An insulated beaker with negligible mass contains 0.250 kg of wat...

Question

An insulated beaker with negligible mass contains 0.250 kg of water at 75.0°C. How many kilograms of ice at -20.0°C must be dropped into the water to make the final temperature of the system 40.0°C?

Accepted Answer

Everyone in this problem, we have a mass of 0.3 kg of water. We add 60°C and it's going to be placed in a massless thermos bottle Were asked how many grams of ice cubes at a temperature of negative 10°C must be poured into the water to get the system's final temperature to 30°C. Alright, So we're given some information about initial and final temperatures or even at some information about um the temperature of the water were given information about putting in ice cubes case we want to find how many grams of ice cubes. That means we want to find a mass. How can we find the mass? Will recall that the mass is related to the quantity of heat. Q. Okay, If we're looking at the equation for Q. We have a mass in there. So let's go ahead and write out what we know about the heat Q. In order to see if that will help us calculate the mass. Alright, so we know that the heat Q. The system, it's going to be equal to zero. Okay, we have energy conserved. Um So the heat is going to be equal to zero. Now the heat of the system is going to be composed of two things. Okay, this is gonna be the heat from the water plus the heat related to the ice. Alright, so let's start with water. Okay, so what happens with the water? We have an initial temperature of the water of 60°C and we have a final temperature of 30°C. Okay, so from 60° to 30° water isn't going to change um states it's not going to go through a phase change but it is going to undergo a change in temperature. So water is going to be cooled and in other words this is a change in temperature. Okay? And so we're we're looking the quantity of heat related to water. Well this is just going to be equal to m the massive water. Okay? And again, we're looking at the quantity of heat for a temperature change. M C delta T. So we have massive water A. C. And we'll put C. W. For water and delta T. All right, well, we know the mass of water. We're told at 0.3 kg. Now, see well for water We can look up this constant and it's gonna be 100 and 90 with a unit of jewels per kilogram health. Now, the change in temperature, it's too big square bracket here. Okay, this is gonna be the final temperature of minus the initial temperature. Okay, the final temperature is 30 degrees Celsius. And we're going to add 273.15 to convert this to Calvin. Okay, we have per kilogram Calvin here. So we want to keep the units consistent. Okay, we're going to subtract the initial temperature which is 60 degrees Celsius. And again add 273.15 to convert this to Calvin. And if we work this out, we're gonna get negative 37,710. The unit of kilogram will cancel, the unit of Calvin will cancel. And we're left with a unit of jewel. Okay. So you figure it out Quantity of heat for the water. Now let's move on to the ice. Okay, what happens to the ice? Well, the ice starts at negative 10°C and we're gonna have a final temperature of 30°C. Okay, so the ice has to warm. Okay. And it's also going to melt because the melting point of water is at 0°C. Okay. So this is going to be a change in temperature and a phase change. Alright, So when we're looking quantity of heat Q. We have to take those into consideration. It's a cue for ice is going to be, we're gonna start with the mass of the ice. Sea ice delta. T just like we did for the water. Okay, we're warming that ice. Okay. And in this case we're going to warm the ice until it gets to the point where it will melt, it's going to undergo a phase change. And so we're going to have the mass of the ice and ice. The latent heat of fusion. Okay, Because we're going from solid to liquid and then once it's melted, it's going to continue to change temperature. Okay? It's going to continue to warm because the final temperature is 30 degrees Celsius. Okay, So that mass that we initially had of ice is going to continue to warm. But in this case we're going to use the C. Value for water because now it has melted case we have the mass of the ice. Seawater, delta T. Okay. Alright, so the mass of the ice, this is what we're trying to find. Right, how many grams of ice cubes do we put in? So this m ice is what we're trying to find. Okay, and we can factor that. It's in every term. So let's go ahead and factor we have em ice and then let's fill in what we know again this constant C for ice, we can look up jewels per kilogram kelvin Time to change in temperature. Now in this case this is taking us from ice until the ice is going to melt. Okay, so the final temperature here is going to be the melting point, which is 0°C. So we're going to have zero and let me put a square bracket in here as well. It's messy. Okay, let me put a big square bracket here as well. So the final temperature again is zero degrees Celsius. Okay, we add 273.15 to convert to Calvin and then we're going to subtract the initial temperature which we're told is negative 10 degrees Celsius. So negative 10 plus 273.15 Calvin. And that takes care of this first part. This first term. Let's move down. So we have some more room. All right. We're gonna add the second part and the second part is gonna be the mass of the ice. So we've factored that out and then it's gonna be the latent heat of fusion L. App which we can look up in the table in our textbook or that your professor provided. It's gonna be 334 times 10 to the three. The unit here is jewel per kilogram Alright. And then the one last term we have a massive ice which we factored out this seawater again, that constant for water. We used in the first part of this Problem. again, you can look that up in your table in your textbook. Okay. Jewel per kilogram Calvin. And then the change in temperature. Now when we're talking about this transition, we're talking about a change in temperature from when the ice melts to its final temperature of 30 degrees. Okay, so this is going to be the final temperature 30 plus to 73.15 to convert to Calvin minus the initial temperature, which is the initial temperature when it melted, which is going to be zero Plus 273.15 Calvin. Okay. Okay. So when we talk about the ice warming up. Okay. We talked about it being warmed and melted. We talk about it warming up. There's 22 parts to that. Okay we have it warming up until the melting point and then once it melts it's going to warm up to that final temperature. Okay. And the reason we break that up so because we have to use different constants for each of those. Okay one when we're dealing with ice one when we're dealing with water. Alright so we work this all out, we're going to have the mass of the ice Times. Okay. Alright so we've found Q. Water we found Q. Ice. Let's go back up and look at what we have here. We know that the total Q. For the system is Q. Water plus que dice que the quantity of heat and we know that that's equal to zero. So let's go back down and set this up. We know that Q. System equals Que ice actually Q. water, let's start with water. Let's cue ice, that's how we wrote it before and we know that this is equal to zero. Okay now queue for water. We found to be negative 37,710 jewels. And we didn't put a unit here on Q. Ice, let me do that right now we have massive ice times all of this in here. We have jewels per kilogram Calvin times Calvin. Okay same thing here. Jewel per kilogram Calvin times Calvin and then we have jewel per kilogram So that's gonna work out to jewel per kilogram. So we have 780, jewel per kilogram. This is going to equal to zero. And so the mass of the ice ice that we're looking for Is going to be equal to we moved the 37,000 to the other side. 37,000 710 jewels Divided by 480,700 jewels per kilogram. This is going to leave us with a unit of kg which is what we want. Okay, can we get 0. kg. Now in the question we were asked how many grams of ice cubes. Okay so we have kilograms. Let's convert to grams. Okay, so we multiply by 1000 to get the mass of the ice in grams is 78. g. Again multiply the kilograms by 1000. And this is what we get. So if we go back up and compare to our answer choices, We see that if we round to the nearest gram, we're going to have answer B we need 78 g of ice cubes in order to get that final temperature to 30°. Thanks everyone for watching. I hope this video helped see you in the next one

An insulated beaker with negligible mass contains 0.250 kg of water at 75.0°C. How many kilograms of ice at -20.0°C must be dropped into the water to make the final temperature of the system 40.0°C?

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