Learn the toughest concepts covered in Physics with step-by-step video tutorials and practice problems by world-class tutors

Table of contents

20. Heat and Temperature

1

concept

5m

Play a video:

Was this helpful?

Hey guys in a previous video, we covered linear thermal expansion which had to do with one 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 gonna talk about a very similar concept, something called the volume, thermal thermal expansion or volumetric thermal expansion. The idea here is the exact same except now we're just gonna apply to three dimensional objects like spheres or cubes. So the idea here is that if you increase the temperature of a three D object, you're going to increase their volume. So the volume is going to increase. Alright, so let's take a look here. The idea is that with linear thermal linear thermal expansion, we're talking about one dimensional objects. So what happens is when you change the temperature, the length increases, that's the only dimension that this thing increase. Now we're talking about volumetric, we're talking about three dimensional objects. What happens is if you take a cube or something like that, 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 gonna expand in all three dimensions and it's gonna change a volume delta V. Now the equation that we use for linear thermal expansion was delta L. And for volumetric it's gonna be delta v. So really these equations are going to look very similar. So let's take a look here. The equation for delta V is gonna be beta times V naught times delta T notice the similarities. We have a coefficient. Then we had some initial length here, we have another coefficient called beta and then the initial volume times delta T. Alright, so go ahead and pause the video. What do you think the equation for V final is going to look like? But hopefully you guys realize that these things are also going to look similar as well. The final is just going to be the initial one plus beta times delta T. Right, So it's basically the same exact setup is just some of the letters that are different. Alright, so what you need to know here is that this beta is a new coefficient. This beta has to do with the volumetric expansion coefficient. Whereas alpha had to do with the linear expansion coefficient. Alright, and so what you need to know about this beta is that for the same material like aluminum and aluminum, that beta is actually equal to three times alpha. Because if you have the linear expansion coefficient, beta is just gonna be the same thing in three dimensions. So it's just gonna be three times that I have a couple of examples up here. So for example, we have aluminum is 2.4 times 10 to the -5. And then beta is going to be three times that these are actually the actual values for some of these um 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 is an initial temperature of 333 and has an initial volume. So we have that tea not is equal to here. We're actually giving it in Calvin 333 and RV not is going to be 50 and this is going to be cm cubed. Now we want to figure out how much the ball shrinks by 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 delta v here. Alright, so we're going to decrease the temperature to 303 sisters, Artie final is going to be 303 kelvin. Alright, so we're also told the last thing is that our expansion of coefficient of linear expansion. This is going to be alpha is going to be 2.9 times 10 to the minus five. So these are all our values here. So what's delta v? We're just gonna use the equation, we're looking for delta v not the final and we're just gonna use this equation over here. So delta v is gonna be this is beta times the initial volume, times delta T. So which variables do I have? What I'm looking for? Delta V and I have the initial volume. I don't have the beta. Remember what I was given is the linear coefficient um the the coefficient of linear expansion. And I'm also not sure what the delta T is as well. So let's go ahead and find those outs. So how do we figure out beta? Well, uh for for aluminum, all that all that we know is this coefficient of linear expansion? 2.9. However, what you have to realize is that for the same material we can always relate beta and alpha together. So beta is equal to three alpha. So because we're dealing with volumetric expansion, we're just gonna do three times 2.9 times 10 to the minus fifth. And your beta coefficient is going to be 8.7, 10% of the minus fifth. Alright, so that's the coefficient. Now, what about delta T. Well, how do we get delta t. Remember? We're changing from temperatures were changing from an initial temperature of and their final temperature is going to be 303. So what this means here is that delta T is t final minus t. Initial. What you're gonna get is negative, 30 Kelvin. Alright, so this is actually we're gonna plug into this term right here. So that means delta V is just gonna be this is gonna be uh sorry, this is gonna be 8.7 times 10 to the minus fifth. That's our coefficients. 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 centimeters cubed and then we have our temperature of negative 30. Alright, if you go out and plug this in, but you're gonna get is negative 0.13 and again, this is going to be centimeters cubed. That's basically the decrease in volume. Once you shrink this, once you decrease the temperature of the ball. All right, So that's it for this one. Guys, let me know if you have any questions.

2

example

4m

Play a video:

Was this helpful?

Alright guys, so hopefully try this on your own. Let's go ahead and work this out together. So we have a geodesic hemispherical dome made out of aluminum. So what does that mean? Hemispherical dome is something like you might see in a greenhouse or something like that. Um So it's kind of just imagine if you took a sphere and you kind of just cut it exactly in half. Right? So this is sort of like 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 m. This r is equal to 25, it's half the distance across the entire thing. Then what happens is you're gonna measure it again on a warmer day where it's 30°C 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, expands in all three dimensions because we have a three dimensional objects. 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 gonna 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 were 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 gonna be delta v is equal to beta times the initial volume times the change in the temperature. 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 of volumetric expansion coefficients. The initial volume. We actually don't have that. That wasn't given to us in the problem. What about the change in the temperature? Well, let's see the we're going from negative 10°C and then we went to 30°C. So, that just means here that the change in the temperature delta T. It is just equal to 30 minus negative 10. This is the difference which is just 40 degrees Celsius now, because we're working delta. Again, it doesn't matter for you Celsius or kelvin, you could convert this to kelvin, which is what you're gonna find is it still just works out to 40. All right. So, we have what the change in the temperature is. Now what 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 gonna draw this out real quick. The volume of a sphere is going to equal four 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 one sphere. So it's one half of v sphere. So we can do is we can just basically cut this fraction of four thirds in half. So we're just gonna do two thirds pi r cubed. So that is the volume of the hemisphere. So that just means that your v not is just gonna equal two thirds pi times the radius. What's the initial radius at negative 10 degrees Celsius? We're told that it's exactly 25 m. So we're just gonna plug in 25. I don't have any conversions. You cube that what you'll get is three points 27 times 10 to the fourth and that's in meters cubed Alright, so now we just take this number and we plug it back into this equation over here. So, your delta v is just gonna equal beta, which is 7.2 times 10 to the minus five times the initial volume, which is just the 3.27 times 10 to the fourth. And then we're gonna multiply by the change in the temperature, which is 40°C or Kelvin doesn't matter. And then we're doing what you're going to get here is the change in volume is 94.2 m cubed. So that's actually a pretty substantial difference, basically what happens is that again, the aluminum is 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.

3

example

4m

Play a video:

Was this helpful?

Hey guys, we've got an interesting problem here for you. So we've got a glass flask that is completely filled with mercury and then we're going to increase the temperature of both the flask and the mercury and we're gonna 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°C. Then what happens is you increase the temperature to 100°C, you have an increase of 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 expands? 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 coefficient 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, you want to call that delta v glass now, for the same change 0-100, you also have the mercury that starts to change in volume. What happens is this delta v is going to be greater than the delta V for the glass. So what happens is if this thing is already completely filled with mercury, if it expands more volume, basically, you're gonna have mercury that's just to leak out of this container and it starts to spill out over the edges. This is really what we want to find here. So, I'm gonna call this V spill, that's really what we're looking at here. And what happens is if the the volume for the mercury changes more than the volume for the glass, then v spill is just gonna be the subtraction of those two. It's going to be the delta V for the mercury minus the delta V for the glass. So really simply here, if the mercury increases by 15, but the glass increases by 10, then the amount that spills over, It's just the difference between them. It's five, Right? 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 delta V equations, these delta V variables, we're gonna be using our delta V equation for volumetric thermal expansion. So basically what I have to do is just replace these equations here with um the correct substance. Right? So I've got delta V for the mercury. So this is going to be the beta for the mercury times the initial volume of the mercury. I'm just going to use H G. That's the chemical symbol for mercury, times the change in temperature of the mercury minus the beta for the glass? Mine times the the initial for the glass times delta T. For the glass. Right, so we just got the uh the coefficients there. Alright, so basically what I've got here is I've got the two coefficients like this. Now what happens is I need to figure out the initial volumes of the glass and also the change in the temperatures. Now, if you think about what's going on here, is that both of these were initially at zero degrees and then they both end up at 100 degrees. So these two delta teas are going to be the same. This delta T 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 then it's completely fooled. The mercury. The mercury also has 250 centimeters cubed. So basically what happens is these two variables on both of the terms here are actually going to be the same. And because of that actually just makes the equation a little bit simpler here. So V spill, which is what I'm looking for here is actually just going to be the initial times delta T remember, it's going to be the same for both times. This is gonna be beta for H G minus beta for the glass. All I've done here is a sort of just grouped together these two variables which are the same and then these to sort of get combined into a parentheses. Alright, so this v spill here. Alright, so now it's time to 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 cm cube so I don't have to do any converting. So the volume initial is going to be 250 cm cubed Times The Delta T, which is just 100. And now the beta coefficients, the one for mercury is 1.8 times 10 to the - -1.2 times 10 to the -5. All right, so you go ahead and work this out and what you're going to get here is that the V spill is just equal to 4.2 cm. So we got a positive number, which just means that we were right selling out of mercury is gonna spill out of the container. And the amount that spills out is just this amounts 4.2 cm cubed. So hopefully that makes sense. Guys, let me know if you any questions

Additional resources for Volume Thermal Expansion

PRACTICE PROBLEMS AND ACTIVITIES (3)

- A brass rod is 185 cm long and 1.60 cm in diameter. What force must be applied to each end of the rod to preve...
- A steel tank is completely filled with 1.90 m3 of ethanol when both the tank and the ethanol are at 32.0°C. Wh...
- A geodesic dome constructed with an aluminum framework is a nearly perfect hemisphere; its diameter measures 5...

© 1996–2023 Pearson All rights reserved.