Hey guys. In a previous video, we covered linear thermal expansion, which had to do with 1-dimensional objects that were changing temperature. Remember the idea was: if you change the temperature of a metal rod or pole or something like that, then the length also changes, and these equations describe the relationship between the changing temperature and the changing length. In this video, we're going to talk about a very similar concept, something called volumetric thermal expansion. The idea here is the exact same, except now we're just going to apply it to three-dimensional objects like spheres or cubes. So the idea here is that if you increase the temperature of a 3D object, you're going to increase their volume, so their volume is going to increase. Alright, so let's take a look here. The idea is that with linear thermal expansion, we're talking about 1-dimensional objects. So what happens is when you change the temperature, the length increases. That's the only dimension that this thing increased. Now when we're talking about volumetric, we're talking about three-dimensional objects. What happens is if you take a cube or something like that, it has some initial volume and now you're going to increase the temperature, then it's going to expand not just along the length but also the width and the height. It's going to expand in all three dimensions and it's going to change a volume, ∆v. Now the equation that we use for linear thermal expansion was ∆l, and for volumetric, it's going to be ∆v. So really these equations are going to look very similar.

So let's take a look here. The equation for ∆v is going to be:
∆v
=
β
V
₀
∆T
Notice the similarities. We had a coefficient, then we had some initial length. Here we have another coefficient called β and then the initial volume times ∆T. Alright? So go ahead and pause the video. What do you think the equation for V final is going to look like? Well, hopefully you guys realize that these things are also going to look similar as well. V final is just going to be V initial times (1 plus β times ∆T). Right? So it's basically the same exact setup, just some of the letters are different.

Alright, so what you need to know here is that this β is a new coefficient. This β has to do with the volumetric expansion coefficient, whereas α had to do with the linear expansion coefficient. This β is actually equal to 3 times α because if you have the linear expansion coefficient β is going to be the same thing in three dimensions, so it's just going to be 3 times that. I have a couple of examples here. So for example, we have aluminum as α = 2.4 × 10^{¯5} and β is going to be 3 times that. These are actually the actual values for some of these, for some of these materials here.

Alright? So let's go ahead and take a look at our example, that's really all we need to know. So a ball of lead has an initial temperature of 333 K and has an initial volume. So we have that T₀ is equal to here we're actually given it in Kelvin 333, and our V₀ is going to be 50 cm³. Now we want to figure out how much the ball shrinks by how much does the ball shrink when you decrease the temperature. So we're actually looking to find here in part A, actually this is the only part here, is we're actually looking to find what is this ∆V here. Alright. So we're going to decrease the temperature to 303, so this is our T final. This is going to be 303 K. Alright? So we're also told the last thing is that our coefficient of linear expansion, this is going to be α, is going to be 2.9 × 10^{¯5}. So these are all our values here. So what's ∆v? We're just going to use the equation if we're looking for ∆v not V final, then we're just going to use this equation over here. So ∆v is going to be this is β times the initial volume times ∆T. So which variables do I have? Well, I'm looking for ∆v and I have the initial volume. I don't have the β. Remember what I was given is the coefficient of linear expansion, the coefficient of linear expansion, and I'm also not sure what the ∆T is as well. So let's go ahead and find those out. So how do I figure out β? Well, for aluminum, all we know is this coefficient of linear expansion to 2.9. However, what you have to realize is that for the same material, we can always relate β and α together. So β is equal to 3α. So because we're dealing with volumetric expansion, we're just going to do 3 times 2.9 × 10^{¯5}, and your β coefficient is going to be 8.7 × 10^{¯5}.

Alright. So that's the coefficient. Now what about ∆T? Well, how do we get ∆T? Remember we're changing from temperatures; we're changing from an initial temperature of 333, and then our final temperature is going to be 303. What this means here is that ∆T is T final - T initial, which is going to be −30 K. Alright? So this is actually what we're going to plug into this term right here. So that means ∆v is just going to be this is going to be 8.7 × 10^{¯5}. That's our coefficient. Then we have the initial volume. It's okay we actually keep it in centimeters cubed because that just means our answer is going to be in centimeters cubed. So we have 50 cm³ and then we have our temperature of −30.

If you go ahead and plug this in, what you're going to get is —0.13 cm³. That's basically the decrease in volume once you shrink this, once you decrease the temperature of the ball.

Alright, so that's it for this one, guys. Let me know if you have any questions.

# Volume Thermal Expansion - Online Tutor, Practice Problems & Exam Prep

### Volume Thermal Expansion

#### Video transcript

### Expansion of a Hemispherical Dome

#### Video transcript

Alright, guys. So hopefully you tried this on your own. Let's go ahead and work this out together. So we have a geodesic hemispherical dome made out of aluminum. What does that mean? A hemispherical dome is something like you might see in a greenhouse or something like that. So just imagine if you took a sphere and you kind of just cut it exactly in half. Right? So this hemisphere goes down like this. Now what happens is it's made of aluminum, and on a certain day where you have a temperature of negative 10, you measure the radius of this hemispherical dome to be 25 meters. This r is equal to 25. It's half the distance across the entire thing. Then what happens is you're going to measure it again on a warmer day when it's 30 degrees Celsius. Now, because we're working with a metal like aluminum, it basically is going to expand, but it doesn't just expand in one dimension; it expands in all three dimensions because we have a three-dimensional object. So basically what's happening here is that on a warmer day everything has kind of swollen up and expanded like this. So now you can kind of imagine that the dome is going to be shaped like this. Right? I'm obviously just exaggerating this, but basically the radius of this has increased a little bit. And now what happens is that this geodesic hemispherical dome has a little bit more interior space inside of it. So basically that's what we're looking for in this problem. How much more interior space, and what that means here is we're trying to figure out well how much volume did you add to the dome just by increasing the temperature. So that's going to be delta v. What is the change in the volume? That's going to represent how much more space you have inside that dome. So now that we know what variable we're looking for, let's go ahead and get started with our volumetric thermal expansion equations.

We just have one for delta v. So this is going to be ΔV = β ⋅ V 0 ⋅ ΔT .

Now if we're looking for delta v, we just have to figure out everything else on the right side of the equation. Now beta is just going to be our aluminum volumetric expansion coefficients. The initial volume, we actually don't have that. It wasn't given to us in the problem. What about the change in the temperature? Well, let's see. We're going from negative 10 degrees Celsius and then we went to 30 degrees Celsius. So that just means here that the change in the temperature delta T is just equal to 30 minus negative 10. This is the difference which is just 40 degrees Celsius. Now because we're working with delta, again, it doesn't matter if we use Celsius or Kelvin. You could convert this to Kelvin, which you're going to find is that it still just works out to 40.

Alright? So we have what the change in the temperature is. Now all we have to do is just find the initial volume of this hemisphere. So let's go ahead and work that out. Right? So the initial volume of this hemispherical dome. How do we do that? Well, you may remember from geometry or trigonometry or whatever, that if you have a sphere like this, I'm going to draw this out real quick. The volume of a sphere is going to equal 4 thirds pi r cubed, but we don't have a sphere that we're working with. We're working with a hemisphere. So the volume of a hemisphere is just going to be half of that. Right? It's half of 1 sphere. So we can do is we can just basically cut this fraction of 4 thirds in half. So we're just going to do 2 thirds pi r cubed. So that is the volume of the hemisphere. So that just means that your V 0 = 2 3 π ⋅ r 3 , where the initial radius at negative 10 degrees Celsius is exactly 25 meters.

So now we just take this number and we plug it back into this equation over here. So your delta v is just going to equal beta, which is 7.2 × 10⁻µ × the initial volume, which is just 3.27 × 10⁴ and then we're going to multiply by the change in the temperature, which is 40 degrees Celsius or Kelvin, doesn't matter. And then what you're going to get here is the change in volume is 94.2 meters cubed. So that's actually a pretty substantial difference. Basically, what happens is that again, the aluminum has kind of expanded on a hot summer day and therefore you have a little bit more space, a little bit more volume inside of that hemispherical dome.

Alright. So that's it for this one, guys. Let me know if you have any questions.

### Overflowing Mercury

#### Video transcript

Hey, guys. We've got an interesting problem here for you. So we've got a glass flask that is completely filled with mercury. Then, we're going to increase the temperature of both the flask and the mercury, and we're going to try to figure out how much mercury is sort of overflowing and spilling out of the flask. Let me just draw this out for you just to make this really clear for you. So I have this sort of glass flask like this. Right? It's filled with mercury at 0 degrees Celsius. Then what happens is you increase the temperature to 100 degrees Celsius. You have an increase in temperature and things start expanding. Now we have a three-dimensional expansion because we have a three-dimensional object like a glass that holds some liquid in it. So what happens here is that the glass and the mercury both expand. So the question is do they expand the same amount? And the answer is no, because remember the expansion depends on these beta coefficients, the coefficients of volumetric expansion, you'll notice that the one for mercury is actually bigger than the one for the glass. So here's what's going on here. You're increasing from 0 to 100, and what happens is the glass changes in volume by some amount. I'm gonna call that ΔVglass. Now for the same change, 0 to 100, you also have the mercury that starts to change in volume. What happens is this ΔV is gonna be greater than the ΔVglass. So what happens is if this thing is already completely filled with mercury if it expands in more volume, basically you're gonna have mercury that starts to leak out of this container and it starts to spill out over the edges. This is really what we wanna find here. So I'm gonna call this Vspill. That's really what we're looking at here. And what happens is if the volume for the mercury changes more than the volume for the glass, then Vspill is just gonna be the subtraction of those 2. It's gonna be the ΔVmercury minus the ΔVglass. So really simply here, if the mercury increases by 15 but the glass increases by 10 then the amount that spills over is just the difference between them. It's 5. Right? So that's what spills out of the container. So that's really what we've got going on here and because we're looking at these ΔV equations, these ΔV variables, we're gonna be using our ΔV equation for volumetric thermal expansion. So basically, what I have to do is just replace these equations here with, the correct substance. Right? So I've got ΔV(Hg), so this is gonna be the beta for the mercury times the initial volume of mercury. I'm just gonna use Hg. That's the chemical symbol for mercury, times the change in temperature of the mercury, minus the beta for the glass, mine times the Vinitial for the glass, times ΔT for the glass. Right? So we just got the coefficients there. Alright. So basically, what I've got here is I've got the 2 coefficients like this. Now what happens is I need to figure out the initial volumes of the glass and also the change in temperatures. Now if you think about what's going on here is that both of these were initially at 0 degrees and then they both end up at 100 degrees. So these two ΔTs are gonna be the same. This ΔT that we're working with here is just gonna be 100. Now what about the initial volume? Well, the initial volume here is just gonna be 250 for both. If the glass holds 250 centimeters cubed and it's completely filled with mercury, the mercury also is 250 centimeters cubed. So basically, what happens is these two variables on both terms here are actually gonna be the same. And because of that, it actually just makes the equation a little simpler here. So Vspill, which is what I'm looking for here, is actually just going to be Vinitial times ΔT, remember it's gonna be the same for both, times, and this is gonna be beta for Hg minus beta for the glass. All I've done here is I've sort of just grouped together these two variables which are the same and then these 2 sort of, get combined into a parenthesis. Alright. So this Vspill here Alright. So now it's time to just go ahead and start plugging in. Now, normally, what I would do is I would convert this, to meters cubed, but it actually asks us to keep it in centimeters cubed, so I don't have to do any converting. So the volume initial is gonna be 250 centimeters cubed times the ΔT, which is just 100, and now the beta coefficients. The one for Mercury is 1.8×10-4 minus 1.2×10-5. Alright. So you go ahead and work this out and what you're gonna get here is that the Vspill is just equal to 4.2 centimeters. So we got a positive number, which just means that we were right. Some amount of mercury is gonna spill out of the container and the amount that spills out is just this amount, 4.2 centimeters cubed. So hopefully, that makes sense guys. Let me know if you have any questions.

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