of water and 0.0180 kg of ice in thermal equilibrium at atmospheric pressure. If 0.750 kg of lead at 255°C is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.

Accepted Answer

everyone in this problem, we have an experiment where a student releases 10 pieces of $1 coins heated to 140 degrees Celsius into a thermally insulating container of mass, 50 g and specific heat capacity 420 jewels for kilogram degrees Celsius. The container is filled with a mixture of ice, five g and water 200 g at the same temperature. Each coin has a specific heat capacity of 390 jules kilogram degrees Celsius and a mass of 8.1 g were asked to determine the final temperature. T. And told that we can assume the container does not allow heat exchange with its surroundings. So we're giving information about mass and about specific heat capacities. And we're asked to find information about temperature. So let's recall that we can relate temperature mass and specific heat capacity through the heat Q. Now we're told that the container does not allow heat exchange with the surroundings of conservation of energy. We know that Q. The quantity of heat for the system is going to be equal to zero. Okay, now what is the system consist of? Well, we're gonna have plenty of heat for the system is equal to that of the coins. What's that, of the water in the container? The heat related to the ice in the container And finally the heat related to the container itself. Okay. And we don't want to forget this. The heat related to the container is something that can often be forgotten. So don't forget about the container itself. Okay. And we know that all of these added together is going to equal zero. Alright let's go through each of these and see what we can determine with the information we're given. Um And try to use this equation to solve for that final temperature. T. Okay so starting with the coins the coins are going to undergo a change in temperature. Okay? They have some initial temperature we're told is 100 and 40 degrees Celsius. And they have the final temperature T. That we're trying to find. Okay so they're just going to undergo a change in temperature and when we have a change in temperature quantity of heat Q. Is going to be equal to M. And in this case M. Of the coins K. C. The specific heat capacity. And again in this case due to the coins. Times delta T. The change in temperature. Alright well the mass of the coins, The massive one point we know is 8.1 g. Okay so we have 8.1 g. We want to put this into kg. So let's divide by 1000 to give us 0. 81 kg. That's for a single coin. But we have 10 coins. Okay so the total mass of the coins is going to be that times 10. The specific heat capacity for the coins we're told is 390 jewel per kilogram degrees Celsius case. We have 390 joules per kilogram degrees Celsius. Then we're gonna multiply by delta T. Now because our specific heat capacity is in joules per kilogram degrees Celsius. We're going to have delta T. In degrees Celsius. And the change in temperature is going to be the final temperature, which is what we're looking for. And we've been told to call that T. Okay, subtracting the initial temperature which is 140°C for the coins. All right. So, we worked this out. We're gonna have 31 0.59. Okay. And the unit here this is the portion that's gonna be multiplied to the T. The unit is going to be kilogram and then we have times jewel per kilogram degrees Celsius. Okay, the kilograms will cancel We're gonna be left with jewel per degree Celsius times T. And then we're gonna subtract 0.6 in our unit here kilogram jewel per kilogram degrees Celsius times degrees Celsius. We're just left with jewel. Okay. Alright, so we have Q. For the coins. Now let's move to the next one which is the water. Okay, So what happens to the water. Okay? The water is again going to go through a change in temperature. All right, let's give ourselves more room now because it's going through a change in temperature. We're gonna have the same equation. So we have Q. For the water is going to be M and in this case the mass of the water see again, specific heat capacity for the water times delta. T. The mass of the water we're told is 200 g. To put this into kg. We divide by 1000 and we get 0.2 kilograms. Okay. The specific heat capacity for water, you can look this up in a table in your textbook or that your professor provided, it's 4190 unit jewel per kilogram. And in this case we're going to use degree Celsius and then delta T. Again our final temperature t minus the initial temperature of the water. Now the initial temperature of the water, we're told that we have a container with the mixture of ice and water and at the same temperature. Okay? In order to have ice and water at the same temperature, we have to be at zero degrees Celsius. Okay. And so we have t minus zero degrees Celsius. And when we multiply we get 838. Okay. And similarly to above our unit is jewel per degree C times that final temperature. T. Alright. We've got the coin, the water moving onto the ice. What happens to the ice? Okay, well the ice is going to change um phase. It's gonna go through a phase change. Okay? Because it's sitting at 0° and it's going to warm up, it's going to first go through a phase change and then it is going to go through a change in temperature to reach that final temperature. Okay. And here we're going to assume that all of the ice melts, we're gonna sue the entire mass of ice we have melts. Okay? And if we're wrong in the end, we're going to end up with a final temperature, that's negative. Okay. And we'll know that our assumption was wrong, but for now we're making that assumption. Um and we'll see later that that does work out. All right. So starting with the phase change, when we have a phase change and we're going from ice to water? Well, this is going to be m the mass and in this case the massive ice times, L. F. The latent heat of fusion. Okay. Alright. And then we have to deal with the change in temperature, which is going to be again the massive ice specific heat capacity. C and in this case the specific heat capacity we're using is going to be for water, Hey, why is that? Well, when our ice melts, we're gonna have water. Okay. And then it's gonna be that water that changes in temperature from zero degrees up to whatever our final temperature is. Okay, So, once it melts, we're dealing with water, so, we have the initial mass of the ice um increasing temperature as water. So, massive ice C specific heat capacity of water times delta T. All right. The massive ice, we're giving us five g. Again, we want kilograms. So, we're going to divide by 1000 We get 0. kg. The latent heat of fusion, we can look this up 334 Times 10 to the three. And this is for water jewels per kilogram. Now the second part, the mass of the ice. Again, 0.005 kg. And this is where it matters that all of that ice is melted. Or that assumption we're making. Because all of this mass is going to then change temperature in the specific heat capacity of water. We use this for the previous calculation, 4190 jewels per kilogram degrees Celsius. Many times delta T. The change in water, That's going to be the sorry, the change in temperature. The final temperature. T. Again, what we're looking for minus the initial temperature. And the initial temperature of water or of ice, sorry, is at zero degrees Celsius. Okay, same as water. We know that water and ice are in the same container at the same temperature. It must be zero degrees. And if we work this out, what we get is 1670 jewels. Okay, we have kilogram times jewel per kilogram. So we end up with just jewels. The unit of kilogram cancels plus 20.95 Unit here. Dual degree C times T. Alright, so, we've done the coins. The water. The ice, we have the container left and then we're gonna put it all together. Now the container, what's going to happen to it, it's going to undergo a change in temperature. And because it doesn't allow heat exchange with its surroundings, it's going to be in thermal equilibrium with whatever's inside of it. Okay, so that means that the change in temperature that eight undergoes is gonna be the same as the water ice. Everything else inside. Okay now we have a change in temperature. So the heat is going to be given by em see Okay the specific heat capacity of the container I'm sell titty. Now the mass of the container we're told Is 50 g. Okay, converting into kg. We divide by 1000 we get 0.05 kg. The specific heat capacity of the container we're told Okay, this is 420 jewels per kilogram degrees Celsius and delta T. We're gonna have the final temperature T. That we're looking for Minour the initial temperature of the container. And again because it's in thermal equilibrium it's gonna be 0°C. And this is going to give us 21 the unit of jewels per degrees C. Just like for the other calculations times that final temperature. T. Alright so we have our Q. Values for each portion. Now let's put it together. Now remember Q. The system which is the heat Q. Of the coins. Plus the heat Q. Related to the water quantity of heat cure related to the ice, that's the quantity if he Q related to the container Is equal to zero. Okay so using the numbers we have we get 0.59 and I'm going to use the units here as well. Okay we have jewels per degrees Celsius times T minus 4422. jewels Plus jules per degrees Celsius times T. Plus 1670 jewels Plus 20.95 jewels per degree Celsius times T. Plus 21 jewels per degree Celsius times T. Is equal to zero. Okay so we're just adding up all of those values we found. Okay if we put everything with the T. On one side, everything without the T. On the other side we get .54 jewels per degree C time's up. Final temperature T. That we're looking for Is equal to And 52.6 jewels. Okay. When we divide the unit of jewel will cancel and we're gonna be left the unit of degrees Celsius which we want. Okay and we're gonna get a final temperature t Equal to 3°C. And so that is the final temperature we were looking for. And I just want to mention again we made the assumption that all of the ice melts we said if that wasn't true we would get a final temperature that's negative. What we find is the final temperature is indeed positive. So that assumption was valid. Case. We're okay with that. Alright, so let's go up to our answer choices And we'll see that we have answer be the final temperature. T. is equal to 3°C. Thanks everyone for watching. I hope this video helped see you in the next one.

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A copper calorimeter can with mass 0.100 kg contains 0.160 kg of water and 0.0180 kg of ice in thermal equilibrium at atmospheric pressure. If 0.750 kg of lead at 255°C is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.

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