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. It's expansion due to a change in temperature. We actually see this every day in everyday life. If you've ever crossed a bridge or walked on a sidewalk, you notice that there are 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 it has to do with. So let's go ahead and check this out here. In this theory, we're specifically going to talk about 1-dimensional objects like a metal pole or a 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 Δt, somehow either by cooling it or heating it, then you're also going to change the length, which is Δl 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 ℓ₀ and some initial temperature T₀, then if I increase the temperature somehow, by heating this up, I'm going to call this ΔT like this, then I'm also going to increase the length of the rod, and this is going to be ΔL. 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 2 times Δt, then you're going to double the change in the length. Basically, this is also going to be doubled, so there's a one to one sort of mapping between the change in temperature and the change in length. Now in some problems you're going to have to calculate this Δl here. So for example, in our problem down here, we're given a bunch of information and we couldn't 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 Δl is going to be αℓ₀Δt. So one way I like to kind of remember this is that, this equation kind of looks like the word a lot. So Δl = allot. I don't know. That's kind of how I just remember this. So this, this Greek letter alpha here has to do with the linear or it's called the linear expansion coefficient, and it basically is a measure, it's 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 1 over Kelvin or 1 over Celsius. And this ℓ₀ here is just the initial length of whatever it is that you're talking about. It could be a rod or something like that. Alright. So that's basically it. We're going to 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 meters at a temperature of 20°C. So we have a length and a temperature. We have ℓ = 50 and a 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, 2.4 times 10 to the minus 5. So in part a, what we're going to do is we're going to calculate 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 going to end up at 35 degrees Celsius. So what that means here is that this is actually our T₀, and therefore the length that we were given this 50 meters is actually going to be our original initial length. Alright? So we want to calculate Δl. Remember the equation for this is going to be alpha, so alpha aluminum times ℓ₀ times ΔT. Alright? So going through our variables, we have the aluminum, linear coefficient and we also have the initial length of the rod. 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 going to use ΔT. Wherever we have to plug in temperature, we're going to do it in Kelvins. But remember here that the ΔT is going to be the same in Kelvin and Celsius. So because ΔT in Kelvin is the same as ΔT in Celsius, then 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 Kelvins. 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, now get started. So our Δl here is going to be we're going to plug in some numbers. This is going to be 2.4 times 10 to the minus 5, then our ℓ₀ is going to be 50, and then our, ΔT is going to be 15. So if you go ahead and plug this in, what you're going to get is a Δl of 0.018 meters, which is about 1.8 centimeters. So obviously, you can tell that the distances are actually very small. This is a 50-meter metal rod and it only lengthens by less than 2 centimeters. And, you know, that kind of makes sense because in everyday life 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. In 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 Δl, remember Δ just means, final - initial, to rewrite an equation and solve for this ℓℕ. So I'm just going to give you the equation real quickly. This is going to be ℓ₀ (1 + αΔt). Alright. So really quickly, I can actually show you where this comes from. Basically, if we rearrange and solve for this ℓℕ, you're going to get that ℓℕ = ℓ₀ + Δl. 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 Δl. What you're going to get is ℓℕ = ℓ₀ + α ℓ₀ Δt. So what you can do is you can just, sort of get this ℓ₀, off to one side and you can get ℓ₀. You could sort of factor it out. This becomes (1 + αΔt). That's where really this equation comes from. Alright? So you're just going to 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 ℓℕ. Right? So we have we're going to look for ℓℕ, but now we're going to actually go from an initial temperature of 20 degrees Celsius to 50 degrees Celsius. Right? So let's go ahead and take a look. We're just going to use our equation now. ℓℕ = ℓ₀ (1 + α Δt). So, this ΔT here, if we're going from 20 to 50, remember it doesn't matter if we're in Celsius or Kelvin, is actually just going to be 30. So that's really all I need to do. So this is just going to be equals, the initial length. Remember the initial length was just 50. We're not going to use this length over here. Remember, this is a Δl. We're going to use the initial length of the rod, it's just 50. (1 + 2.4 × 10⁻⁵ⁱ Δt = 30). So this is going to be the ℓℕ. If you go ahead and plug this in, what you're going to get is 50.036 meters, and that is the answer. Alright? So let me go ahead and let me know if you guys have any questions. Let's go ahead and get some practice.

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

Linear thermal expansion describes how materials expand in size with temperature increases. For one-dimensional objects like rods, the change in length (ΔL) is directly proportional to the change in temperature (ΔT) and the initial length (L_{0}). The equation is expressed as $\mathrm{\Delta L}=\alpha L\u2080\mathrm{\Delta T}$, where α is the linear expansion coefficient. Understanding this concept is crucial for applications like construction, where materials must accommodate thermal changes without damage.

### Linear Thermal Expansion

#### Video transcript

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}.

### Expanding Steel Measuring Tape

#### Video transcript

Hey, guys. So let's get started with this problem here. Hopefully, you took a shot at it 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. We have some steel measuring tape and it's calibrated for measurement accuracy at 20 degrees Celsius. What does that mean? I like drawing stuff out. Basically, what it just means is that if you have a ruler, if you were to be at 20 degrees Celsius, the markings on the ruler would show 50.000 meters. Right? So you measure this thing out. I'm going to call this *l _{ruler}*. The measurements on the marking would be 50.00 meters. And if you were to actually take a different measurement instrument like a laser pointer or something and you were to measure a line, then that would also be exactly 50.000 meters. Basically, the measurement that it's showing you is exactly equal to the real-life measurement of 50.00 meters. That's what it means to be calibrated at a certain temperature. Then what happens is that your measuring tape is going to increase; you're going to sort of increase the temperature to 40 degrees Celsius. Right? So when I copy and paste this, so then at 40 degrees Celsius what happens? Well, the actual distance of 50.00 meters doesn't change. Right? If you were to measure with a laser pointer or something like that, it 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; it expands a little bit like this. Now what happens is the measurements, the markings on the ruler don't change so they'll still show exactly 50.00 meters at this distance here but the real measurement won't be 50.00 meters anymore. This is point 50.000. 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. Alright? So the distance on the ruler is still going to be 50.000 but the

*l*is what we're trying to find. Alright. So that's kind of what's going on in this problem. Hopefully that kind of makes sense. Now which equation are we going to use? Well, basically what happens is that this

_{actual}*l*is kind of like our length initial. Right? So this is kind of like our

_{ruler}*l*and this

_{naught}*l*is really kind of like our

_{actual}*l*. 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 going to use this second equation over here. Right? So this guy. So we're going to have Lf=Lnots1+αdeltat. Alright? So basically, this is

_{final}*l*equals

_{actual}*l*and then one plus alpha times delta t. Alright? So all you have to do here is just go ahead and plug and chug. So we've got

_{ruler}*l*equals, this is going to be

_{actual}*l*which is going to be 50.000 times 1 plus then we've got 1.2 times 10 to the minus 5. That's the linear expansion coefficient for steel which is given to us, and then we have to figure out the delta T, right, the change in temperature. Now remember delta T can be in Celsius or Kelvin. It doesn't matter which one you use. So basically, the difference between 20 and 40 Celsius, 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 is you should get 50.012 meters. So what happens here is that at a higher temperature of 40 degrees Celsius, even though your ruler will say 50.000, 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.

_{ruler}## Do you want more practice?

More sets### Here’s what students ask on this topic:

What is linear thermal expansion and how is it calculated?

Linear thermal expansion refers to the increase in length of a material when its temperature is increased. The change in length (ΔL) is directly proportional to the change in temperature (ΔT) and the initial length (L_{0}). The formula to calculate this is:

$\mathrm{\Delta L}=\alpha {L}_{0}\mathrm{\Delta T}$

where α is the linear expansion coefficient, which varies for different materials. This coefficient is typically given in units of 1/K or 1/°C. Understanding this concept is crucial for applications like construction, where materials must accommodate thermal changes without damage.

What is the linear expansion coefficient and how is it used?

The linear expansion coefficient (α) is a material-specific constant that quantifies how much a material expands per degree change in temperature. It is used in the formula for linear thermal expansion:

$\mathrm{\Delta L}=\alpha {L}_{0}\mathrm{\Delta T}$

where ΔL is the change in length, L_{0} is the initial length, and ΔT is the change in temperature. The units of α are typically 1/K or 1/°C. This coefficient helps predict how much a material will expand or contract with temperature changes, which is essential for designing structures that can withstand thermal stress.

How do you calculate the final length of a rod after thermal expansion?

To calculate the final length (L_{f}) of a rod after thermal expansion, you can use the formula:

${L}_{f}={L}_{0}(1+\alpha \mathrm{\Delta T})$

where L_{0} is the initial length, α is the linear expansion coefficient, and ΔT is the change in temperature. This formula accounts for the initial length and the additional length gained due to thermal expansion. It is useful in engineering and construction to ensure that materials can accommodate temperature changes without causing structural issues.

Why is it important to consider linear thermal expansion in construction?

Considering linear thermal expansion in construction is crucial because materials expand and contract with temperature changes. If this expansion is not accounted for, it can lead to structural damage, such as cracks in concrete, warping of metal beams, or failure of joints. For example, gaps are often left between sections of sidewalks and bridges to allow for expansion on hot days. By understanding and planning for thermal expansion, engineers can design structures that remain safe and functional under varying temperature conditions.

Can you use Celsius and Kelvin interchangeably in thermal expansion calculations?

Yes, you can use Celsius and Kelvin interchangeably in thermal expansion calculations when dealing with temperature changes (ΔT). This is because the size of one degree Celsius is equal to one Kelvin. Therefore, the change in temperature (ΔT) will be the same whether measured in Celsius or Kelvin. However, this interchangeability only applies to temperature differences, not absolute temperatures.

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