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Thermodynamics: C l i t Calorimetry, H t Heat • Physics 1 Internal Energy Internal energy (also called thermal energy) is the energy an object j or substance is due to the kinetic and potential p energies g associated with the random motions of all the particles that make it up. The kinetic energy is, of course, due to the motion of the particles. i l To T understand d d the h potential i l energy, imagine i i a solid lid in i which all of its molecules are bound to its neighbors by springs. As the molecules vibrate vibrate, the springs are compressed and stretched. (Liquids and gases are not locked in a lattice structure like this.)) The hotter something is, the faster its molecules are moving or vibrating, and the hi h its higher i temperature. Temperature T i is proportional to the average kinetic energy of the atoms or molecules that make up a substance. Internal Energy vs. Heat The term heat refers is the energy gy that is transferred from one body or location due to a difference in temperature. This is similar to the idea of work, which is the energy that is transferred from one body b d to another h due d to forces f that h act between b them. h H Heat iis internal energy when it is transferred between bodies. Technically, a hot potato does not possess heat; rather it possesses a good deal of internal energy on account of the motion of its molecules If that potato is dropped in a bowl of cold water, molecules. water we can talk about heat: There is a heat flow (energy transfer) from the hot potato to the cold water; the potato potato’ss internal energy is decreased, while the water’s is increased by the same amount. Isolated, Closed and Open Systems 1 • A system is the portion of the physical world being studied. • The system plus surroundings comprise a universe. • The boundaryy between a system y and its surroundings g is the system wall. • If heat cannot p pass through g the system y wall,, it is termed an adiabatic wall, and the system is said to be thermally isolated or thermally insulated. • If heat can pass through the wall, it is termed a diathermal wall. • Two systems connected by a diathermal wall are said to be in thermal contact. Isolated,, Closed and Open p Systems y 2 • An isolated system cannot exchange mass or energy with its surroundings. di • The wall of an isolated system y must be adiabatic. • A closed system y can exchange g energy, gy, but not mass,, with its surroundings. • The energy exchange may be mechanical (associated with a volume change) or thermal (associated with heat transfer through a diathermal wall). • An open p system y can exchange g both mass and energy gy with its surroundings. Isolated, Closed and Open Systems 3 Isolated System Neither energy nor mass can be exchanged. Closed System Open System Energy, but not mass can be exchanged. Both energy and mass can be exchanged. Units ffor Heat Like any type of energy, the SI unit for heat is the Joule. Another common unit is the calorie, which is approximately the amount of heat energy needed to raise one gram one degree Celsius. 1000 calories are in a Calorie, which is used to measure the energy in foods (that the human body can make use of) of). The British thermal unit (BTU) is approximately the energy needed to raise one pound of water one degree Fahrenheit. a e e t. 1 cal = 4.186 J 1 BTU = 1055 J = 252 cal Internal vs. “External” Energy Suppose a 1 kkg bl S block k off iice is i sliding lidi att 7 m/s. / This is the speed of the center of mass of the block, not the speed of each individual water molecule. To calculate the total kinetic energy of the water molecules of the block directly, we would have to know the speed of each molecule as it vibrates, all 33.4 trillion trillion of them! (In practice we would just measure the temperature & mass of the ice.) ice ) The internal energy of the ice does not depend on the motion of the whole body relative to Earth. What matters is the motion of the molecules in the reference frame of the block. O Otherwise,, it would be impossible for a cold object to move quickly or a hot one to move slowly. Note: If friction is present, it could do work on the ice and convert some of the “uniform” kinetic energy of the block into “random” kinetic energy of its molecules (internal energy). Regardless the total energy of the block is the kinetic energy of the Regardless, center of mass + the internal energy: Ktotal = Kcm + Eint vcm = 7 m/s / Temperature Scales F h h i water ffreezes at 32 °F; Fahrenheit: °F bboils il at 212 °F Celsius: water freezes at 0 °C; boils at 100 °C Kelvin: water freezes at 273.15 K; boils at 373.15 K A change h off 100 °C corresponds d to a change h off 180 °F. °F Thi This means 5 C° = 9 F° or 1 C° = 1.8 F° Note that the degree symbol is on the opposite side of the letter, letter indicating that we we’re re talking about temperature differences. In other words, five steps on the Celsius q to nine steps p on the Fahrenheit scale, but 5 °C is scale is equivalent certainly not equal to 9 °F. Since these scales are linear, and they’re offset by 32 °F, we get the conversion formula: F = 1.8C + 32 One step on the Kelvin scale is the same as one step on the Celsius scale. These scales are off by 273.15 K, so: K = C + 273.15 Room temperature is around 293 kelvins, which is 20 °C, or 68 °F. Absolute Zero & the Kelvin Scale The Kelvin Th K l i scale l is i setup t so that th t its it zero point i t is i the th coldest ld t possible ibl temperature--absolute zero, at which point a substance would have zero internal energy. This is -273.15 273.15 °C, C, or -459.69 459.69 °F. F. Absolute zero can never be reached, but there is no limit to how close we can get to it. Scientists have cooled substances to within 10-5 kelvins of absolute zero. How do we know how cold absolute zero is, if nothing has ever been at that temperature? The answer is by graphing Pressure vs. Temperature for a variety of gases and extrapolating. extrapolating P A gas exerts no ppressure when at absolute zero. T (°C) -273.15 °C 0 °C Thermal Equilibrium Two bodies T b di are said id to t be b att thermal th l equilibrium ilib i if they th are att the th same temperature. This means there is no net exchange of thermal energy between the two bodies. The top pair of objects are in contact, but since they are at different temps, they are not in thermal equilibrium, and energy is flowing from the hot side to the cold side. heat hot 26 °C cold 26 °C No net heat flow The and, are in Th two t purple l objects bj t are att the th same temp t d therefore th f i thermal equilibrium. There is no net flow of heat energy here. Heat Transfer Processes Heat energy can be transferred from one body to another in three different ways. • Conduction: Energy is transferred when two objects are in direct contact Molecules of the hotter object bump into molecules of the contact. colder object and cause them to speed up, warming the colder object. • Convection: Energy is transferred from one body to a cooler one via currents in a fluid (a gas or liquid). • Radiation: R di ti All objects, bj t att any temperature, t t radiate di t electromagnetic l t ti radiation (light of visible and invisible wavelengths). Unlike conduction & convection, convection no medium (matter of any type) is necessary for heat transfer through radiation. Objects absorb radiation as well. At thermal equilibrium it will absorb as much as it radiates. Thermal Conductivity, k Heatt transfer H t f via i conduction d ti was described d ib d a few f slides lid back. b k Thermal Th l conductivity, k, refers how easily heat can move through a material. Metals have high thermal conductivity, meaning heat passes through them readily. Wood is a fairly good insulated of heat, and styrofoam is even better. These materials have low thermal conductivities. k is very low for air as well. (Attic insulation and styrofoam cups trap air, making them good insulators.) Heat from a boiler passes through all sides of its metal enclosure. enclosure The rate at which heat is transferred is given by: H = k A ( T2 - T1 ) L T2 T1 heat eat A = area of side wall L = thickness thi k off wall ll k = thermal conductivity of the metal T2 - T1 = temperature difference H is simply power, and its SI unit is the Watt. SI Units ffor Thermal Conductivityy k A (T - T ) H= 2 1 L k must have units that cancel out all the units on the right, leaving only the units for H. The units are: W m·K m K or equivalently, W m·°C m C Since one kelvin is as big a change in temp as one degree Celsius, these units are equivalent. Note: k for thermal conductivity is not the same as the k in Hooke’s Law in which it represents the spring constant! Thermopane Windows heat heat In a house we often want to prevent heat from getting in or getting out. Windows can be problematic. Thermopane windows have two or more panes of glass with air or some other gas between the panes. Which type of window window, a double pane or a thick single pane, is better for minimizing heat transfer, if the total thickness is the same? answer: There is more glass in the single pane window to block the heat, but the air in between the panes of the do ble pane window double indo has thermal conductivity cond cti it that is about 35 times lower than that of the glass itself. So much more heat would be transferred through the single pane. Triple pane vs. Double pane If they are of the same total thickness and pane thickness,, which is better at minimizingg heat transfer,, a double or triple pane window? heat answer: heat The double pane window has more air between the outer panes panes, so its thermal conductivity is lower. lower However, air is a mobile medium, and convection currents can shuttle warm air from the warm side to the cold side. On the warm side the air rises, moves across the the cold side, and sinks, moving in a loop andd carrying i its i energy from f the h warm side id to the h cold side. The middle pane in the triple pane window reduces the energy transfers due to convection and is the better window (but probably more expensive). R Value The R value of a material is its “thermal resistance” and refers to how good an insulator is. Here’s how it’s defined: L R = k As in previous equations: L = the thickness of the material k = thermal conductivity of the material Note that the R value is inversely proportional to thermal conductivity, meaning good heat conductors have a low R value and are poor insulators. Also, the R value is directly proportional to the thickness of the material, meaning the thicker it is, the better it insulates. i l t Thus, Th more insulation i l ti in i the th attic tti can save energy. Wind & Heat Loss A breeze can cool us off in the summer, and wind can make us feel colder in the winter. Why is this? answer: When we sweat the perspiration absorbs body heat, and when it evaporates, it takes this heat with it. This is called evaporative cooling. A steaming cup of hot chocolate cools in the same way. p us warm in the winter is because it traps p air The reason a coat keeps that is heated by our bodies. (Wearing layers is like having a triple pane window.) A thin layer of stagnant air also surrounds the outside walls ll off buildings b ildi andd helps h l insulate i l them. h Wind i d tends d to blow bl this hi warm air away, along with its heat.. The windier it is, the colder it feels to us, us and the greater the heat loss from a building Trees around you home can save energy in two ways: blocking wind in the winter; inter; and shielding your o r home from excess e cess solar radiation in the summer. Laws of Thermodynamics (examples upcoming) • Zeroth Law: If object A is in thermal equilibrium with object B, and if object B is in thermal equilibrium with object C, C then objects A and C are also in equilibrium. This is sort of a “transitive pproperty p y of heat.” • First Law: Energy is always conserved. It can change forms: kinetic potential, kinetic, potential internal etc., etc but the total energy is a constant. constant Another way to say it is that the change in thermal energy of a system is equal to the sum of the work done on it and the amount of heat energy transferred to it. • Second Law: Duringg any y natural pprocess the total amount of entropy in the universe always increases. Entropy can be defined informally as a measure of the randomness or disorder in a system. Heat flows naturally from a hot to cooler surroundings as a consequence of the second law. Change in Entropy Equation Because most systems are many up off so many particles, i l calculating l l i entropy via probabilities would be very difficult. Fortunately, we are normally concerned only with changes in entropy entropy. If we have a system in which energy is not changing forms, the change in entropy is defined as: ∆Q ∆S = T ∆S = change in entropy ∆Q = change in internal energy (heat flow) T = absolute temperature The 2nd Law of Thermodynamics says that during any process: ∆Suniverse = ∆Ssystem + ∆Ssurroundings ≥ 0 Change in Entropy Example A glass l rodd is i heated h t d andd then th blown bl by b a glassblower. When it is at 185°C it is brought outside to cool. 3200 J of heat are transferred from the glass to the air, which is at 18°C. Find the change in entropy of the universe: ∆Suniverse = ∆Ssystem + ∆Ssurroundings = ∆S Sglass + ∆S Sair ∆Qglass ∆Q Qair + = T Tair glass -3200 3200 J 3200 J = + 458 K 291 K = -7 J/K + 11 J/K = +4 J/K Change in Entropy Example (cont.) A th As the glass l cooled l d we assumedd that th t the th air i temp t didn’t did ’t go up appreciably due after the heat transfer, which would have complicated the problem. Important points: • The temps were converted to kelvins. • The Th glass l lost l as muchh thermal h l energy as air i gained, i d as the h 1stt Law L requires. • ∆Q Qglass is negative since the glass lost thermal energy so ∆S Sglass is also negative. • ∆Qair is positive since the air gained thermal energy so ∆Sair is also positive. • Even though the ∆Q’s are the same size, the ∆S’s aren’t, since the temps are different. • The positive ∆S is greater than the negative ∆S, as the 2nd Law requires. Second Law Consequences • Heat H t will ill nott flow fl from f a cold ld body b d to t a hot h t body. b d • “Reverse diffusion” is a no-no (such as smoke from a fire isolating itself in a small space). • An object or fluid of uniform temperature (no matter how hot) cannot do useful work. (There must be temperature difference so that there will be a heat flow, which can be used to do work.) • The various forms of energy tend to degrade over time to thermal energy. This represents useful, low probability forms of energy converting into an unusable, high probability form. • Without input p of energy, gy, bodies tend to reach thermal equilibrium. (We can maintain temperature differences via refrigerators or heating units, but this requires energy.) continued on next slide Second Law Consequences (cont.) • Any time we do something that decreases the entropy of a system the energy we expend in doing it increases the entropy system, of the surroundings even more. • A perpetual motion machine is impossible to make. make A perpetual motion machine is a device that would absorb thermal energy from a hot body and do as much work as the energy it absorbed. • During any process the entropy of the universe cannot decrease. Expending energy to decrease the entropy of a system will lead to an increase in entropy for the surrounding by a greater amount. Thermal Equilibrium and the Zeroth Law • If warm a a and d coo cool objects a are ep placed aced in tthermal e a co contact, tact, energy, known as heat, flows from the warm to the cold object until thermal equilibrium is established. • Zeroth Law of Thermodynamics Two systems systems, separately in thermal equilibrium with a third system, are in thermal equilibrium with each other. • The property which the three systems have in common is known as temperature θ. • Thus the zeroth law may be expressed as follows: if θ1 = θ2 and θ1 = θ3, then θ2 = θ3. 25 Thermodynamic Variables • Thermodynamic variables are the observable macroscopic variables of a system system, such as P P, V and T T. • If the are used to describe an equilibrium state of the system, they are known as state variables variables. • Extensive variables depend on the size of the system; e.g. mass, volume, l entropy, t magnetic ti moment. t • Intensive variables do not depend on size; e.g. pressure, temperature magnetic field temperature, field. • An extensive variable may be changed to an intensive variable, known as a specific f value, by dividing it by a suitable extensive variable, such as mass, no.of kmoles, or no. of molecules. molecules • Example: the specific heat is normally (heat capacity)/(mass). Equilibrium States • An equilibrium q state is one in which the p properties p of the system do not change with time. • In many cases, an equilibrium state has intensive variables which hi h are uniform if th throughout h t th the system. t • A non-equilibrium state may contain intensive variables which vary in i space and/or d/ titime. • An equation of state is a functional relationship between the state t t variables; i bl e.g. if P P,V V and d T are th the state t t variables, i bl th then the equation of state has the form f(P, V, T) =0. • In I 3-dimensional 3 di i lP P-V-T V T space, an equilibrium state is represented by a point, and the equation of state is represented by a surface surface. 27 Processes 1 • A process refers to the change of a system from one equilibrium state to another. • The initial and final states of a process are its end-points. end points • A quasistatic process is one that takes place so slowly that th system the t may be b considered id d as passing i th through ha succession of equilibrium states. • A quasistatic process may be represented by a path (or line) on the equation-of-state surface. • If it is non-quasistatic, only the end-points can be shown. • A reversible process is one the direction can be reversed by an infinitessimal change of variable. • A reversible process is a quasistatic process in which no dissipative forces, such as friction, are present. • A reversible change must be quasistatic, but a quasistatic process need not be reversible; e.g. if there is hysteresis. 28 Processes 2 • An isobaric process is one in which the pressure is constant. • An isochoric process is one in which the volume is constant constant. • An isothermal process is one in which the temperature is constant. • An adiabatic process is one in which no heat enters or leaves y i.e. Q = 0. the system; • An isentropic process is one in which the entropy is constant. • It is a reversible adiabatic process. • If a system is left to itself after undergoing a non-quasistatic process, it will reach equilibrium after a time t much longer than the longest relaxation time τ involved; i.e. t » τ. • Metastable equilibrium occurs when one particular relaxation time τ0 is much longer than the time ∆t for which the system is observed; i.e. τ0» ∆t . 29 Three Types of Process Isothermal process System Adiabatic process P Adiabat Isotherm Heat bath or reservoir Adiabatic free expansion V P ● 1 End points ● 2 V 30 Boyle’s Law and the Ideal Gas Scale h P P = PA + gρh P = a(TC + 273.15) • Boyle’s Law • At sufficiently ffi i tl low l pressure, the product PV was found to be constant for gases held at a given temperature θ; i.e. PV = f(θ) for P → 0 0. • Fixed points (prior to 1954) • The Th iice and d steam t points i t were defined to be 0oC and 100oC exactly exactly. • The ideal gas (or kelvin) scale l was d defined fi d as TK = TC + 273.15. 31 The Ideal Gas Law • Fixed point (1954) • The triple point of water is Ttr = 273.16 K, Ptr = 6.0 x 10–3 atm. • Ideal gas law PV = nRT or Pv = RT, • where n is the no. of kmoles, v is the volume per kmole, T is the absolute temperature in K K, and the gas constant R = 8.314 8 314 x 103 J/(K.kmol). J/(K kmol) • For a constant quantity of gas, P1V1/T1 = P2V2/T2. P P V T increasing V increasing V T P increasing 32 T Ideal g gas: Macroscopic p description p • Consider a gas in a container of volume V, at pressure P, and at temperature T • Equation of state – Links these quantities q – Generally very complicated: but not for ideal gas l Equation of state for an ideal gas Collection of atoms/molecules moving randomly No long-range forces Their size (volume) is negligible PV = nRT RT R iis called ll d the h universal i l gas constant In SI units, units R =8.315 =8 315 J / mol·K mol K n = m/M : number of moles Boltzmann’s constant l Number of moles: n = m/M m=mass M=mass of one mole One mole contains NA=6.022 X 1023 particles : Avogadro’ss number = number of carbon atoms in 12 g of carbon Avogadro • In terms of the total number of particles N PV = nRT = ((N/NA ) RT PV = N kB T kB = R/NA = 1.38 X 10-23 J/K kB is called the Boltzmann’s constant • P, P V, V and d T are the h thermodynamics h d i variables i bl Th Ideal The Id l G Gas L Law pV V = nRT RT Whatt iis th Wh the volume l off 1 moll off gas att STP ? T = 0 °C = 273 K 5 p = 1 atm = 1.01 1 01 x 10 Pa V RT = n P 8.31 J / (mol ⋅ K ) 273 K = 1.01 x 10 5 Pa = 0.0224 m 3 = 22.4 l Specific Heat Specific heat is defined as the amount of thermal energy needed p Its to raise a unit mass of substance a unit of temperature. symbol is C. For example, one way to express the specific heat of water is one calorie per gram per degree Celsius: C = 1 cal / (g·ºC), or 4.186 J / (g·ºC). This means it would take 20 cal of thermal energy to raise 4 grams of water 5 ºC. Water has a veryy high g specific p heat,, so it takes more energy gy to heat up water than it would to heat up most other substances (of the same mass) by the same amount. Oceans and lake act like “h sinks” “heat i k ” storing i thermal h l energy absorbed b b d in i the h summer andd slowing releasing it during the winter. Large bodies of water thereby help to make local climates less extreme in temperature from season to season. Specific Heat Equation Q = mC ∆T Q = thermal energy m = mass C = specific heat ∆T = change in temp Ex: The specific p heat of silicon is 703 J / ((kg· g ºC). ) How much energy is needed to raise a 7 kg chunk of silicon 10 ºC? answer: Q = 7 kg · 703 J ·10 ºC = 49 210 J kg· ºC Note that the units do indeed work out to be energy units. Calorimetry Schmedrick S h d i k takes k another h hhorseshoe h out off the h fire fi when h it’s i ’ at 275 ºC, drops in his bucket of water, and this time covers the bucket. The bucket and cover are made of an insulating material. material The bucket contains 2.5 L of water originally at 25 ºC. The 1.9 kg shoe is made of iron, which has a specific p heat of 448 J / ((kg·ºC). g ) Let’s find the temp p of the horseshoe and water once equilibrium is reached. Let’s assume that the container allows no h to escape. Then heat h the h 1stt Law implies i li that all heat the shoe loses is gained by the water Since one milliliter of water has a water. mass of one gram, the bucket contains 2.5 kgg of water. At thermal equilibrium q the water and shoe are at the same temp. The total thermal energy in the bucket d does not change, h b it but i is i redistributed. di ib d continued on next slide Calorimetry (cont.) Let T = the equilibrium temperature. Q lost byy iron = Q ggained byy water miron Ciron ∆Tiron = mwater Cwater ∆Twater (1.9 kg)(448 J / kg·ºC)(275 ºC - T) = (2.5 kg)(4186 J / kg·ºC)(T - 25 ºC) Note N t hhow the th ∆T terms t are written itt so that th t each side is positive. We’ve got a simple linear equation with T on both sides. sides Solving it gives us T = 43.8 ºC. This is the equilibrium temp--the final temp for both the shoe and water. If T had come out over 100 ºC, the answer would have b been invalid, i lid since i the th specific ifi heat h t for f steam is different than that of water. Latent Heat The word “latent” comes from a Latin word that means “to lie hidden.” When a substance changes phases (liquid ↔ solid or gas ↔ liquid) energy is transferred without a change in temperature. This “hidden energy” is called latent heat. For example, l to t turn t water t ice i into i t liquid li id water, t energy mustt be b added to bring the water to its melting point, 0 ºC. This is not enough however enough, however, since water can exist at 0 ºC C in either the liquid or solid state. Additional energy is required to change 0 ºC ice into 0 ºC C water. water The energy increases the internal energy of the water but does not raise its temp. When frozen, water molecules are in a crystalline y structure,, and energy gy is needed to break this structure. The energy needed is called the latent heat of fusion. Additional energy is also needed to change water at 100 ºC to steam at 100 ºC, and this is called the latent heat of vaporization. Latent Heat Formula Q = m Lf or Q = m Lv Q = thermal energy m = mass L = heat of fusion or vaporization L is the energy per unit mass needed to change the state of a substance from solid to liquid or from liquid to gas. Ex: Lf (the latent heat of fusion) for gold is 6440 J/kg. J/kg Gold melts at 1063 ºC. 5 grams of solid gold at this temp will not become liquid until additional heat is added. The amount of heat needed is: (6440 J/kg) J/k ) (0.005 (0 005 kkg)) = 32 JJ. Th The li liquid id gold ld will ill still till be b at 1063 ºC. Latent Heat / Specific Heat Example Superman vaporizes S i a 1800 kkg iice monster t with ith his heat ray vision. The ice monster was at -20 20 ºC C. After being vaporized he is steam at 135 ºC. How much energy did Superman expend? Substance Specific Heat (in J / kg kg·ºC) C) ice 2090 liquid water 4186 steam 1970 For water: Lf = 3.33 ·105 J / kg; Lv = 2.26 ·106 J / kg Q = (1800 kg)(2090 J / kg·ºC)(20 ºC) heating ice to melting pt. + (1800 kg)(3.33 ·10 105 J / kg) ice to water, const. temp of 0 ºC C +(1800 kg)(4186 J / kg·ºC)(100 ºC) heating water to boiling pt. + (1800 kg)(2.26 ·106 J / kg) water to steam, const. temp of 100 ºC + (1800 kg)(1970 J / kg·ºC)(35 ºC) heating steam to 135 ºC = 5.62 ·109 J total energy expended by Superman Latent Heat & Entropy A 10 g ice cube at 0 ºC C falls to the ground and melts. melts The temp outside is 10 ºC. Calculate the change in entropy of the universe due to the meltingg of the ice only. y answer: For the cubie: Q = m Lf = (0.01 kg)(3.33 ·105 J / kg) = + 3330 J. Thi is This i the th energy absorbed b b d by b the th ice i from f the th surroundings. di ∆Sice = ∆Qice /Tice = +3330 J / 273 K = +12.198 J/K. For the h surroundings: di Q = -3330 J, since i the h surroundings di lost l as much thermal energy as the cubie gained. The temperature of the backyard does not decrease significantly significantly, though though, with such a small energy loss. ∆Ssurr = ∆Qsurr /Tsurr = -3330 J / 283 K = -11.767 J/K. For the universe: ni erse: ∆Suniv = ∆Ssurr + ∆Sice = 12.198 12 198 J/K - 11.767 11 767 J/K = +0.431 J/K. Thus, the 2nd Law is satisfied. Thermal Expansion As a material heats upp its atoms/molecules move or vibrate more vigorously, and the average separation between them increases. This results in small increases in lengths and volumes. Buildings, railroad t k bridges, tracks, b id andd highways hi h contain t i thermal th l expansion i joints j i t to t prevent cracking and warping due to expansion. The amount of expansion depends p on the original g length/volume, g , the type yp of material,, and the change in temp. L is length, V is volume, T is temp, α is the coefficient of linear expansion, and β is the coef. of volume expansion. Wh a solid When lid off a single i l material t i l expands, d it does d so proportionally ti ll in i all directions. Since volume has 3 dimensions and length is only 1, β = 3 α. L th expansion: Length i ∆L = α ∆T L Volume expansion: cold solid hot solid ∆V = β ∆T V Thermal expansion • In most liquids or solids, when temperature rises – molecules have more kinetic energy • they are moving faster, on the average – consequently, things tend to expand (works for a gas) • amount of expansion ∆L depends on… – change in temperature ∆T – original length L0 L0 ∆L – coefficient of thermal expansion • L0 + ∆L = L0 + α L0 ∆T • ∆L = α L0 ∆T (linear expansion) V • ∆V = β V0 ∆T ((volume expansion) p ) V + ∆V • α ≅ 3β Thermostats Bimetallic strips are used in thermostats, at least in some older ones. When the temperature changes, the strip bends, making or b ki an electrical breaking l t i l circuit, i it which hi h causes the th furnace f to t turn t on or shut off. In this model when the strip bends it tilts a bulb of mercury mercury, which then bridges two wires and allows current to flow. Thermal Expansion & The Concorde The Concorde Th C d is i a supersonic i jet j t made d off a heat h t tolerant t l t aluminum l i alloy. Its nose tilts down on takeoff and landing so the pilot can see the runway runway. In flight the nose comes up to reduce drag, drag but at a speed of around 1,350 mph, friction with the air causes significant heatingg of the plane, p , enough to make the Concorde grow in length by 7 inches! (To maintain this speed d ffor one hhour, the th Concorde must burn over 6,700 6 700 gallons of fuel.) Thermal Expansion and Teeth Thermal Expansion and Teeth Coefficients of linear expansion: p Enamel: 11.4 x 10-6 °C Dentin: 8.3 x 10-6 °C If you quickly switch from eating/drinking something hot to somethingg cold,, the brittle enamel will contract more than the dentin, and develop small cracks called crazes. Crazing: Thermal expansion matching is important in choosing materials for fillings fillings. Exercise Thermal Expansion • On a cold winter day with T = -10 10 °C C, the Eiffel Tower is 300 m tall. • How tall is it on a hot summer day when T = 40 °C? • The Eiffel Tower is made from steel with a thermal expansion coefficient α =11 x 10-6/°C R ll ∆L/L = α ∆T Recall: (A) 300.001 m (B) 300.16 300 16 m (C) 316 m (D) 420 m