. A vessel whose walls are thermally insulated contains 2.40 kg of water and 0.450 kg of ice, all at 0.0°C. The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to 28.0°C? You can ignore the heat transferred to the container.

Video transcript

Hey everyone in this problem, we have a mixture of 0.225 kg of ice and 1.2 kg of water in a thermally insulating tank and that has a temperature of 0°C waste steam is exhausted from a cooking pot and is conveyed into the tank. And were asked to determine the massive steam that must be delivered to the tank to generate water at 50 degrees Celsius. We're told to assume that the tank absorbs negligible heat and that all processes occur at atmospheric pressure. Now, because the tank absorbs negligible heat, Okay, we have conservation of energy and the quantity of heat Q. Of the system is going to be zero. Okay, no. Net he is gained or lost. Now, what what? That heat? That Q. System. What is that composed of? Okay, well, the heat quantity of heat of the system is composed of the quantity of heat related to the ice, the quantity of heat related to the water and finally the quantity of heat related to the steam. And we have to consider all three. Alright, let's start with the ice. And if we look at what we have here, we're given some information about mass. Were given information about temperature. We want to find information about mass. And so when we're looking at Q. This is going to help us relate mass and temperature and the specific heat capacity. C. Okay, So, let's start with this information and see if we can figure out that mass that we're looking for. All right, So when we have ice, we're starting at zero degrees. Okay, we're adding steam so that the temperature is going to increase, we know to 50 degrees Celsius. So the ice because it's at zero degrees is first going to melt. Okay, this is going to be a phase change and then it is going to be a temperature change as well. Okay. It's going to warm up. Alright. So, what does this look like in terms of the equation? So, we get Q the quantity of heat for ice. Okay. When we're talking about melting, Okay, this is gonna be mass and in this case it's the ice times L. F. K. The latent heat of fusion. When we're talking about a phase change, this is the equation we use and we're using LF the latent heat of fusion because we're melting We're going from solid to liquid. All right. And then if we're talking about the temperature change again, we're gonna have the mass m. Okay? And then we're gonna have see the specific heat capacity times delta T. Okay. Now, we have to figure out which specific heat capacity we're going to use in this case. Now, the temperature change. Okay, we're starting at 0°C. So that ice is gonna melt first and then it's going to increase up to 50°C. That temperature change happens. It's actually happening with water because ice is melted turned into water and then it's going to keep increasing the temperature. So we're going to use the specific heat capacity of water. Alright, substituting this in. We get 0.225 kg. Okay, the massive ice that we were given. Now we have the latent heat of fusion 334 times 10 to the three jewels per kilogram Plus the mass of the ice. Again. 0.225 kg times Specific heat capacity for water. Now this is a value that you can look up in the table in your textbook or that your professor provided. It's going to be 4190 jewels per kilogram and we can have per kilogram Calvin or we can have per kilogram Celsius. We're getting degrees Celsius. So let's do per degree Celsius. And we have a change in temperature. So the final temperature, 50 degrees Celsius minus the initial temperature which is zero degrees Celsius. All right. And if we work this out, okay, the first time we have kilogram times jewel per kilogram. So, this is gonna give us a unit of jewel. We get 75,150 jewel. Okay? Plus the second we have kilogram times jewel times degrees Celsius divided by kilogram degrees Celsius case with a kilogram and degrees Celsius cancel. And we're left with the unit of jewel again. aN:aN:000NaN 47137. jewels. Okay, Which gives a final quantity of heat for ice. 122,287. jewels. Okay. Alright so we figured out Q. Ice now let's go to water. What happens with the water? Well the water starts at 0°C and then we're rising up to 50°C. Okay so water is just going to heat up. So this is gonna be a temperature change, it's not going to undergo any phase changes. And when we're looking at a temperature change, the quantity of heat is going to be equal to M in this case massive water K. C. The specific heat capacity. And again in this case we're talking about water times delta T. The change in temperature. Alright well the mass is 1.2 kg. This was given in the problem. The specific heat capacity for water again Oops, not kilograms jewels per kilogram degrees Celsius. And again this isn't a table in your textbook or that your professor provided. Okay. And delta T. We have the same delta T as with the ice, we finish at 50 degrees Celsius and our initial temperature is zero degrees Celsius. When we multiply all of this together we get 251, jewels. Okay so we found Q. Ice we found water. What we're left with is Q. Steam. Alright, now the steam, what's going to happen? The steam is going to start at 100°C. Okay and it's gonna initially condense. Okay so this is gonna be a phase change and then it is gonna cool and so this is gonna be a temperature change. Alright, So when we're looking at the equation, when we're talking about a phase change, we're going to have m and it's going to be M. F. Steam, massive steam times L And in this case we're talking about going from a gas to a liquid and so we're going to use the latent heat of vaporization LV. Okay, now, when we did this before with our latent heat of fusion for the ice, the ice was melting. Okay, that means that the ice, the heat is entering the material, the ice is warming up. And so we had positive ml in this case we're condensing steam. So the steam is losing heat, the heat is leaving the material, and so we have a negative. Okay, so we always want to check the sign and it depends on whether the heat is entering the material or leaving the material. Alright, so this accounts for the phase change and we want to account for the temperature change. Okay, so we have em again the massive steam K. C. And delta T. Now, in this case, when we're talking about, see again, the steam condenses first, it's going to condense into water and then it's going to go through the rest of the temperature change from degrees Celsius to 50 degrees Celsius. So the specific heat capacity we use is again C. W the specific heat capacity for water because when that temperature change happens the steam has been changed into water. Alright. Now, m steam the mass of the steam is what we're trying to find. Okay, so let's factor that out. The mass m the steam. I'm just gonna highlight it to indicate that is what we're looking for. The latent heat of vaporization. You can look this up for water negative. 20 2000 sorry, 256 times 10 to the three jewels per kilogram. And again, that's going to be either at a table in your textbook or that your professor provided. And then we have plus K. C. W. Let me actually put a big square bracket here. So we don't mix up. Okay, now, specific heat capacity for water, like we've used in the other calculations jewels per kilogram and we can say degrees Celsius here, times delta T. Now, in this case the final temperature is 50 degrees Celsius. And the initial temperature the temperature that the steam started out as degrees Celsius. Okay, So you have 15 -100. All right, so if we're looking at units we have jewel per kilogram here, we have jewel per kilogram degrees Celsius times degrees Celsius. This is going to leave us with jewel per kilogram as well. And so we get the mass of the steam em steam times negative. And it's 2,465,500 and again the unit jewel per kilogram. Alright, so we found the quantity of heat que for the ice, The water and the steam. Okay, we're looking for the mass of the steam em steam. Let's just go back up for a second and remember where we started again, we're looking for the mass of the steam and we have this heat for the system that we know is equal to zero. Okay, so let's add those things up and set them to zero and see where that gets us. So we know that Q. Ice plus cute water plus Q. Steam Is equal to zero. Okay, the heat quantity of heat for the ice was 122,287. jewels. For water it was 251, jewels. Okay and for steam we have the mass of the steam em steam which we're looking for Times negative. 2,465,500 in the unit there is dual per kilogram okay, and this is all equal to zero. Okay, if we move this over to the other side and keep the numbers on one side, we get 2,465, jewels per kilogram times the mass of the steam em steam is going to be equal to 373,687 0. jewels. Okay, When we divide the unit of jewels will cancel and we're left with, the mass of the steam is equal to 0.1516 kg. Okay, we want this ingram's, the possible answers are in grams. So we want this in grams. We're gonna multiply by 1000 And this is going to give us 151.6 g. And so that is the mass the steam that we were looking for. And if we go back up to our potential answer choices, We see that the mass of the steam that must be delivered to generate that 50°C is C- 151.6 g. Thanks everyone for watching. I hope this video helped see you in the next one.