A 65 cm^3 block of iron is removed from an 800°C furnace and immediately dropped into 200 mL of 20°C water. What fraction of the water boils away?

Video transcript

Hey everyone. So this problem is dealing with specific heat capacity. Let's see what it's asking a 400 g piece of copper initially at a temperature of 500 degrees Celsius is placed in g of water at 60 degrees Celsius. We are asked to calculate the mass of water that boils away. Multiple choice answers here are a 6.12 times 10 to the third, sorry 10 to the negative three kg. E 5.1 times 10 to the negative three kg. C 8.21 times 10 to the negative three kg or D 9.34 times 10 to the negative three kg. OK. So this is a pretty straightforward conservation of energy problem where we have the heat or Q of the copper plus the heat of the water is equal to zero because they're an eclipse system. We can recall that the heat of the copper is given by equation MC delta T and those are all sub C for copper. Now, the tricky part here is that we are asked to calculate the mass of the water that boils away. So we're dealing with a phase change. So the heat of the water is going to be the mass of the water, specific heat capacity of the water multiplied by delta T of the water. Plus that phase change term, which is the mass of the steam or the water that boils away multiplied by. And so, and that equals zero. OK. And so an L is our latent um heat, constant water, everything the point of ization for water. So when we go through these terms, we are, we have the mass of the copper, you can look up the specific heat capacity, the copper, our delta T terms are interesting. So we, so delta T right is simply T final minus T initial. And so in this phase change, our tea final is actually going to be the same for copper and water because both are going to get to 100 degrees Celsius for the boiling point of water. And then water is going to boil away. So we can write Delta T for copper at T F minus T initial copper or 100 g or sorry, 100 degrees Celsius minus 500 degrees Celsius, the initial temperature of the copper. And then, so that's our Delta tea copper. Our Delta tea water is going to be similar where the final temperature is also 100 degrees Celsius and minus 60 degrees Celsius. And so let's look at our other terms, we have massed the water specific heat capacity of the water. We are looking for solving for this mass of the steam and then like our specific heat capacity, this latent heat term is also something that we could look up a constant for water. All right. So let's plug in these values and solve for our as of steam, a mass of copper. We were given 400 g of the problem. We're going to rewrite that and um standard unit. So that's gonna be 0.4 kg. Our specific heat of copper, we can look that up. That's 386 Jews per kilogram cel degree Celsius. And then our delta T we've calculated as negative 400 degrees C plus our mass of water or that is 0.300 kg multiplied by our specific heat capacity of water. It's 41 86 Jews per kilogram seed and then our um delta of water is 40. Let me see. And then that equals will move that um mask steam and that specific uh or sorry, that latent heat turn to the other side. So that equals negative s multiplied by 2.26 times 10 to the sixth Jews per kilogram. And so when we plug this into our calculator, and we divide to isolate um our massive steam, we get a massive steam equal to 5.1 times 10 to the negative three kg. And that is the answer to this problem. We look at our multiple choice answers and it aligns with answer choice B So that's all we have for this one. We'll see you in the next video.