: ClCalor ime try, HtHeat

• Physics 1 Internal Energy Internal energy (also called thermal energy) is the energy an objjpgect or substance is due to the kinetic and potential energies 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 partilicles. T o und erstand dh the potenti ilal energy, i magi ne a solid lidi in which all of its molecules are bound to its neighbors by springs. As the molecules 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 high er i ts . Temperature 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 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 bdbody to anot her d ue to forces th at act b etween t hem. Heat is 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 , we can talk about heat: There is a heat flow (energy transfer) from the hot potato to the cold water; the potato ’s 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 boundaryyy between a system and its surroundin gs is the system wall. • If heat cannot ppgy,ass through the system wall, it is termed an , 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,,py Closed and Open Systems 2

•An isolated system cannot exchange mass or energy with its surroundings. • The wall of an isolated system must be adiabatic.

• A closed system can exchange ener gy, but not mass , with its surroundings. • The energy exchange may be mechanical (associated with a volume change) or thermal (associated with through a diathermal wall).

• An oppyen system can exchangggye both mass and energy with its surroundings. Isolated, Closed and Open Systems 3

Isolated Closed Open System System System Neither energy nor mass Energy, but not mass can can be exchanged. Both energy and mass can be be exchanged. exchanged. Units for 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). The British thermal unit (BTU) is approximately the energy needed to raise one pound of water one degree Faaehrenhe et.it.

1 cal = 4.186 J 1 BTU = 1055 J = 252 cal Internal vs. “External” Energy Suppose a 1 kg block of i ce i s slidi ng at 7 m/ s. vcm = 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 .) 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. OO,therwise, 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 center of mass + the internal energy: Ktotal = Kcm + Eint Temperature Scales FhFahren hiheit: water freezes at 32°F; bo ils 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 c hange o f 100°C correspond s to a ch ange o f 180°FThiF. This means 5C° = 9F° or 1C° = 1.8F° Note that the degree symbol is on the opposite side of the letter , indicating that we ’ re talking about temperature differences. In other words, five steps on the Celsius scale is eqqpuivalent to nine steps on the Fahrenheit scale, but 5°C is 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 K el vi n scal e i s set up so th at it s zero poi nt i s th e cold est possibl e temperature--absolute zero, at which point a substance would have zero internal energy. This is -273.15°C, or -459.69°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 . P A gas exerts no pressure when at absolute zero.

T (°C) -273.15°C 0°C Thermal Equilibrium Two b odi es are said t o b e at th ermal equilib ri um if th ey are at th e 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.

hot heat cold

26°C 26°C

No net heat flow The t wo purpl e obj ect s are at th e same t emp and , th eref ore are i n 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 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). • Radi ati on: All obj ect s, at any t emperat ure, radi at e el ect romagneti c radiation (light of visible and invisible wavelengths). Unlike conduction & 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 Hea t t ransf er vi a cond ucti on was d escrib ed a f ew slid es b ack . Th ermal 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 . The rate at which heat is transferred is given by: H = kA (T - T ) L 2 1 A = area of side wall

T2 T1 L = thick ness of wall k = thermal conductivity of the metal

heat T2 - T1 = temperature difference H is simply power, and its SI unit is the Watt. SI Units ffyor Thermal Conductivity

kA H = (T2 - T1) L k must have units that cancel out all the units on the right, leaving only the units for H. The units are:

W or equivalently, W m·K 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 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 heat glass with air or some other gas between the panes. Which type of 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 double pane windo w has thermal cond ucti vit y that is heat 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,,g, which is better at minimizing heat transfer, a double or triple pane window? heat answer: The double pane window has more air between the outer panes, so its thermal conductivity is 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 heat andiid carrying its energy f fhrom the warm sid idhe to the 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:

R = L 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 i nsul at es. Th us, more i nsul ati on i n th e atti c 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. The reason a coat keeppps us warm in the winter is because it traps air 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 of b uildi ngs and h el ps i nsul ate th em. Wi nd tend s to bl ow thi s warm air away, along with its heat.. The windier it is, the colder it feels to 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; and shielding yo ur home from e xcess solar radiation in the summer. (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 , then objects A and C are also in equilibrium. This is sort of a “transitive ppproperty of heat.” • First Law: Energy is always conserved. It can change forms: kinetic, potential , internal etc ., but the total energy is a 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: Duringgy any natural process 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 ofilllif so many particles, calculating entropy via probabilities would be very difficult. Fortunately, we are normally concerned only with changes in 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 g lass rod i s h eat ed and th en bl own b y 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

= ∆Sglass + ∆Sair

∆Qglass ∆Qair = + Tglass Tair -3200 J 3200 J = + 458 K 291 K = -7 J/K + 11 J/K = +4 J/K Change in Entropy Example (cont.) As the g lass cool ed we assumed th at th e ai r t emp did n’t go up appreciably due after the heat transfer, which would have compli- cated the problem. Important points: • The temps were converted to kelvins. • The gl ass l ost as much th ermal energy as ai r gai ned , as th e 1 st Law requires.

• ∆Qglass is negative since the glass lost thermal energy so ∆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 • Hea t w ill not fl ow f rom a cold b od y t o a h ot b od y. • “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 inppgy,ut of energy, 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 of the surroundings even more. • A perpetual motion machine is impossible to 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 an d cool objects ar e pl aced in th erm al con tact, energy, known as heat, flows from the warm to the cold object until thermal equilibrium is established. • Zeroth Law of Thermodynamics Two 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, such as P, V and T. • If the are used to describe an equilibrium state of the system, they are known as state variables. • Extensive variables depend on the size of the system; e.g. mass, voltlume, entropy, magne tittic moment. • Intensive variables do not depend on size; e.g. pressure, temperature, magnetic field. • An extensive variable may be changed to an intensive variable, known as a specific value, by dividing it by a suitable extensive variable, such as mass, no.of kmoles, or no. of molecules . • Example: the specific heat is normally (heat capacity)/(mass). Equilibrium States

• An equilibrium state is one in which the ppproperties of the system do not change with time. • In many cases, an equilibrium state has intensive variables whic h are un iform throug hou t the sys tem. •A non-equilibrium state may contain intensive variables which vary id/tiin space and/or time. •An equation of state is a functional relationship between the stttate vari iblables; e.g. ifPVif P,V an dTd T are the s tttate vari iblables, then the equation of state has the form f(P, V, T) =0. • I3In 3-dimens iona l P-V-T space, an equilibrium state is represented by a point, and the equation of state is represented by a surface. 27 Processes 1 •Aprocess 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. •A quasistatic process is one that takes place so slowly that the syst em may b e consid ere d as pass ing throug h a 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 28 process need not be reversible; e.g. if there is hysteresis. Processes 2

•An isobaric process is one in which the pressure is constant. • An isochoric process is one in which the volume is constant. •An isothermal process is one in which the temperature is constant. •An adiabatic process is one in which no heat enters or leaves the system; i.e. Q = 0. •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 Adiabatic process P System Adiabat

Isotherm

Heat bath or reservoir V

Adiabatic free expansion P

●1 End points

●2 30 V Boyle’s Law and the Ideal Gas Scale

• Boyle’s Law • At su ffici entl y l ow pressure, h the product PV was found to be constant for gases held at P a given temperature θ; i.e. PV = f(θ)forP) for P → 0. • Fixed points (prior to 1954)

PPP = PA +g+ gρh • The i ce an d s team po in ts were defined to be 0oC and 100oC exactly. P = a(TC + 273.15) • The ideal gas (or kelvin) scaldfidle was defined as 31 TK = TC + 273.15. The Ideal Gas Law • Fixed point (1954) –3 • The triple point of water is Ttr = 273.16 K, Ptr = 6.0 x 10 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, and the gas constant R=8314x10R = 8.314 x 103 J/(K. kmol).

• For a constant quantity of gas, P1V1/T1 = P2V2/T2.

P PPVV

T increasing

V increasing P increasing32 V T T Ideal ggppas: Macroscopic description

• Consider a gas in a container of volume V, at pressure P, and at temperature T • Equation of state – Links these quantities – 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 = n RT RiR is call llded thihe universal gas constant

In SI units , R =8 .315 J / 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

23 ™ One mole contains NA=6.022 X 10 particles : Avogadro’s number = number of carbon atoms in 12 g of carbon • In terms of the total number of particles N

PV = nRT = (N/NA ) RT

PV = N kB T

-23 kB = R/NA = 1.38 X 10 J/K

kB is called the Boltzmann’s constant

• PVP, V, and T are thethdhermodynam ics var ibliables The Id ea l Gas Law

pVV = nRT

Wha t is the vo lume of 1 mol of gas at STP ? T = 0 °C = 273 K p = 1atm1 atm = 101x101.01 x 105 Pa V RT = n P 8.31 J(/ mol ⋅K) 273 K = 5 1.01x10 Pa 3 = 0.0224 m = 22 .4 l Specific Heat

Specific heat is defined as the amount of thermal energy needed to raise a unit mass of substance a unit of temperature. Its 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 veryygp high specific heat, so it takes more energy 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 “heat si nk s” stori ng th ermal energy ab sorb ed i n th e summer and 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= mass C = specific heat ∆T = change in temp

Ex: The specific heat of silicon is 703 J/((gkg·ºC). How much energy is needed to raise a 7 kg chunk of silicon 10 ºC? answer: 703 J Q = 7 kg · ·10 ºC = 49 210 J kg·ºC Note that the units do indeed work out to be energy units. Calorimetry ShSchme dikdrick ta kes anot her horses hoe out o fhf the fi re wh en i i’t’s 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 . 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 heat of 448 J/((gkg·ºC))p. Let’s find the temp of the horseshoe and water once equilibrium is reached. Let’s assume that the container allows no hhhheat to escape. Then the 1st Law iliimplies that all heat the shoe loses is gained by the water. Since one milliliter of water has a mass of one gram, the bucket contains 2.5 kggq of water. At thermal equilibrium the water and shoe are at the same temp. The total thermal energy in the bucket dhbiidiibddoes not change, but it is redistributed. continued on next slide Calorimetry (cont.) Let T = the equilibrium temperature. Q lost by iron = Q ggyained by 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)

NtNote how the ∆T terms are writt en so that each side is positive. We’ve got a simple linear equation with T on both 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 been i nvalid , si nce th e specifi c h eat f or 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 examplttle, to turn wat tier ice i itliidtnto liquid water, energy must tb be added to bring the water to its melting point, 0 ºC. This is not enough, however, since water can exist at 0 ºC in either the liquid or solid state. Additional energy is required to change 0 ºC ice into 0 ºC water . The energy increases the internal energy of the water but does not raise its temp. When frozen, water molecules are in a cryy,gystalline structure, and energy 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 = mLf or Q = mLv 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. 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 /k)kg) (0. 005 kg ) = 32 J. The liqu id go ld w ill still b e at 1063 ºC. Latent Heat / Specific Heat Example Superman vapor izes a 1800 kg ice mons ter with his heat ray vision. The ice monster was at -20 ºC. After being vaporized he is steam at 135 ºC. How much energy did Superman expend? Substance Specific Heat (in J/kg·ºC) ice 2090 liquid water 4186 steam 1970

5 6 For water: Lf = 3.33 ·10 J/kg; Lv = 2.26 ·10 J/kg Q = (1800 kg)(2090 J/kg·ºC)(20 ºC) heating ice to melting pt. + (1800 kg) (3.33 ·105 J/kg) ice to water, const. temp of 0 ºC +(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 falls to the ground and melts . The temp outside is 10 ºC. Calculate the change in entropy of the universe due to the meltinggy of the ice only. answer:

5 For the cubie: Q = m Lf = (0.01 kg)(3.33 ·10 J/kg) = + 3330 J. This i s th e energy ab sorb ed b y th e i ce f rom th e surroundi ngs.

∆Sice = ∆Qice /Tice = +3330 J / 273 K = +12.198 J/K. For thdihe surroundings: Q = -3330 J, since th e surroundi ngs l ost as much thermal energy as the cubie gained. The temperature of the backyard does not decrease significantly, though, with such a small energy loss. ∆Ssurr = ∆Qsurr /Tsurr = -3330 J / 283 K = -11.767 J/K.

For the uni verse: ∆Suniv = ∆Ssurr + ∆Sice = 12.198 J /K - 11. 767 J/K = +0.431 J/K. Thus, the 2nd Law is satisfied. Thermal Expansion As a material heats up 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 track s, b bidridges, and dhih highways cont tiain th ermal expansi on j jitoints t o prevent cracking and warping due to expansion. The amount of expan- sion depends on the ori ginal len gth/volume , the t ype of material , and the change in temp. L is length, V is volume, T is temp, α is the coef- ficient of linear expansion, and β is the coef. of volume expansion. When a so lid of a s ingl e mat eri al expand s, it d oes so proporti onall y i n all directions. Since volume has 3 dimensions and length is only 1, β = 3α. LthLength expans ion: ∆L = α ∆T L Volume expansion: ∆V = β ∆T cold solid hot solid 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 L ∆T (linear expansion) 0 V • ∆V = β V0 ∆T ((p)volume expansion) • α≅3β V + ∆V Thermostats Bimetallic strips are used in thermostats, at least in some older ones. When the temperature changes, the strip bends, making or bkibreaking an el ltiliithihectrical circuit, which causes th thfe furnace t tto turn on or shut off. In this model when the strip bends it tilts a bulb of mercury, which then bridges two wires and allows current to flow. Thermal Expansion & The Concorde The C oncord e i s a supersoni c j et mad e of a h eat t ol erant al umi num alloy. Its nose tilts down on takeoff and landing so the pilot can see the runway. In flight the nose comes up to reduce drag , but at a speed of around 1,350 mph, friction with the air causes significant heatinggp, of the plane, enough to make the Concorde grow in length by 7 inches! (To maintain this speedfd for one hour, th e Concorde must burn over 6 ,700 gallons of fuel.) Thermal Expansion and Teeth Thermal Expansion and Teeth

Coefficients of linear expansion:

Enamel: 11.4 x 10-6 °C Dentin: 8.3 x 10-6 °C

If you quickly switch from eating/drinking something hot to something 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. Exercise Thermal Expansion • On a cold winter day with T =- 10 °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 RllRecall: ∆L/L = α ∆T (A) 300.001 m (B) 300 .16 m (C) 316 m (D) 420 m