Ch 19: Heat and TemperatureWorksheetSee all chapters
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Ch 01: Intro to Physics; Units
Ch 02: 1D Motion / Kinematics
Ch 03: Vectors
Ch 04: 2D Motion (Projectile Motion)
Ch 05: Intro to Forces (Dynamics)
Ch 06: Friction, Inclines, Systems
Ch 07: Centripetal Forces & Gravitation
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Ch 09: Conservation of Energy
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Ch 14: Rotational Equilibrium
Ch 15: Angular Momentum
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Ch 18: Fluid Mechanics
Ch 19: Heat and Temperature
Ch 20: Kinetic Theory of Ideal Gasses
Ch 21: The First Law of Thermodynamics
Ch 22: The Second Law of Thermodynamics
Ch 23: Electric Force & Field; Gauss' Law
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Ch 25: Capacitors & Dielectrics
Ch 26: Resistors & DC Circuits
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Ch 28: Sources of Magnetic Field
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Ch 30: Alternating Current
Ch 31: Electromagnetic Waves
Ch 32: Geometric Optics
Ch 33: Wave Optics
Ch 35: Special Relativity
Ch 36: Particle-Wave Duality
Ch 37: Atomic Structure
Ch 38: Nuclear Physics
Ch 39: Quantum Mechanics

Concept #1: Linear Thermal Expansion


Hey guys, in this video we're going to talk about these things called linear thermal expansions. Whenever a substance is heated up, it's temperature is increased, it has a tendency to increase its length and the same is true or sorry the opposite is true for when you decrease the temperature, it has a tendency to decrease its length. Alright let's get to it. Typically substances increase their size right with an increase in temperature. This is known as, let me do this little less sloppily, they increase their size. This is just known as a thermal expansion for obvious reasons, right? It's an expansion due to a change in temperature, thermal expansion. Thermometers exploit thermal expansion of mercury to measure the temperature of a fluid. You stick a thermometer in your mouth, your mouth which is hotter than the room temperature of the thermometer is putting heat into that mercury. That mercury's length is going to increase until your mouth and the thermometer or the mercury are in thermal equilibrium so no more temperature so sorry no more heat's going to flow into the mercury, temperature's constant, length is fixed, you can read the thermometer. For substances that are roughly one dimensional, meaning that they're very very thin compared to how they are long like a thin rod. That's the world's worst thin rod by the way. The expansion is an increase in the substance's length. So we said size but what size are we particularly talking about for roughly 1D objects? Length. And this is known as a linear expansion. Linear, one dimensional. Substances that are not linear, substances that are not roughly one dimensional can also have changes to other dimensions right we said changes in size. What size means depends on the structure of your substance. We also have area expansions for objects that are roughly two dimensional like a flat disk. So if you add heat and increase the temperature, that disc's area is going to change. It's going to increase its area. Let me minimise myself. There are also volume expansions for objects that are three dimensional like a cube. If you have a cube and you add, you increase the temperature then that cube can get bigger. Its volume can increase. Whenever you increase the temperature, you increase the size and what we mean by size depends on whether the object is a one dimensional, two dimensional or three dimensional object. Now substances have a natural resistance to changes in their size. They typically do not want to increase their size some much more so than others. Now we can define something that's called an allowance right I put it in parenthesis, the allowance that a substance has for expansions so the opposite of the resistance. The allowance that a subject has for expansion is known as the expansion co-efficient. So as physicists, we choose not to measure the resistance to increases in length or area or volume increases in size. We choose to measure how easily it is for a substance to increase its size. So for solids, really quickly by the way these are due to let me just read the next bulletpoint here these are due to forces between the particles. So the substance has forces holding the particles together, the stronger those forces are the higher the resistance is so the lower the allowance is or the lower the co-efficient. Solids have very very tight bonds between their molecules that make them up, so they really don't want to stretch all that much to allow a change in size but gases have very very weak bonds between or, I shouldn't say bonds because they're not bound chemically, they're technically associated they're very very weak associations or weak interactions between one another so they allow changes in size very very easily. So the stronger the forces, the stronger the resistance but remember we measure the opposite of the resistance, the allowance, which we call the coefficient. So the stronger the forces, the lower the coefficient, the harder it is to increase the length or the size and the units of any expansion coefficient they're are all the same are inverse temperature so Kelvin inverse and the resistance to linear expansion which we call alpha is known as the linear expansion coefficient for obvious reasons. Now let's consider an object that's roughly linear right like a thin rod that I've shown with an initial length of L not, so we have some initial length. The change in this length which I called delta L, that change due to a linear expansion so due to some increase in temperature is given by Alpha the coefficient that allowance for the change to happen times the initial length times the change in temperature. So if you think about it, this equation makes perfect sense. The higher the allowance, the higher the alpha, the greater the change in the temperature should be, the higher sorry the greater the change in length should be. The higher the change in temperature, the higher the change in length should be. When you add more temperature, you add more energy into that system and it should grow longer. Now it turns out that you need the initial length in here because linear expansion is actually a proportional process. So the length doesn't increase an absolute amount, the length actually increases a proportional amount. If you multiply alpha times delta T it will tell you the percentage increase. So you need to know the initial length you know the absolute increase. If have something that's very very very small like one millimeter, then the increase is also going to be very very very small. But if you have something that's very very long like 10 meters then the increase is going to be much much much bigger than the increase for this tiny object. That increase is proportional so you need to know that initial length as well. And then what happens is you have your initial, you add in some heat, you increase the temperature so the temperature goes up and then you arrive at a final state which has a final length of L that's increased by delta L. So your final length is just L not plus delta L which simplifies to one plus alpha delta T L not. So we have two equations here. If you want to know the change in the length based on the change in temperature, use the first equation. If you want to relate the final length to the initial length, use the second equation. But the equations tell you the exact same amount of information because the second equation is literally just L equals L not plus delta L. So those equations tell you the exact same information but typically if you have problems that want to relate L sorry delta L to delta T, use the first one because it's just easier. If you want to relate L to L not use the second one. Let's do a quick example. A thermometer is designed so that at zero degrees Celsius, the substance in the thermometer is 10 centimeters long. When the temperature rises to 30 degrees Celsius, the substance increases its length to 15 centimeters. What is the expansion coefficient of the substance? So what did they give us? They gave us a change in temperature right here and they give us a change in length right here. So I told you if you want to use a change in temperature and relate it to a change length, use that first equation so delta L is Alpha L not delta T and we absolutely do know L not. If we want to solve for alpha we just have to divide this over and this means that alpha is delta L over L not delta T. So what's delta L, the change in the length? Well it started at 10 went to 15, so the change is 5 but it's 5 centimeters so that's 0.05 meters. You always want to keep everything in SI units so your answer is in SI units. Now what was the initial length? 10 centimeters which was 0.1 meters. I don't you want you guys to rely on this but if you happen to notice you have length in the numerator and length and the denominator so it actually doesn't matter what unit you used as long as they're the same unit they would cancel so we could've stuck with centimeters, we could've used millimeters, we could have used atronomical units if we wanted but for a guide for those of you who are weak problem solvers always stick with SI units because you can never go wrong. Your answer will always be in SI units. So that's what I tend to do because it's a way to ensure that your answer will always be right. Now the change in temperature. We're told that the change in temperature is 30 degrees Celsius. Something that's very very very important to note and I'm going to probably say this over and over and over throughout the subsequent videos. It was something I already mentioned in a previous video on temperature scales. The change in temperature in degrees Celsius equals the change in temperature in a units of Kelvin. For changes in temperature not for absolute temperatures. This is so incredibly important to remember because you need that delta T in Kelvin, you need it in SI units. Since we're given the change in temperature in degrees Celsius, it's 30 degrees Celsius, we can simply say that that's 30 Kelvin right away because of this we know that whatever the change was in Celsius is the same in Kelvin. And that's so incredibly important because you don't have to convert anything. That's not true for Fahrenheit if you were given an Fahrenheit you have to convert it to Celsius or to Kelvin first and then find the difference. If we plug this into your calculator we get 0.017 inverse Kelvin as our answer for that linear expansion coefficient. Alright guys, that wraps up our discussion on linear thermal expansion. Thanks for watching.

Practice: Train tracks are composed of 10 m segments of steel rails. If the temperature outside is 30°C on average, and the temperature in the factory that produces the rails is 20°C, what length must the steel rails be produced at so that the fit perfectly end-to-end when laid outside? The linear expansion coefficient of steel is 1.2 x 10-5 K-1.

Concept #2: Volume Thermal Expansion


Hey guys, We already talked about linear thermal expansion which is thermal expansion applied to objects that were roughly one dimensional but what if objects aren't? What if they're just like a chunk of some material like a cube which is better approximated as a three dimensional object, you can't really consider one dimensional, then we have to talk about volume thermal expansion. Alright and that's what we're gonna talk about now. Let's get to it. Remember, with an increase in temperature comes an increase in the size of a substance. With a decrease in temperature, there's a decrease in the size as well. An increase in size truly means an increase in the volume. So even what we considered those linear, sorry even when we considered thin objects that were long to be linear, we approximate thermal expansions to being linear for substances that were roughly one dimensional like a thin rod. Even though we consider those as linear expansions and only the length changed, actually the width of it also changed. There was also a very very small change in the width. This was still truly a volume expansion. All thermal expansion are really volume expansions. The only thing was that this change in the width was so small compared to the change in the length, that we could just ignore it. We could just say that was basically the same width because the object was basically one dimensional. These expansions in volumes are all obviously referred to as volume expansions, for the obvious reasons. Volume thermal expansion is volume expansion. Just as a substance has a resistance to increases in its length, substances also have resistance to increases in their volume and just like for resistance to increases in their length, resistances to increases in their volume are still due to interactions between particles in the substance. Solids really really don't want to increase their bond length, gases which only have weak associations with one another are fine with increasing the distance between the molecules. And just like with linear expansions, some substances have a large allowance for expansions. We as physicists don't like to talk about the resistance through expansions, we like to talk about the allowance to expansions and this allowance which for volume expansions we give as the Greek letter beta, are known as volume expansion coefficients. As with linear expansion, the units of volume expansion are also inverse Kelvin. The units of any expansion coefficient is the same. They're all going to be inverse Kelvin. Now for a substance with some initial volume V not, the change in the volume delta V due to a volume expansion, due to an increase in the temperature, is going to be in beta V not not delta T. Notice this is the exact same equation as it was for a linear expansion. A linear expansion we said delta L was equal to alpha L not delta T. So it's basically the change in the size is equal to the expansion coefficient times the initial size times the change in temperature regardless of what you're considering size. If it's linear, you use length and you use the linear coefficient. If it's volume, you use volume and you use the volume coefficient but the equations are the exact same and so is the equation for the final volume it's one plus beta delta T V not, which is also identical to that for linear expansion. So what's happening is you're inputting temperature, sorry, you're inputting heat which is raising the temperature. That volume is, stupid program, technical difficulty. That volume is expanding by some delta V and then you are finishing with some final volume V. Let's do a quick example to illustrate this. A glass cylinder of 1 and a half centimeter radius and 10 centimeter height holds 8 centimeters of mercury. If the initial temperature of the mercury is zero degrees Celsius, what temperature does the mercury have to be heated to to reach the brim of the cylinder? The volume expansion coefficient of mercury is this and we want to ignore any thermal expansion of the glass. So first let's draw what's actually happening here. We have a glass, cylindrical container that's holding mercury in it which is going to be green in this case even though mercury is not green mercury is actually silver but this mercury is 8 centimeters high in this glass. So we have 8 centimeters of mercury and the glass itself is 10 centimeters tall, so we have 2 centimeters left over and the radius of this glass is 1 and a half centimeters. We want to know how much temperature is required to increase the volume of this mercury so that it reaches the brim. So the mercury is first occupying some initial volume here and we want to expand it to occupy this new volume so that it occupies the totality of the volume of this glass cylinder. Now we were specifically told to ignore any expansion of the glass that means that as the mercury is expanding we're not considering the glass to expand too because that would mean that this new volume that the mercury has expand into is growing as the glass gets bigger. We can ignore this because the expansion coefficient for glass is much much smaller than the expansion coefficient for mercury, so glass resists changes in size much more than mercury does. So whatever change in volume the glass actually has is going to be very very small, we're just going to ignore it. What we are going to do is we want to find this final temperature based on how much volume the mercury has to occupy to reach the brim. So what we're doing is we're taking a change in volume this new fall in that it has to occupy and we're relating it to a change in temperature. Finding that change in temperature is going to tell us what the final temperature is because we know the initial temperature. So that's what we want to solve for. We want to solve for the change in temperature which will then lead us to the final temperature. This is delta V over beta V not. Remember guys that both of these equations for expansions both of them tell you the same thing because of how they are related, but there's the convenience to which one you want to choose. You want to choose the first one when you're relating a change in volume to a change in temperature, use the second one when you want to relate the final volume to the initial volume. There is that convenience depending on which one you choose. So we don't know the change in volume yet but we have the geometry to figured it out. We don't know the final volume, I'm sorry, the initial volume but we have the geometry to figure it out and we absolutely do know beta. So let's find this change in volume. Well this change in volume is clearly just going to be the volume of that cylinder that's 2 centimeters tall. The volume of a cylinder is always the area of the face of the cylinder, the circular area, times the height of the cylinder. So circular area's pi R squared and the radius is 1 and a half centimeters or 0.015 meters pi R squared. The height of this cylinder is 2 centimeters sorry not 2 centimeters but 0.02 meters. That is the volume that the mercury has to expand into. Plugging this into a calculator we get 1.4 times 10 to the -5 cubic meters. So now we know the change in volume the last thing we need to find is the initial volume and that initial volume is once again going to be the circular area of the face of the cylinder times the height of the cylinder. Now this is the initial volume of the mercury so the height of that cylinder is 8 centimeters, the same radius 0.015 meters squared times 0.08 now and this turns out to be 5.7 times 10 to the -5 cubic meters. Now we know everything we need to solve this problem alright. So this is going to be 1.4 times 10 to the -5 cubic meters, that's the change in volume, beta we're told 1.8 times 10 to the -4, and the initial volume we found to be 5.7 times 10 to the -5 meters, and this whole thing is going to be 1365 Kelvin. Now I've said this several times now and I'm going to say it again because of how incredibly important it is to remember this. Your change in temperature in Kelvin equals a change in temperature in degrees Celsius. A degree Celsius and a unit Kelvin have the same size, so their changes are the same. This means that this equals this right here it equals 1365 degrees Celsius because this is a change in temperature, it's a change in temperature. Now what we want to do is we want to find the final temperature. Since the initial temperature was zero degrees Celsius we know that the final temperature is just going to be that change. So that is what you have to heat the mercury up to to get it to expand to the brim of this glass cylinder. Now we said that this guy and this guy were equivalent. Does that mean that my final answer would also be 1365 Kelvin? Absolutely not. My final answer would not be 1365 Kelvin. That's because my initial temperature would not be zero Celsius, it would actually be 273 Kelvin. So my final temperature would be 1365 Kelvin plus 273 Kelvin and that would be a completely different number than what we've got. So the final temperature, this absolute temperatures, are different in units of Kelvin and Celsius but their change in temperature are equivalent in Kelvin and degrees Celsius. Alright guys, that wraps up our discussion on volume thermal expansion. Thanks for watching.

Practice: A cube and a sphere are made of the same material. Initially, the sphere just fits inside of the cube; that is, the length of the cube is twice the radius. If the sphere and cube are both heated up some amount, will the sphere still fit inside the cube?