Alright, guys. So welcome back. Let's take a look at this problem together. In this problem, we're going to add some heat energy to water that's already initially at 90 degrees. We want to calculate how much of the water vaporizes. Now remember, the first thing I like to do here is draw the t versus skew graph for water. I'm going to just do that really fast here. It's going to look something like this. Alright. So here's what's going on. So I'm going to add some amount of heat energy. That's going to be my q to some amount of liquid water. So this is going to be my m. My initial temperature here is 90 degrees Celsius, so this is my t initial here. Alright? So what I'm going to do here is mark on the graph where I'm starting off. So this is the boiling point, which is 100 and if I'm at 90, that's a little bit underneath that. So here's 90 degrees. This is my initial. What's going on here is I'm going to add some heat energy. So basically, I'm going to go up the graph like this. And then what happens is when I reach the boiling point, I'm not going to keep increasing in temperature, then the water starts to vaporize into steam. So once you reach this temperature here, you're actually going to start moving along this line. What we want to figure out is how much of the water vaporizes. If we go all the way across then that means that all of the mass has vaporized, but if we only go partially across, we only go sort of halfway, then that means that some amount of the water has vaporized and some hasn't. That's really what we're trying to find here. When we're going across this way, what we're really trying to find here is how much of the water vaporizes. So I'm going to call this mv. Now what's really important here is that this mv is not equal to this m. This m here is not equal to mv, and that's because of the equations that describe or govern these different sections. Remember, we use m cΔT for the diagonals and mL for the flat parts. So we have m cΔT and mL. Now what happens is if you go all the way across, then these 2 m's are going to be the same. But if you only go some of the way across, and that means that only some of the mass has vaporized not the full 0.6. So these things are not going to be the same. Alright? Hopefully, that makes sense. Alright. So then how do we actually figure this out here? Well, we're going to need an equation. We have some amount of heat that is added to this water here, so let's go ahead and start there. So this is really just a total amount of heat that's added to the water. The reason I say total is because you have 2 different steps that are going on here. From the initial here to the final, wherever that is. I've already I don't even know if it's right here. I've got some temperature change and then I've got some phase change here. So I'm going to have to use both of my equations. I'm going to have to use m c for water times Δt to get to the boiling point, plus the mass that vaporizes times the latent heat of vaporization. Remember use Lv for this one and Lf for this one. So really this variable, this mv is what I'm looking for. So let's get started here with what I know. Well, I know that the total amount of heat here is Q = 5.89 *10^5. Now this m here is going to be 0.6. Right? I'm going to use this m here. That's the total amount of mass. The specific heat for water is going to be c = 4186. Now what about the change in the temperature? Well, initially, I'm going from 90 degrees, and then if I hit the phase change, that's going to be at 100. So what happens here is that Δt, my t initial is 90, but my Δt here is going to be 10 degrees. As I'm going up here, I'm changing a temperature of 10. So let me just go ahead and put it right there. Alright? So my ΔT is 10. Now I'm going to add it to this mv here and then the latent heat of vaporization. So this is going to be just from my table over here Lv = 2.256 * 10^6. That's in joules per kilogram. Okay? So we've actually pretty much had all of numbers. All we need to do here is just work our way down to this mv. Alright. So, basically, what happens here is you end up with Q = 5.89 * 10^5, and this is going to equal, actually, when you subtract everything, when you subtract this over, you're going to get -2.5 * 10^4. This is going to equal mv * 2.256 * 10^6. And then what happens is now you just divide by the latent heat. Right? You're just going to move this over. So what this becomes here is this becomes 5.64 * 10^5. You divide this by the latent heat of vaporization 2.256 * 10^6, and that's going to give you the mass that vaporizes. When you work this out what you're going to get is 0.25 kilograms. Notice how this number here is 0.25 is less than the total amount of liquid water that you have. That just means that sort of confirms that we actually didn't vaporize all of the water. We only vaporized only a little bit less than half of it. It was about 0.25 kilograms. So that's the answer there. Hopefully that makes sense. Let me know if you have any questions.
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Latent Heat & Phase Changes
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