WEBVTT physics/ap-physics-1-2/fullerton
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Hi everyone and welcome back to Educator.com.
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I am Dan Fullerton and I am thrilled to be opening up our unit today on thermophysics.
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We are going to start with heat temperature and thermal expansion.
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Our objectives are going to be to calculate the temperature of an object given its average kinetic energy, to describe the temperature of a system in terms of a distribution of molecular speeds, and describe thermal equilibrium as a probability process where energy is typically transferred from high to low energy particles.
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We will also explain heat as the process of transferring energy between systems at different temperatures, and finally calculating the linear and volume metric expansion of a solid as a function of its temperature.
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Let us talk about thermophysics.
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Thermophysics explores the internal energy of objects due to the motion of the atoms and molecules comprising the objects.
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It explores the transfer of this energy from object to object, known as heat, a transfer mechanism for energy.
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Let us start with temperature.
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The internal energy of an object, known as its thermal energy is related to the kinetic energy of all the particles comprising the object.
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The more kinetic energy the constituent particles have as they move in their vibrations as part of that object, the greater the objects thermal energy.
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For most systems, the kinetic energy of the constituent particles is not the same, it is a distribution, therefore the system is modeled as a distribution of kinetic energies, typically using Maxwell Boltzmann's statistics.
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As we talk about temperature and phases of matter, in solids the particles comprising the solids are held together very tightly, therefore their motion is limited to just vibrating back and forth in their given positions.
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In liquids, the particles can move back and forth across each other, but the object itself does not have a defined shape.
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In gases, the particles move throughout the entire volume available, but in all cases the total thermal energy is the sum of the kinetic energies of the constituent particles.
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Average kinetic energy and temperature -- actual kinetic energies of individual particles may vary significantly and the average kinetic energy we can find by taking 3/2 times this constant Kb, known as Boltzmann's constant times the temperature and that temperature should be in Kelvins (K), our si unit of temperature.
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If Kb is Boltzmann's constant, that is 1.38 × 10^-23 J/K -- the temperature is in Kelvins, the si unit of temperature again, not Celsius, not Fahrenheit, but Kelvins.
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Now it is important to note that even though two objects can have the same temperature and therefore the same average kinetic energy, they may have different internal energies, depending on what those particles that are moving are.
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Let us take a look at some temperature scales.
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We have Fahrenheit (F) on the left, Celsius (C) in the center, and Kelvins (K) on the right.
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Now they all work in the same basic way, but they have different values at different key temperature readings.
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The Fahrenheit scale has water freezing at 32 ° F and water boiling at 212 ° F and if you extrapolate that back, you get to what is known as absolute 0 at about -459.7 ° F, where absolute 0 is a theoretical minimum temperature.
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It is the point on a volume vs. temperature graph on a gas where the extended curve would hypothetically reach 0 volume.
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It is not specifically the absolute lack of motion of particles, it is a theoretical minimum, but for our purposes, really cold and you do not get any colder than that.
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For Celsius, water boils at 100 ° C, freezes at 0 ° C, and absolute 0 would be -273.15 ° C.
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Now Kelvins, the scale we are going to use here in physics -- and Kelvins has the same size of its main unit, a Kelvin -- 100 ° between water freezing and boiling, but the only difference is we are going to start 0 at absolute 0.
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That means that water freezes at 273.15 K and it boils at 373.15 K.
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Converting between temperature scales is fairly straight forward.
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If you know Celsius, to get Kelvins, you add 273.15.
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If you know Celsius and you want Fahrenheit, multiply the Celsius temperature by 9/5 and add 32.
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If you know Fahrenheit and you want Celsius, take the Fahrenheit temperature, subtract 32 and multiply by 5/9.
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Now let us talk about heat as the transfer of thermal energy from one object to another object due to their difference in temperature.
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That is typically accomplished through some sort of particle interactions or collisions in which momentum is transferred from one object to another.
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That is conduction.
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Now energy is typically transferred from higher energy to lower energy particles and after many collisions, both systems of particles likely have the same average temperature.
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That does not mean that every collision works this way, but on the average it goes in that general direction.
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Because the particles comprising objects have a distribution of particles, velocities, and energies, on the microscopic scale, this transfer of energy is a probabilistic process.
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So as you look at it more and more closely, you have to get more and more into statistics of distribution.
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So methods of heat transfer -- We can transfer heat from one object to another by three different methods.
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Conduction is the transfer of energy along an object to the particles comprising the object colliding.
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Think of sticking an iron rod in the fire.
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Okay, the fiery end is going to get hot real quick, but if you hold that long enough, the other end that started off cool is going to get pretty hot.
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That is by the transfer of the energy from particle to particle to particle in that object.
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Convection is the transfer of energy as a result of energy or heated particles moving from one place to another, like the convection ovens -- heated air molecules move from one place to another.
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And radiation is the transfer of energy through electromagnetic waves.
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Now as we try and quantify heat transfer in conduction, we can get a look at the rate of heat transfer, (H) in J/s or also watts (W).
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Heat is K × A × Δt/L, where Δt is your temperature gradient, the difference in temperature; (A) is the cross-sectional area, (L) is the length of your object, and (K) is a thermal conductivity depending on the material, typically something you would look up, a material property.
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Now, I have put in here a table of some thermal conductivities of selected materials and on the left we have materials such as aluminum, concrete, copper, glass, stainless steel, and water and on the right they are thermal conductivities in J/(s-m-K).
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For example, copper has a much higher thermal conductivity than something like water.
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Let us see how we can put this into practice.
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What is the average kinetic energy of the molecules in a steak at a temperature of 345 K?
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Well the average kinetic energy is given by 3/2 times Boltzmann's constant times the temperature.
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So that will be 3/2 × 1.38 × 10^-23 (Boltzmann's constant) and a temperature of 345 K is going to give us an average kinetic energy of about 7.14 × 10^-21 J.
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Let us take a look at another example this time dealing with body temperature.
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Normal canine body temperature is 101.5 F. What is normal canine body temperature in degrees C and K?
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Well let us convert temperature in degrees C -- is 5/9 times the temperature in degrees F minus 32.
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So that will be 5/9 × 101.5 - 32 = 38.6 ° C.
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Now let us convert that to K.
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Temperature in K is the temperature in degrees C plus 273.15, so that will be 38.6 + 273.15 = 311.75 K.
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Let us look at the temperature of space.
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The average temperature of space is estimated as roughly -270 ° C, that is cold.
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What is the average kinetic energy of the particles in space?
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Well, first thing we are going to do is convert to K, so the temperature in K is the temperature in degrees C plus 273.15, so that will be -270 + 273.15 = 3.15 K.
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Average kinetic energy then, is going to be 3/2 times Boltzmann's constant times our temperature, or 3/2 × 1.38 × 10^-23 × 3.15 K (temperature) or 6.5 × 10^-23 J.
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All right, let us look at the temperature of the sun.
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Given the average kinetic energy of the particles comprising our sun is 1.2 × 10^19 J.
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Find the temperature of the sun in K.
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Well, if average kinetic energy is 3/2 Kbt, then that means the temperature must be 2 times the average kinetic energy divided by 3 times that Boltzmann's constant, Kb.
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Or 2 × 1.2 × 10^19 J/3 × 1.38 (Boltzmann's constant) × 10^-23, or 5800 K.
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Let us take a look at a heat transfer problem.
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Let us find a rate of heat transfer through a 5 mm thick glass window with a cross-sectional area of 0.4 m² if the inside temperature is 300 K and the outside temperature is 250 K.
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Well the rate of heat transfer (h) is KaΔt/L, where if we look up (K) for glass, we can find that the thermal conductivity of glass is about 0.9.
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So that is going to be 0.9 times the cross-sectional area (0.4) times Δt, the change in temperature, is 50 K (temperature gradient) divided by our length (5 mm or 0. 0.005 m).
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So our heat transfer rate is going to be about 36 J/s or 3600 W.
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Let us look at heat transfer across a rod.
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One end of a 1.5 m stainless steel rod is placed in an 850 K fire.
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The cross-sectional area of the rod is 1 cm and the cool end of the rod is at 300 K.
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Calculate the rate of heat transfer through the rod.
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Well, first let us figure out that cross-sectional area.
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Area is πr², so that is going to be π times our radius (1 cm), so that is 0.1 m² or 3.14 × 10^-4m².
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Now we are also going to need the thermal conductivity of steel and there are different conductivities depending on the types of steel, but let us just assume an average thermal conductivity of steel, rough estimate of about 16.5.
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Our rate of heat transfer (h) is KaΔt/L, where (K) for steel is 16.5.
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Our cross-sectional area, we just determined was 3.14 × 10^-4m² and our temperature gradient from 850 K to 300 K is 550 K...
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...divided by the length of our rod 1.5 m or 1.9 J/s or 1.9 W.
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All right, so you know when objects are heated, they tend to expand and when they are cool, they tend to contract and at higher temperatures, objects have higher average kinetic energies so their particles vibrate more.
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At those higher levels of vibrations those particles are not bound as tightly to each other, so the object expands -- exact opposite, as it cools down, they do not vibrate as much and they are bound a little bit more tightly, so they contract.
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This is why if you have a stuck jar of pickles or something and you are trying to open it and you cannot quite untwist it, go try and run it under hot water because if you run it under hot water, the lid is going to start expanding.
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It is going to expand at a faster rate than the glass, so if you run it under hot water, you give yourself a little bit more room and you loosen it up so hopefully, now you are strong enough to undo the lid.
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Linear expansion -- the amount of material expands is characterized by the materials coefficient of expansion.
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One-dimensional expansion, we use the linear coefficient of expansion which gets the symbol α.
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So the change in an object's length due to linear expansion is this -- linear coefficient of expansion times its initial length times its change in temperature (Δt).
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For volumetric expansions, the amount of material that expands is again characterized by the coefficient of expansion, but if it is three-dimensional expansion, you use the volumetric coefficient of expansion, which gets the symbol β.
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The change in volume is that coefficient of expansion, the volumetric coefficient of expansion, β times the initial volume times the change in temperature.
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In most cases the volumetric coefficient of expansion is roughly 3 times the linear coefficient of expansion and that change in temperature can be provided in either ° C or K because the sign of the individual units are the same and we are looking at a relative change, not an absolute C or K -- it does not really matter for these problems.
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So, some coefficients of thermal expansion again.
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We have aluminum, concrete, diamond, glass, stainless steel, and water and we have the linear coefficient of expansion and the volumetric coefficient of expansion.
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Now water is a little bit tricky here.
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Although I have included it here, it actually expands when it freezes, so calculations near the freezing point of water require a little more detailed analysis than is provided here.
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There is a window of a couple of degrees in water, that make it a little bit more complicated, so just keep that in mind, that this is not the full story for water.
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Let us take a look at a contracting railroad tie.
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A concrete railroad tie has a length of 2.45 m on a hot sunny 35 ° C day.
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What is the length of the railroad tie in the winter when the temperature dips to -25 ° C?
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Well, if it is a concrete railroad tie, let us find the linear coefficient of expansion for concrete and for concrete, that just so happens to be about 12 × 10^-6.
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So, ΔL is equal to α L-initial ΔT.
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That is 12 × 10^-6 × 2.45 m (initial length) × -60 (temperature change) or a total of -0.0018 m.
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So what is the new length of the railroad tie?
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Well, ΔL is equal to (L) - L-initial -- Δ anything is the final value minus the initial.
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Therefore the final value is going to be ΔL + L-initial...
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...which will be -0.0018 m + 2.45 m, so its new length will be about 2.448 m.
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All right, pretty straightforward.
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Let us take a look at the expansion of an aluminum rod.
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An aluminum rod has a length of exactly 1 m when it is at 300 K.
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How much longer is it when placed in a 400 ° c oven?
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Well, a couple of things I am going to need to know here.
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First I am going to need to convert this temperature to K, because I start at K and then I am crossing over to C, that is kind of tough to tell the temperature difference between the two.
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First, let us convert that -- our temperature in K is our temperature in ° C + 273.15, so that is going to be 400 ° C +273.15 or 673.15 K.
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For dealing with the expansion of aluminum, I am also going to have to know that α, the linear coefficient of expansion for aluminum is about 23 × 10^-6, so now I can find that shift in length.
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Δ L is α L-initial ΔT or 23 × 10^-6 × 1 m (initial length) × 673.15 (change in temperature) to 300 K...
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...is going to be about 373.15 or a total change in length of about 0.0086 m.
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How about looking at some volumetric expansion.
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A glass of water with volume 1 liter is completely filled at 5 ° C.
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How much water will spill out of the glass when the temperature is raised to 85 ° C?
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Well, we have to realize here that both the glass and the water are going to expand, so let us see how much each expands and find the difference between those two.
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If we start with the water, the change in volume is going to be β, the volumetric coefficient of expansion times its initial volume times our difference in temperature and the volumetric coefficient of expansion for water (β) is 207 × 10^-6, so that is 207 × 10^-6 × 1 L × 80 ° or about 0.0166 L.
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The glass is a slightly different story.
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Change in volume is (β)(V0ΔT) again, but the volumetric coefficient of expansion for the glass is 27 × 10^-6 × 1 L × 80 ° or about 0.0022 L.
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We have considerably more expansion from the water than the glass, so how much is going to spill out?
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We are going to take the difference of these two, 0.0166 L - 0.0022 L to find out that we have 0.0144 L spilling out.
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All right, let us take a look at an example problem where we are looking at some graphs of average kinetic energy vs. temperature.
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Which graph best represents the relationship between the average kinetic energy of the random motion of the molecules of an ideal gas in its absolute temperature.
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Well, first of all, let us write down that relation.
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The average kinetic energy is 3/2 Boltzmann's constant times (t).
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Notice that we have a direct linear relationship between the average kinetic energy and the temperature.
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As temperature goes up, average kinetic energy goes up.
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There it is -- our direct linear relationship.
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Let us take a look at one more.
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Jodie cannot remove her wedding ring.
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If she runs the entire ring under hot water, what is going to happen to the hole in the middle>?
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Will it expand, contract, or stay the same? Well, here is how we are going to treat this.
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We are going to find what happens if we treat this as two rings, an outer ring and an inner ring.
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Let us treat it as a circle, a bigger circle and a more little circle.
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The big circle is going to expand and the inner circle is also going to expand.
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Let us expand them both and then we are going to recombine them and when we do that what we are going to find is if the inner one has expanded and the outer one has expanded, of course this is where the finger goes inside that one and that one is expanded as well, therefore, they both expand.
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In linear expansion, every linear dimension of an object changes by the same fraction when it is heated or cooled.
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That is a good way to get the ring off -- run it under hot water, hopefully it expands -- maybe try a little bit of dish soap or some lubricant there as well.
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Hopefully that gets you a good start on temperature, heat, and thermal expansion.
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Thank you so much for your time and make it a great day!
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We will see you soon.