1

concept

## Linear Thermal Expansion

7m

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Hey guys in this video, I'm going to introduce a new concept called linear thermal expansion. Now, this might sound scary at first, but all thermal expansion really means is that if you increase the temperature of most materials, then their size is also going to increase. In other words, they expand. That's why we call it linear thermal expansion. Its expansion due to a change in temperature. We actually see this everyday and everyday life. If you've ever crossed a bridge or walk down a sidewalk, you notice that there's gaps in between them and that's basically to allow these materials to expand on a hot summer day without breaking. So that's basically what has to do with So let's go and check this out here. And this year, we're specifically going to talk about one dimensional objects, like a metal pole or rod, like a thin rod or something like this. And the idea here is that if you change the temperature right, if you change this temperature delta T. Somehow either by cooling it or heating it, then you're also going to change the length, which is delta L. Of the rod. And that's because the length is going to be the most significant dimension of the size of the rod. Let's go and check this out here. The idea here is that if I have a rod at some initial length, l not and some initial temperature. T. Not, then if I increase the temperature somehow, by heating this up, I'm gonna call this delta T. Like this, then I'm also going to increase the length of the rod and this is gonna be delta L. And these two things are related to each other. The change in the temperature is equal to and proportional to the change in the length of the rod. So what this means is that if you double the change in temperature, if you have plus two times delta T, then you're going to double the change in the length, basically this is also going to be double. So there's a 1 to 1 sort of mapping between the change in temperature and the change in length. Now, in some problem, we're gonna have to calculate this delta L here. So, for example, in our problem down here, were given a bunch of information and we're going to calculate how much the length of the rod increases. That's going to be the change in the length. So, let me show you the equation real quickly. The delta L is going to be alpha, times L, knots, times delta T. So one way I like to kind of remember this is that this equation kind of looks like the word a lot. So delta L equals a lot. I don't know. That's kind of how I just remember this. So this uh this greek letter alpha here has to do with the linear, it's called the linear expansion coefficient and it basically is a measure is just a number of how easy it is for objects to expand if you ever need this, it's going to be given to you and its units are going to be one of the kelvin or one of her Celsius. And this will not here is just the initial length of whatever it is that you're talking about. It could be a rod or something like that. All right, so that's basically it. We're gonna get back to this equation here in just a second. So, let's go ahead and start our example. So, we have some information about an aluminum metal rod. So it has a length of exactly 50 m 50 m at a temperature of 20 C. So we have a length and a temperature we have L equals 50. And the temperature of 20 c. And we have the linear coefficient expansion as well. So this linear coefficient is going to be 2.4 for aluminum, uh 2.4 times 10 to the minus five. So, in part a what we're gonna do is we're gonna calculate how the or how much the length of the rod increases. So that is going to be the change in the length of the rod. If you heat it up to 35 degrees Celsius. So, the idea here is that we're going from 20 degrees Celsius and then we're gonna end up at 35 degrees Celsius. So what that means here is that this is actually our t knots and therefore the length that we were given this 50,000,050 m is actually gonna be our original initial length. All right. So, we want to calculate delta L. Remember the equation for this is going to be alpha. So alpha aluminum times L knots, times delta T. Alright, so, going through our variables, we have the aluminum linear coefficients and we also have the initial length of the broad. All we have to do now is figure out what the change in the temperature is. So, if we take a look here, we're told that we're going from 20 degrees Celsius to 35 degrees Celsius. Now, what happens is in most equations, we're gonna use delta T where we ever have to plug in temperature. We're gonna do it in kelvin's. But remember here that the delta T. Is gonna be the same in kelvin and Celsius. So because delta T in kelvin is the same as delta T. And Celsius, that it actually doesn't matter which one we use. If you're given 20 to 35 That's a change of 15. And it would be the same exact thing in Kelvin's. So we can actually just use either one of them in these equations. You can only do this if you're asked for a delta, right? If you're plugging in the delta. So let's go ahead and get started. So our delta L here is going to be we're going to plug in some numbers. This is gonna be 2.4 times 10 to the minus five, then r L not is going to be 50 and then our delta T. Is going to be 15. So if you go ahead and plug this in, what you're gonna get is a delta L. of 0.018 m, which is about 1.8 cm. So obviously you can tell that the distances are actually very small, this is a 50 m metal rod and it only lengthens by less than two cm and you know that kind of makes sense because in every day we don't see things just like magically getting longer and smaller. These distances are actually very short. Alright, so let's go ahead and take a look at now part B and part B. We're asked for the final length of the rod if we continue heating up to another temperature of 50 degrees. So here in this problem, we're actually not calculating the change in the length, we're actually calculating what the final length of the rod is going to be. So that actually brings me to the second equation we're going to use. So we can actually use this relationship that delta L. Remember delta L just means uh delta just means final minus initial to rewrite an equation and solve for this L. F. So I'm just gonna give you the equation real quickly. This is gonna be L notes and this is gonna be parentheses, One plus alpha times delta T. Alright, so really quickly, I can actually show you where this comes from, basically if we rearrange and solve for this L. F. You're gonna get that L final is going to equal L. Notes plus delta L. And then all you have to do is you just have to substitute this first equation. You just have to plug this guy into this delta L. What you're gonna get is el final equals L knots plus. And then you're gonna get alpha L. Not times delta T. So what you can do is you can just sort of get this l not off to one side and you can get L notes, you can sort of factor it out, this becomes one plus alpha times delta T. That's where the really this equation comes from. Alright, so I'm just gonna use this equation whenever you ask for the final length of the rod, not the change so that so to finish things off, we're going to do the el final, Right? So we're gonna look for el final, but now we're actually gonna go from an initial temperature of 20 degrees Celsius to degrees Celsius. Right? So let's go ahead and take a look, we're just going to use our equation. Now, El final is equal to this is gonna be al initial one plus alpha times delta T. So this delta T. Here, if we're going from 20 to 50 remember it doesn't matter if we're in Celsius or kelvin is actually just gonna be 30. So that's really all I need to do. So this is just gonna be equals the initial length. Remember the initial length is just 50. We're not gonna use this length over here, but this is a delta L. We're gonna use the initial length of the rod, it's 2 50 one plus, and then this would be 2.4 times 10 to the minus five. That's the coefficient. And then our delta T. Is going to be 30. So this is gonna be the L final if you go ahead and plug this in, what you're gonna get is 50.36 m. And that is the answer. All right, So let me go ahead and let me know if you guys have any questions. Let's go ahead and get some practice.

2

Problem

On a very cold day at a temperature of –12°C, a power line made of aluminum between two support towers measures exactly 150.56m. You go out on a hot day and measure the power line to be exactly 150.71m. What is the temperature (in °C) outside? The linear expansion coefficient of aluminum is 2.4×10^{-5}.

A

29.5 °C

B

–11.8°C

C

41.7 °C

3

example

## Expanding Steel Measuring Tape

3m

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Hey guys, so let's get started with this problem here. Hopefully took a shot at, on your own. It's a little tricky to kind of understand the language of this problem. And I think this really benefits from a drawing. So let's get started here. So we have some steel measuring tape and it's calibrated for a measurement accuracy at 20°C. What does that mean? So I like drawing stuff out basically what it just means is that if you have a ruler If you were to be at 20°C, The markings on the ruler would show 50 m. Right? So you measure this thing out, I'm gonna call this L ruler. The measurements on the market would be 50-50.000 m. And if you were to actually take a different measurement instrument, like a laser pointer or something and you were to measure out a line, then that would also be exactly 50.0 m. Basically the measurement that it's showing you is exactly equal to the real life measurement of 50.0 m. That's what it means to be calibrated at a certain temperature, Then what happens is that your measuring tape is going to increase your gonna sort of increase the temperature to 40°C, right? So when a copy paste this? So then a 40°C, what happens? Well, the actual distance of 50 m doesn't change, right. If you were to measure with a laser pointer or something like that would still be the same. But what happens is the steel ruler has increased its length a little bit right? It's made of steel and expands a little bit like this. Now, what happens is the measurements, the markings on the ruler don't change though. There's still show exactly 50 m at this distance here. But the real measurement won't be 50.0 m anymore. This is 0.50 50.0. The actual measurements will actually be this line over here. That's really what we're trying to find here. So, this is what the actual distance is and that's what we're trying to find. All right, So, the distance on the ruler still gonna be 50.0. But the l actual is what we're trying to find. All right, So, that's kind of what's going on this problem. Hopefully, that kinda makes sense. Now, which equation we're gonna use? Well, basically what happens is that this l ruler is kind of like our length initial, right? So this is kind of like r L knots. And this l actual is really kind of like R L final, Right? So we have some initial distance, it expands to some final distance and this is what we're trying to find. So because of that, we're actually gonna use this second equation over here. Right? So this guy, So we're gonna have L F equals L knots and then one plus alpha times delta T. Alright, so basically this is el actual equals L. Ruler And then one plus alpha times delta T. Alright so all we have to do here is just go ahead and plug and chug. So we've got el actual equals uh this is gonna be L ruler which is going to be 50.1 plus. Then we've got the 1.2 times 10 to the minus five. That's the linear expansion coefficient for steel which is given to us. And then we have to figure out the delta T. Write the change in the temperature number delta T. Can be in Celsius or kelvin. It doesn't matter which one you use. So basically the difference between 20 and 40 C, this delta T. Here is just 20. Right? That's so you can plug in. So we've got 20 like this. So when you plug all of this in, what you should get you should get 50. m. So what happens here is that at a higher temperature of 40°C, even though your ruler will say 50. The actual distance that it's measuring Is going to be the length of this line which is 50.012. So your ruler is going to be inaccurate because it's calibrated for a certain temperature. That's kind of what's going on in this problem. Hopefully that makes sense and let me know if you have any questions

Additional resources for Linear Thermal Expansion

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