Latent Heat & Phase Changes - Video Tutorials & Practice Problems

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concept

Latent Heat & Phase Changes

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Hey guys. So in previous videos we talked about the specific heat and how that heat dealt with temperature changes in this video, we're gonna talk about latent heats and how those deal deal with phase changes in the material. Now, a lot of problems are going to combine specific heats and latent heat. So I'm also gonna show you a step by step process to navigate these kinds of problems. Let's go ahead and check it out. So let's first talk about the phase of a material. So if you've ever taken another science course, like chemistry phase of the material is really just like the state of matter. We use three commonly in physics solids, liquids and gasses and these phases usually just depend on the temperature of the material. So for example, the compound H20 which we all commonly known as water, water is a liquid form. But if you cool it down, if you put water in the freezer and then it becomes ice. It's the same thing. It's just in a solid form. And if you heat water up enough, it becomes steam still the same compound mouth. And that was just a gaseous form. Now, remember when a material absorbs or loses heats plus or minus Q. It changes temperature like heating or cooling water, but it can also change phase. But the most important thing is they can only do one or the other and not both. To show you what I mean by this, I'm going to show you a graph in which we're plotting the temperature versus heat For water. The idea here is we're gonna have an ice cube that starts at 0° Kelvin and we're gonna see how the temperature responds as we inject more heat into the system. Let's go ahead and check this out. So imagine we have this ice cube that's here at zero and you stick a flame underneath it and you start to heat it up. Well, basically what happens is you're gonna start raising the temperature of that ice. The sort of graph for this is going to look like a straight line and the equation that's going to sort of the equation that this is going to obey is Q equals and Cats, this is just the our temperature change equation. So we're gonna use the C. For ice times delta T. So the idea here is that we're only just dealing with one phase of matter, just ice. And so in these temperature change sort of phases, you're just gonna use Q equals M cat. And what happens here is that the temperature is going to be changing, but the phase remains constant. Still ice. But now your temperature is rising and then eventually gonna get to this one point right here, which is the melting point of water to 73 kelvin or zero degrees Celsius. And what happens here is that actually something different because now what happens is that the temperature is not going to rise anymore. Instead what you're going to notice is that now the ice instead of raising its temperature starts to transform into water. So what's going on here is you've actually hit this part of the diagram which is a phase change. The idea here is that now the temperature actually remains constant, but now the phase is going to be changing. All the additional heat that you're putting into the ice is not going into raising the temperature, it's actually going to breaking the chemical bonds of ice so that this thing can transform into water. So that's the difference between these two and one, the temperature is changing and the phase is constant and the other, it's the opposite, the temperature is constant at the phase is changing. The equation that we use for these phase change. Parts of these diagrams is Q equals M. L. So that's the equation that we use for these things here. This l here is called the latent heat. Some books will also call it the heat of transformation. This is a constant, that's very similar to the specific heat. It just depends on the material per kilogram, but it also depends on the phase change. So it depends on what material you're talking about, like water or mercury or whatever. But it also it also depends on which two states you're traveling between. So we actually have a couple of different names. If we're going from a solid to a liquid, like we were going here, ice to water, we call this the heat of the latent heat of fusion. So this is LF. Alright, so if you keep now injecting more heat basically, what happens is that at this point all of the ice is transformed into water and now you're just gonna have another diagonal section like this. So now the only phase that you have here is water and you're just gonna have another Q. Equals M. Cat equation. This is going to be M. C. For water times delta T. And then what happens is you're gonna hit this other temperature 100 degrees Celsius or 3 kelvin, that's the boiling point of water. Now you have another phase change here, what happens is that all the water stops heating up and now it just becomes steam and so now you have another Q equals Ml Except now this l here, when you're changing from liquid to gas is called the latent heat of vaporization, we call this L. V. Alright, so this is another sort of phase change portion. And then finally what happens is that once all this stuff has turned into steam, now you can actually continue increasing this thing basically on to infinity here. So this is steam and you're gonna have one final Q equals M. Cat equation. Use LSC for steam. Alright, so that's sort of these sort of diagrams work. Now a lot of problems are actually gonna combine specific and latent heat in your problems, you might have one or the other, you might actually even a combination of both. So I'm gonna show you where to actually uh to to sort of step by step process for solving these kinds of problems. So let's take a look at our first one here, we're supposed to calculate the heat required for the following heating processes, assuming we have 400 g of water, that's 0.4 kg. And the first part here, we're just gonna heat this water up from the 80 to 100°. So let's go ahead and take a look at the steps. The first thing we do is always draw out these T vs. Q graphs. These are really gonna help us visualize exactly what it is that we're trying to do and how much heat is gonna be required. So we're trying to go from 8200 degrees Celsius. So basically for water we're going from 0 to 100 so 80 would be similar around here, doesn't actually have to be to scale this kind of just to help us visualize what's going on. So really what happens is we're going from here on the diagram and we're trying to get it to but we're actually not trying to boil the water. So all we're trying to do here is we're trying to go from here to here. So the first thing we're gonna do is draw the T VS. Q. Graph identify your initial and final temperatures and the second thing you're gonna do is you're gonna draw the path from the initial to the final which is what we did here. So here is our T. Initial and here is our T. Final. The next thing you're gonna do is you're gonna write out your Q. Total equation. So the Q total here, the total heat is really just gonna be the heat required to raise the temperature from 80 to 100. Remember the equation that we're gonna use for these diagonal sections here is gonna be Q equals M. Cats. So we're really just gonna use a Q equals M. Cat equation. Um So that's really all there is to it. So we're gonna use Q. Equals delta T. And this is really just gonna be M cats like this. So we actually got all of our constants like this, we've got the mass, we've got the specific heat and the temperature change. So you're just gonna go ahead and plug and chug. So the Q total Is just gonna be the mass which is 0.4 Times 41 86 times the delta T. Which is 20. If you go out and plug this in, you're gonna get 33,500 jewels. And that's the answer to the first part. Let's move on to the second one. Now we're gonna do is we're actually gonna boil exactly half of the water that's already starting from 100 degrees Celsius. Let's go through the steps, we've got a diagram here is our initial temperature, but now we're trying to do is not trying to raise the temperature, we're trying to boil it, that's going to be a phase change. So what happens here is we're trying to boil this water here, we know that this between these two points here, we're gonna have basically a phase change. So but this this this would be um the total heat required to do to boil all of this would be this entire line segment right here. This would be Q equals ml assuming that we had all 400 g of water to boil. But remember we're only gonna boil half of the water. So what happens is we're basically gonna go from here and not all the way to here, We're actually gonna stop exactly halfway in between. So really this is the part that we're kind of working with here, and we're actually gonna ignore all of this other stuff. So the idea here is that the exactly half what that means is that we're only going to boil half of the mass of the water, which is going to be 0.2 kg. Alright, so we've drawn the path where we're traveling to, we're going to use Q. Equals M. L. So that's gonna be our Q. Total equation. So Q. Equals uh this is gonna be Q. For the phase change. And this is just gonna be the M. L. And we're gonna use the latent heat of, but we're going from 100 degrees Celsius. That's gonna be from liquid to gas water to steam. So we're going to use the latent heat of vaporization. Alright, so this Q. Total here is just gonna be uh the mass that we have which is 0.2 times the latent heat, which is going to be 2.256 times 10 to the sixth power. And I just actually just pulled that from this table over here, That's where I got that value from. So if you go ahead and work this out, what you're gonna get is that Q. Is equal to, I get a value of 451,800. Actually no I'm sorry, just 451,000 jewels. That's it. Alright, let's do the last one very quickly. So the last one is gonna be, we're gonna completely boil all of the water, starting from 80°C. So I'm starting from 80 over here and that's gonna be over here but now I want to completely boil it. Remember that point is going to be over here. So if you take a look there's actually two things I need to do the first line segment I need to travel on. Is this one over here and we know that on this equation. We're in this segment here, we're gonna use Q. Equals M. Cats. Then we're gonna actually travel and go on this line. Line segment in which we're gonna boil all of the water. That's going to be Q. Equals M. L. So in this case we actually have heats that are for both of them. It costs some energy in order to raise it to 100 degrees first and then it costs more energy in order to boil all of it so that it turns to steam. All right, so that means that the Q total equation is really just going to be the cue from the delta T. Plus the queue of the phase change. So, I'm going to call this the water too the steam. So, really what we're gonna do is we're just gonna combine our equations Q equals M. Cat plus M times L. V. Right? We have the masses, we have all that stuff. So we're just gonna go ahead and plug this in. So let me go ahead and scoot down real quick. So this is going to be 0.4 times the specific heats, 41 86 times the temperature, which is gonna be 20. And then we're gonna add it to the ml which is gonna be uh 0.4 here, we're using all of the mass, all 400 g of it. And then times the latent heat of vaporization. So 2.256 times 10 to the sixth. Power you go ahead and put all the stuff into your calculator and your Q total the total amount of heat is going to be 9.36 Times 10 to the fifth jewels. And that's the answer, guys. So hopefully this makes sense. This is a very, very important sort of type of problem for you to get really familiar with. So let's go ahead and take a look at some practice problems.

2

Problem

Problem

How much heat must be removed from 0.7 kg of water at 23°C to cool it to 0°C and completely freeze it?

A

-5.31 × 10^{6} J

B

1.67 × 10^{5} J

C

- 1.65 × 10^{6} J

D

- 3.01 × 10^{5} J

3

example

Finding Amount of Water Vaporized

Video duration:

5m

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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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