A copper pot with a mass of 0.500 kg contains 0.170 kg of water, ...

Question

A copper pot with a mass of 0.500 kg contains 0.170 kg of water, and both are at 20.0°C. A 0.250-kg block of iron at 85.0°C is dropped into the pot. Find the final temperature of the system, assuming no heat loss to the surroundings.

Accepted Answer

Hey everyone in this problem. A student put 100 g of metal heated to 150°C and a 250 g copper calorie meter filled with 250 g of water. The initial temperature of the calorie meter and water is 30°C. And were asked what would the final temperature T. F. Of the system be? If the metal specific heat capacity was 498 jewels per kilogram degrees Celsius. And we're told to assume that there is no heat transferred to the environment. Alright. And if there's no heat transferred to the environment, what this tells us is that quantity of heat Q. Of the system, It's going to be equal to zero. Okay. No heat gained or lost. Now, what is Q. System going to be equivalent to what is it composed of? Well, we have to consider the heat cube related to the middle. Did he Q. Related to the copper, the calorie emitter and the heat Q related to the water. All right. What we're going to assume here is that our final with our final temperature, we're not gonna have any phase changes. Okay. We're starting at 30 degrees. We're gonna assume that we're not going above zero degrees Celsius or sorry, above 100 degrees Celsius in order to say boil water or something like that. Okay, So we're gonna start off by assuming that we just have changes in temperature and not phase changes. Okay. And we'll see when we get our final temperature whether or not that assumption was accurate. Okay, so let's start there. So we know that Q system is zero. Okay. And we've just said that in all of these values were considering a change in temperature and when we have a change in temperature, we have the mass M. Okay. This first term we have the mass of the metal. M. Okay, We have the specific heat capacity C. And again in this case of the metal times DELTA T. The change in temperature. And we have the same for the other terms the massive copper specific heat capacity C. For copper, Delta T. The change temperature of copper plus massive water specific heat capacity for water. There's a change in temperature of the water. And let's just add a subscript to these changing temperatures because they are going to be different for different materials. Okay. They're starting at different temperatures. Alright. So filling in the information we know. Okay, we know we have 100 g of metal. Okay, in order to convert from grams to kilograms, recall that we divide by 1000. Okay, so if we divide by 1000 we get 0. kilograms. Okay. The specific heat capacity C of this metal we're told is 500 or sorry, 498 joules per kilogram degrees Celsius times delta T. The change in temperature. Now Delta T. Is going to be the final temperature T. F. Which we're looking for minus the initial temperature and the initial temperature of this metal is 100 and 50 degrees Celsius. And you'll notice that the heat capacity we're given is in jewels per kilogram degrees Celsius. So when we write our temperatures were going to write those in degrees Celsius. Okay. Alright, so that's the first term. Then we have the second term. We have the copper. Okay, now the copper 250 g. Again converting into kilograms. We divide by 1000 250 divided by 1000 gives us 0.25 kg. Okay, Times the specific heat capacity of copper. Okay, this is something you can look up in a table in your textbook or that your professor provided. It's going to be 390 joules per kilogram degrees Celsius. Times a change in temperature. Okay. And the copper has an initial temperature of degrees Celsius. Okay. And the final temperature is T. F. Which we're looking for. So we get T F minus 30 degrees Celsius. Okay. And then our final term we're going to add down below because we're out of room. We have the massive water just like copper is 250 g. Again converting two kg, divide by 1000 250 divided by 1000 gives us 0.2 five kg. The specific heat capacity of water sea. We can look this up 4190 jewels per kilogram degrees Celsius. Times the change in temperature. Okay, this is the same as the copper. Okay, the water and the copper have the same temperature. So we have T F. The final temperature minus the initial temperature of 30 degrees Celsius. Okay. All right. So, what we want to find is T. F. This final temperature and we have three TF terms we see here. So we want to do is we want to simplify, we want to expand, combine like terms. So we get all the T. F. Terms together and all of the number terms on the other side. Six. All right. So if we do that, okay, on the left hand side We're gonna have 49. jewels per degree Celsius Plus 97. jewels per degree Celsius plus 1047.5 jewels per degree Celsius all times TF the final temperature. And these came from this multiplication. So if you look at this 1st 1, 49.8 jewels per degree Celsius. So we had 0.1 kg, we multiplied by 498 jewels per kilogram degrees Celsius, the unit of kilogram cancels. And we're left with 49.8. Okay? And similarly for the others, just with the other values relating to the copper in the water. Alright, And then on the right hand side we're gonna get all of those terms that don't have tfs Okay, we're gonna have 7470 duels Plus 2925 jewels plus 31,425 jewels. And we have just units of jewels here because when we do the multiplication we have kilogram times jewel per kilogram degrees Celsius times degrees Celsius. So the kilogram degrees Celsius divide out and we're left with just a unit of jewel. Okay And so our final temperature T. F. It's going to be equal to the sum of this right hand side, which is gonna be 41, and 20 jewels divided by this left hand side. 1194.8 jewels per degree Celsius. The unit of jewel cancels were left with degrees Celsius. Okay And so that is the final temperature of the system. Okay, it's gonna be a 35°C. Thanks everyone for watching. I hope this video helped see you in the next one.

A copper pot with a mass of 0.500 kg contains 0.170 kg of water, and both are at 20.0°C. A 0.250-kg block of iron at 85.0°C is dropped into the pot. Find the final temperature of the system, assuming no heat loss to the surroundings.

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