20. Heat and Temperature

Latent Heat & Phase Changes

# Finding Amount of Water Vaporized

Patrick Ford

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Alright guys, so welcome back, let's take a look at this problem together. So in this problem we're gonna add some heat energy to water that's already initially at degrees and we want to calculate how much of the water vaporizes. And I remember the first thing I'd like to do here is draw the T. Vs. Q. Graph for water. I'm gonna do that really fast here. It's gonna look something like this. All right, so here's what's going on? So, I'm gonna add some a lot of heat energy. That's going to be my cue to some amount of liquid water. So this is gonna be my m my initial temperature here is 90 degrees Celsius. So this is my T. Initial here. Alright, So what I'm gonna do here is marked on the graph where I'm starting off. So this is the boiling point which is 190, that's a little bit underneath that. So here is 90°. This is my initial what's going on here is I'm gonna add some heat energy. So basically I'm gonna go up the graph like this. And then what happens is when I reached the boiling point, I'm not gonna keep increasing in temperature, then the water starts to vaporize into steam. So once you reach this temperature here, you're actually gonna start moving along this line, 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 masses 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. What we're going across this way, we're really trying to find here is how much of the water vaporizes. So I'm gonna call this Mv. Now, what's really important here is that this M. V. Is not equal to this, M this M here is not equal to Mv. Then that's because of the equations that describe or govern these these different sections. Remember we use em Cats for the diagonals and mls for the flat parts. So um Cats and mls. Now, what happens is if you go all the way across, then these two M. S are gonna be the same. But if you only go some of the way across and that means that this 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? But we're gonna 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, and the reason I say total is because you have two different steps that are going on here from the initial here to the final, wherever that is. 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 gonna have to use both of my equations. I'm gonna have to use M. C. For water times delta T. To get to the boiling point, plus the mass that vaporizes times the latent heat of vaporization. Remember use LV for this one and L. F. 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 5.89 times 10 to the fifth. Now this M here is going to be 0.6 right, I'm gonna use this M. Here, that's the total amount of mass. The specifically for water is gonna be 41 86. 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 delta T, Mighty initial is 90 but my delta T. Here is going to be 10 degrees as I'm going up here. I'm changing a temperature of tents, Let me just go ahead and put it right there. Alright, so my delta T is 10 and I'm gonna add it to this M. V. Here and then the latent heat of vaporization. So this is gonna be just from my table over here. 2.25, 6 times 10 to the sixth, joules per kelvin. Okay so we've actually pretty much had all of the 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 5. Times 10 to the 5th and this is gonna equal um actually when you subtract everything, when you subtract this over, you're gonna get minus 2.5 times 10 to the four. This is going to equal Mv Times 2.25, 6 times 10 to the 6th. And then what happens is now you just divide by the latent heat, right? You're just gonna move this over. So what this becomes here is this becomes 5.64 times 10 to the fifth. You divide this by the latent heat of vaporization, 2.50 to 56 times 10 to the sixth. And that's gonna give you the mass that vaporizes. When you work this out, What you're gonna get is 20.25 kg. Notice how this number here, 0.25 is less than the total amount of liquid water that you have. And 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 kg. So that's the answer there. Hopefully, that makes sense. Let me know if you have any questions.

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