Thermodynamics:C l i t H tCalorimetry, Heat

• Physics 1Physics 1

Internal EnergyInternal energy (also called thermal energy) is the energy an object or substance is due to the kinetic and potential energies j p gassociated with the random motions of all the particles that make it up. The kinetic energy is, of course, due to the motion of the

i l T d d h i l i i lid iparticles. To understand the potential energy, imagine a solid in which all of its molecules are bound to its neighbors by springs. As the molecules vibrate the springs are compressed andAs 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 hi h i T ihigher its temperature. Temperature is proportional to the average kinetic energy of the atoms or molecules that make up aof the atoms or molecules that make up a substance.

Internal Energy vs. HeatInternal Energy vs. Heat

The term heat refers is the energy that is transferred from one gybody or location due to a difference in temperature. This is similar to the idea of work, which is the energy that is transferred from

b d h d f h b h H ione body to another due to forces that act between them. 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 wemolecules. 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 ishot 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 boundary between a system and its surroundings is the y y gsystem wall.

• If heat cannot pass through the system wall, it is termed an p g y ,adiabatic wall, and the system is said to be thermally isolatedor thermally insulated.

• If heat can pass through the wall, it is termed a diathermalwall.

• Two systems connected by a diathermal wall are said to be in thermal contact.

Isolated, Closed and Open Systems 2, p y

• An isolated system cannot exchange mass or energy with its disurroundings.

• The wall of an isolated system must be adiabatic.y

• A closed system can exchange energy, but not mass, with y g gy, ,its surroundings.

• The energy exchange may be mechanical (associated with aThe energy exchange may be mechanical (associated with a volume change) or thermal (associated with heat transfer through a diathermal wall).

• An open system can exchange both mass and energy with p y g gyits surroundings.

Isolated, Closed and Open Systems 3

Isolated System

ClosedSystem

OpenSystemSystem

Neither energy nor mass can be exchanged.

System

Energy, but not mass can be exchanged.

System

Both energy and mass can be exchanged.

Units for Heatf

Like any type of energy, the SI unit for heat is the Joule.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 theuse 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” EnergyS 1 k bl k f i i lidi t 7 // Suppose a 1 kg block of ice is sliding at 7 m/s. This is the speed of the center of mass of the block, not the speed of each individual water

vcm = 7 m/s

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 motionthe 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. Otherwise, it would be O ,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 theRegardless, the total energy of the block is the kinetic energy of the center of mass + the internal energy: Ktotal = Kcm + Eint

Temperature ScalesF h h i f 32 °F b il 212 °FFahrenheit: water freezes at 32 °F; boils at 212 °FCelsius: water freezes at 0 °C; boils at 100 °CKelvin: water freezes at 273.15 K; boils at 373.15 K

A h f 100 °C d h f 180 °F ThiA change of 100°C corresponds to a change of 180°F. This means5C° = 9F° or 1C° = 1.8F° Note that the degree symbol is on the opposite side of the letter indicating that we’re talking aboutopposite side of the letter, indicating that we re talking about temperature differences. In other words, five steps on the Celsius scale is equivalent to nine steps on the Fahrenheit scale, but 5 °C is q pcertainly 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 + 32One 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 + 273.15scale. These scales are off by 273.15 K, so: K C 273.15Room temperature is around 293 kelvins, which is 20°C, or 68°F.

Absolute Zero & the Kelvin ScaleTh K l i l i t th t it i t i th ld t iblThe Kelvin scale is setup so that its zero point is the coldest possible temperature--absolute zero, at which point a substance would have zero internal energy. This is -273.15 °C, or -459.69 °F. Absolute zerozero 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 extrapolatingTemperature for a variety of gases and extrapolating.

P

A gas exerts no pressure when at p

absolute zero.

T (°C)-273.15°C 0°C

Thermal EquilibriumT b di id t b t th l ilib i if th t thTwo bodies are said to be at thermal equilibrium if they are at the same temperature. This means there is no net exchange of thermal energy between the two bodies. The top pair of objects are inenergy 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 coldheathot cold

26°C 26°C

No net heat flowTh t l bj t t th t d th f iThe two purple objects are at the same temp and, therefore are in 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 thecontact. 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• Convection: Energy is transferred from one body to a cooler one via currents in a fluid (a gas or liquid).

R di ti All bj t t t t di t l t ti• Radiation: All objects, at any temperature, radiate electromagnetic radiation (light of visible and invisible wavelengths). Unlike conduction & convection no medium (matter of any type) isconduction & 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, kH t t f i d ti d ib d f lid b k Th lHeat transfer via conduction was described a few slides back. Thermal conductivity, k, refers how easily heat can move through a material. Metals have high thermal conductivity, meaning heat passes throughMetals 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 issides of its metal enclosure. The rate at which heat is transferred is given by:

Lk A (T2 - T1)H = L (T2 T1)H

T TA = area of side wallL thi k f llT2 T1

heat

L = thickness of wallk = thermal conductivity of the metalT2 - T1 = temperature differenceeat T2 T1 temperature difference

H is simply power, and its SI unit is the Watt.

SI Units for Thermal Conductivityf y

kA (T2 - T1)H = L(T2 T1)H =

k must have units that cancel out all the units on the right,k must have units that cancel out all the units on the right, leaving only the units for H. The units are:

m·KW or equivalently,

m·°CW

m K m C

Since one kelvin is as big a change in temp as one degreeSince 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 kNote: k for thermal conductivity is not the same as the kin 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 glass with air or some other gas between the panes. Which type of window a double pane or a thick

heat

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 do ble pane indo has thermal cond cti it that isheat double pane window has thermal conductivity that is about 35 times lower than that of the glass itself. So much more heat would be transferred through themuch 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 minimizing heat transfer, , g ,a double or triple pane window?

answer:heat

answer:

The double pane window has more air between the outer panes so its thermal conductivity is lowerouter 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

d i i f h id hand carrying its energy from the warm side to the cold side. The middle pane in the triple pane window reduces the energy transfers due to convection and is

heat

reduces the energy transfers due to convection and is the better window (but probably more expensive).

R ValueThe R value of a material is its “thermal resistance” and refers to how good an insulator is. Here’s how it’s defined:

R = kLk

As in previous equations:L = the thickness of the materialk = 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 l t Th i l ti i th ttiit insulates. Thus, more insulation in the attic can save energy.

Wind & Heat LossA breeze can cool us off in the summer, and wind can make us feel colder in the winter. Why is this?

answer: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 keeps us warm in the winter is because it traps air p pthat is heated by our bodies. (Wearing layers is like having a triple pane window.) A thin layer of stagnant air also surrounds the outside

ll f b ildi d h l i l h i d d bl hiwalls of buildings and helps insulate them. Wind tends to blow this 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 buildingto us, and the greater the heat loss from a building

Trees around you home can save energy in two ways: blocking wind in the inter; and shielding o r home from e cess solar radiation inin the winter; and shielding your home from excess 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 then objects

(examples upcoming)

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 property of heat.”p p y

• First Law: Energy is always conserved. It can change forms: kinetic potential internal etc but the total energy is a constantkinetic, 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: During any natural process the total amount of g y pentropy 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 Equationf i l l l iBecause most systems are many up of so many particles, calculating

entropy via probabilities would be very difficult. Fortunately, we are normally concerned only with changes in entropy If we have anormally 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:

T∆S = ∆Q T

∆S = change in entropy∆S change in entropy∆Q = change in internal energy (heat flow)

T = absolute temperatureT absolute temperature

The 2nd Law of Thermodynamics says that during any process:

∆Suniverse = ∆Ssystem + ∆Ssurroundings ≥ 0

Change in Entropy ExampleA l d i h t d d th bl bA glass rod is heated and then blown by a glassblower. When it is at 185°C it is brought outside to cool. 3200 J of heat are transferredoutside 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 S= ∆Sglass + ∆Sair

∆Qglass ∆Qair

3200 J

Tglass

glassTair

Qair+=

3200 J291 K3200 J

458 K= -3200 J +

= -7 J/K + 11 J/K = +4 J/K

Change in Entropy Example (cont.)A th l l d d th t th i t did ’tAs the glass cooled we assumed that the air temp didn’t go up appreciably due after the heat transfer, which would have compli-cated the problem. Important points:cated the problem. Important points:

• The temps were converted to kelvins.

Th l l h h l i i d h 1 t L• The glass lost as much thermal energy as air gained, as the 1st Law requires.

Q S• ∆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 ConsequencesH t ill t fl f ld b d t h t b d• Heat will not flow from a cold body to a hot body.

• “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 of energy, bodies tend to reach thermal p gy,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 entropysystem, the energy we expend in doing it increases the entropy of the surroundings even more.• A perpetual motion machine is impossible to make A• 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 itenergy from a hot body and do as much work as the energy it absorbed. • During any process the entropy of the universe cannot• 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 awill lead to an increase in entropy for the surrounding by a greater amount.

Thermal Equilibrium and the Zeroth LawThermal Equilibrium and the Zeroth Law

• If warm and cool objects are placed in thermal contact, a a d coo objects a e p aced t e a co tact,energy, known as heat, flows from the warm to the cold object until thermal equilibrium is established.

• Zeroth Law of ThermodynamicsTwo systems separately in thermal equilibrium with a thirdTwo 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 θ = θ and θ = θ then θ = θ

25

if θ1 = θ2 and θ1 = θ3, then θ2 = θ3.

Thermodynamic VariablesThermodynamic Variables• Thermodynamic variables are the observable macroscopic

variables of a system such as P V and Tvariables 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 variablesthey are known as state variables.

• Extensive variables depend on the size of the system; e.g. l t ti tmass, volume, entropy, magnetic moment.

• Intensive variables do not depend on size; e.g. pressure, temperature magnetic fieldtemperature, magnetic field.

• An extensive variable may be changed to an intensive fvariable, known as a specific value, by dividing it by a

suitable extensive variable, such as mass, no.of kmoles, or no. of moleculesof molecules.

• Example: the specific heat is normally (heat capacity)/(mass).

Equilibrium StatesEquilibrium States

• An equilibrium state is one in which the properties of the q p psystem do not change with time.

• In many cases, an equilibrium state has intensive variables hi h if th h t th twhich are uniform throughout the system.

• A non-equilibrium state may contain intensive variables which i d/ tivary in space and/or time.

• An equation of state is a functional relationship between the t t i bl if P V d T th t t i bl thstate variables; e.g. if P,V and T are the state variables, then

the equation of state has the form f(P, V, T) =0.

I 3 di i l P V T• In 3-dimensional P-V-T space,an equilibrium state is represented by a point,and the equation of state is represented by a surface

27

and the equation of state is represented by a surface.

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• The initial and final states of a process are its end-points.

• A quasistatic process is one that takes place so slowly that th t b id d i th hthe system may be considered as passing through a succession of equilibrium states.

• A quasistatic process may be represented by a path (or line)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.

28• A reversible change must be quasistatic, but a quasistatic

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 constantAn 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.y

• 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 τ is much longer than the time ∆t for which the system is

29

time τ0 is much longer than the time ∆t for which the system is observed; i.e. τ0» ∆t .

Three Types of ProcessThree Types of Process

PPIsothermal processIsothermal process Adiabatic processAdiabatic processAdiabatAdiabat

PPSystemSystem

IsothermIsotherm

VVHeat bath or reservoirHeat bath or reservoir

Adiabatic free expansionAdiabatic free expansion PP

●●11

22

End pointsEnd points

30VV

●●22

Boyle’s Law and the Ideal Gas Scale

• Boyle’s LawAt ffi i tl l• At sufficiently low pressure, the product PV was found to be constant for gases held at

hbe constant for gases held at a given temperature θ; i.e.

PV = f(θ) for P → 0P

PV f(θ) for P → 0.

• Fixed points (prior to 1954)Th i d t i tP P + gρh • The ice and steam points were defined to be 0oC and 100oC exactly

P = PA + gρh

100 C exactly.

• The ideal gas (or kelvin) l d fi d

P = a(TC + 273.15)

31

scale was defined asTK = TC + 273.15.

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 and the gas constant R = 8 314 x 103 J/(K kmol)temperature in K, and the gas constant R = 8.314 x 103 J/(K.kmol).

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

PP PP VV

T increasingT increasing

32VV TT TT

V increasingV increasing P increasingP increasing

Ideal gas: Macroscopic descriptiong p p

• Consider a gas in a container of volume V, at pressure P, and at temperature Tand at temperature T

• Equation of state– Links these quantitiesq– Generally very complicated: but not for ideal gas

l Equation of state for an ideal gasl Equation of state for an ideal gas

Collection of atoms/molecules moving randomly

PV RT R i ll d h i l

No long-range forces

Their size (volume) is negligiblePV = nRT R is called the universal gas constant

In SI units R =8 315 J / mol·K n = m/M : number of molesIn SI units, R =8.315 J / mol K n m/M : number of moles

Boltzmann’s constantl Number of moles: n = m/M m=mass

M=mass of one mole

One mole contains NA=6.022 X 1023 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

Avogadro s number = number of carbon atoms in 12 g of carbon

( A )

PV = N kB T

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

P V d T h h d i i bl

kB is called the Boltzmann’s constant

• P, V, and T are the thermodynamics variables

Th Id l G L

RTV

The Ideal Gas Law

Wh t i th l f 1 l f t STP ?

nRTpV =

What is the volume of 1 mol of gas at STP ?T = 0 °C = 273 Kp = 1 atm = 1 01 x 105 Pap 1 atm 1.01 x 10 Pa

RTV=

( ) K273Kmol/J31.8Pn

⋅=

=

( )

l422m02240

Pa10x01.13

5

==

=

l4.22m0224.0 ==

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 psymbol is C.

For example, one way to express the specific heat of water isFor 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 very high specific heat, so it takes more energy to y g p , gyheat 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 i k ” i h l b b d i h d“heat sinks” storing thermal energy absorbed in the summer and slowing releasing it during the winter. Large bodies of water thereby help to make local climates less extreme in temperaturethereby help to make local climates less extreme in temperature from season to season.

Specific Heat Equation

Q = mC∆TQ = thermal energym = massm = massC = specific heat∆T = change in temp∆T change in temp

Ex: The specific heat of silicon is 703 J / (kg· ºC). How much p ( g )energy is needed to raise a 7 kg chunk of silicon 10 ºC? answer:

703 Jkg· ºCQ = 7 kg · ·10 ºC = 49 210 J

Note that the units do indeed work out to be energy units.

CalorimetryS h d i k k h h h f h fi h i ’Schmedrick takes another horseshoe out of the fire when it’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 bucketThe 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 / (kg·ºC). Let’s find the temp p ( g ) pof the horseshoe and water once equilibrium is reached.

Let’s assume that the container allows no h h h t i liheat to escape. Then the 1st Law implies that all heat the shoe loses is gained by the water Since one milliliter of water has awater. Since one milliliter of water has a mass of one gram, the bucket contains 2.5 kg of water. At thermal equilibrium g qthe water and shoe are at the same temp. The total thermal energy in the bucket d h b i i di ib ddoes not change, but it is redistributed.

continued on next slide

Calorimetry (cont.)

Let T = the equilibrium temperature.Q lost by iron = Q gained by watery g y

mironCiron ∆Tiron = mwaterCwater ∆Twater

(1.9 kg)(448 J /kg·ºC)(275 ºC - T) = (2.5 kg)(4186 J /kg·ºC)(T - 25 ºC)

N t h th ∆T t itt th tNote how the ∆T terms are written so that each side is positive. We’ve got a simple linear equation with T on both sideslinear 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 b i lid i th ifi h t fbeen invalid, since the specific heat for steam is different than that of water.

Latent HeatThe 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

l t t t i i t li id t t bexample, to turn water ice into liquid water, energy must 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 liquidenough, however, since water can exist at 0 C 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 theinto 0 C water. The energy increases the internal energy of the water but does not raise its temp. When frozen, water molecules are in a crystalline structure, and energy is needed to break this y , gystructure. 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 = mLvQ = thermal energym = mass

L is the energy per unit mass needed to change the state

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: L (the latent heat of fusion) for gold is 6440 J/kgEx: 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 istemp will not become liquid until additional heat is added. The amount of heat needed is: (6440 J/k ) (0 005 k ) 32 J Th li id ld ill till b(6440 J/kg) (0.005 kg) = 32 J. The liquid gold will still be at 1063 ºC.

Latent Heat / Specific Heat ExampleS i 1800 k i t ithSuperman vaporizes a 1800 kg ice monster with his heat ray vision. The ice monster was at -20 ºC After being vaporized he is steam at

Substance Specific Heat (in J /kg·ºC)

20 C. After being vaporized he is steam at 135 ºC. How much energy did Superman expend?

Substance Specific Heat (in J /kg C)ice 2090liquid water 4186steam 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 ·105 J /kg) ice to water, const. temp of 0 ºC

f v

+ (1800 kg)(3.33 10 J /kg) ice to water, const. temp of 0 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 & EntropyA 10 g ice cube at 0 ºC falls to the ground and melts The tempA 10 g ice cube at 0 C falls to the ground and melts. The temp outside is 10 ºC. Calculate the change in entropy of the universe due to the melting of the ice only. answer:g y

For the cubie: Q = m Lf = (0.01 kg)(3.33 ·105 J /kg) = + 3330 J. Thi i th b b d b th i f th diThis is the energy absorbed by the ice from the surroundings.∆Sice = ∆Qice /Tice = +3330 J / 273 K = +12.198 J/K.

h di Q i h di lFor the surroundings: Q = -3330 J, since the surroundings lost as much thermal energy as the cubie gained. The temperature of the backyard does not decrease significantly though with such a smallbackyard does not decrease significantly, though, with such a small energy loss. ∆Ssurr = ∆Qsurr /Tsurr = -3330 J / 283 K = -11.767 J/K.

For the ni erse: ∆S = ∆S + ∆S 12 198 J/K 11 767 J/KFor the universe: ∆Suniv = ∆Ssurr + ∆Sice = 12.198 J/K - 11.767 J/K = +0.431 J/K. Thus, the 2nd Law is satisfied.

Thermal ExpansionAs a material heats up its atoms/molecules move or vibrate more pvigorously, and the average separation between them increases. This results in small increases in lengths and volumes. Buildings, railroad t k b id d hi h t i th l i j i t ttracks, bridges, and highways contain thermal expansion joints to prevent cracking and warping due to expansion. The amount of expan-sion depends on the original length/volume, the type of material, and the p g g , yp ,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. Wh lid f i l t i l d it d ti ll iWhen a solid of a single material expands, it does so proportionally in all directions. Since volume has 3 dimensions and length is only 1, β = 3α. L th iβ 3α.

∆LL = α ∆T

Length expansion:

L

Volume expansion:

cold solid hot solid∆VV = β ∆T

Thermal expansionThermal 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 ∆T0 0 0

• ∆L = α L0 ∆T (linear expansion)• ∆V = β V0 ∆T (volume expansion)

Vβ 0 ( p )

• α ≅ 3β V + ∆V

ThermostatsBimetallic strips are used in thermostats, at least in some older ones. When the temperature changes, the strip bends, making or b ki l t i l i it hi h th f t tbreaking an electrical circuit, which causes the furnace to 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 toof mercury, which then bridges two wires and allows current to flow.

Thermal Expansion & The ConcordeTh C d i i j t d f h t t l t l iThe Concorde is a supersonic jet made of a heat tolerant aluminum 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 athe 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 heating of the plane, g p ,enough to make the Concorde grow in length by 7 inches! (To maintain this

d f h thspeed for one hour, the Concorde must burn over 6 700 gallons ofover 6,700 gallons of fuel.)

Thermal Expansion and TeethThermal Expansion and Teeth

Thermal Expansion and TeethThermal Expansion and Teeth

Coefficients of linear expansion:p

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

If you quickly switch from eating/drinking something hot tosomething cold, the brittle enamel will contract more than theg ,dentin, and develop small cracks called crazes.

Crazing:Crazing:

Thermal expansion matching is important in choosing materials for fillingsmaterials for fillings.

ExerciseExerciseThermal Expansion

• On a cold winter day with T = -10 °COn 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

R ll ∆L/L ∆TRecall: ∆L/L = α ∆T(A) 300.001 m (B) 300 16 m(B) 300.16 m (C) 316 m (D) 420 m(D) 420 m