Honors Physics

Notes Nov 16, 20

Persans 1 Properties of solids

Persans 2 Persans 3 Vibrations of atoms in crystalline solids Assuming only nearest neighbor interactions (+Hooke's law)

Fs = C(us+1 ! us ) + C(us!1 ! us ) us is the displacement of the sth plane. The equation of motion of the sth plane is thus: 2 d us M = C (us+1 + us!1 ! 2us ) dt 2 We look for solutions where all atoms move as a travelling wave e!i"t eikx . 2 !M" us = C (us+1 + us!1 ! 2us ) !M" 2eiksa = C (eika + e!ika ! 2)eiksa

2 # 2C $ # 4C $ 2 ka " = % &(1! cos ka)= % &sin ' M ( ' M ( 2 1/ 2 # 4C $ ka " = % & sin ' M ( 2 Persans 4 Normal Mode Frequencies (Continued)

Persans 5 Topics Heat and temperature Kinetic model Properties of materials Thermal expansion Transfer of heat First Law of thermodynamics: Heat and

Persans 6 What is temperature? • Temperature is a measure of the average kinetic energy of the particles in a space. It is proportional to the internal energy of the system.

What is heat? • Heat is the energy that flows between a system and its environment by virtue of a temperature difference between them .

Persans 7 Temperature scales • Common scales are based on temperatures at which everyday things happen. • The scale is based on the freezing point of brine (~00F) and the temperature of the human body (~1000F). • The Centigrade or scale assigns the freezing point of pure water to 00C and the boiling point to 1000C. Celsius is the metric unit and is used worldwide, except in the US. • The Kelvin scale is based on thermodynamics and kinetic theory. 9 5 TF = TC + 32; TC = (TF ! 32); TK = TC + 273 5 9Persans 8 Units of Heat • Since heat is a form of energy, the SI unit is the . • Another common unit of heat is the kilocalorie, which is the amount of heat energy needed to raise the temperature of 1 kg of water by 10C. 1kcal=4185 J • Some US engineers insist on using the BTU (British Thermal Unit) 1 BTU is the amount of heat needed to raise the temperature of 1 lb of water by 10F. 1BTU=0.252 kcal

Persans 9 Kinetic Theory of Gases • A series of experiments on dry air led to the following equation of state for an ideal gas: pV = NkBT p = pressure; V = volume; N = number of gas molecules in that volume; !23 !5 kB = Boltzman's constant=1.38"10 J/K=8.65"10 eV/K pV = nRT n = number of moles of gas

R = universal =N AkB

Persans 10 Kinetic theory: a model 1. A gas consists of molecules 2. The molecules are in random motion and obey Newton’s laws. 3. The total number of molecules is very large (so we can use statistical averages) 4. The average distance between molecules is large compared to the size of the molecules. 5. Molecules experience only collisions forces. 6. Collisions are elastic and of short duration.

Persans 11 Deducing pressure from kinetic energy

• Find the average change in momentum when a molecule bounces elastically between two massive walls separated by distance L. • Assume that all molecules have the same average speed. • Compare the pressure we deduce with the ideal gas law.

Persans 12 2mvx1 !momentum = "2mvx1; Fx1 = (2L / vx1) F mv2 p = x1 = x1 ; 1 A L3 m m For many atoms: p = # v2 = N v2 L3 xi L3 x 2 2 2 2 2 v = vx + vy + vz = 3vx

1 m 2 2 N $ 1 2 % p = N v = & mv ' 3 V 3 V ( 2 )

2 $ 1 2 % pV = N & mv ' and comparing to pV = NkBT 3 ( 2 ) 1 3 we conclude that : mv2 = k T 2 2 B Persans 13 Properties of common matter • Phases – solid – liquid – gas – (plasma) • Phase changes – melting – vaporization (boiling) – dissociation of valence electrons from atoms • Thermal expansion

Persans 14 • Anharmonic Forces Interatomic forces include higher order terms than linear term in atomic displacement.

Figure 6-2 (a) Harmonic and (b) Anharmonic potential energy functions for the interaction of two atoms. R is the atomic

separation , while E1, and E2 represent two possible vibrational energies. For (a) an increase in energy does not result in a change in the average atomic separation, given by the midpoints of constant energy lines. For (b) an increase in energy results in an increase in average separation. Persans 15 Thermal Expansion • To a good approximation over moderate temperature ranges, the change in length of a solid increases proportionately to the change in temperature. "L # !L"T L = original length, != linear expansion coefficient

material α (10-6/C) Al 25 Steel 12 Glass 3 Concrete ~12

Persans 16 Thermal expansion of area and volume For solids, the amount by which the area of a plate or the volume of a block changes can be easily estimated from the linear expansion:

2 !L !A(simple square) = (L + !L) " L2 = 2L!L + !L2 # 2L!L = 2A L !A !L # 2 A L 3 3 !V (L + !L) " L !L = # 3 V L3 L

Persans 17 Volume Thermal Expansion • The change in volume with temperature can be defined for solids, liquids and gases. #V $ !V #T ! $ 3" V = original volume, != volume expansion coefficient material β (10-6/C) Al 75 Steel 35 Glass 9 Concrete ~35 Water 210 Air 3400

Persans 18 Thermal Expansion: The special case of ideal gases • For many gases, it is found that the pressure, volume, and temperature are related in a simple way: pV = nRT p = pressure; V = volume; n = number of moles of gas; R= a constant

R = 8.314 J/(mol K) mass (grams) n = molecular mass (g/mol)

Persans 19 Exercise: 16 Nov 06 ______• An ideal gas is cooled from 300C to 00C. By what ratio does its volume change?

Persans 20 Exercise: 16 Nov 06 • Two aluminum plates are held in place by a steel bridge as shown. The temperature for the figure on the left is 0oC. The plates heat in the sun to 40oC and expand. This causes a buckle as shown in the figure on the right.

20 m

a) Find the differential expansion of Al and steel. b) Estimate the height of the buckle.

Persans 21 a) Find the differential expansion of Al and steel. !(Steel) =11#10-6 ; !(Al) = 23#10"6 $$L = L$!$T = 20m#(12#10"6 K "1 )# 40K = 9.6x10"3 meters b) Estimate the height of the buckle. Use L=10 m for one side. 2 % $$L ( 2 2 & L + $LS + ) = (L + $LS ) + h ' 2 * 2 2 $$L L2 + $L + + 2$L L + L$$L + $L$$L S 4 S 2 2 $$L = (L + $LS ) + + L$$L + $L$$L 4 h2 + L$$L =10m#10"2 m = 0.1m2 , h + 0.3m

Persans 22 Heat capacity and specific heat • When heat energy is added to or subtracted from an object, its temperature changes (unless it undergoes a phase change). In many cases, near room temperature, the temperature change is linear in energy added. !Q = C!T C = heat capacity • The specific heat takes the mass of material being heated into account. c = C / m ! "Q = mc"T

Persans 23 Some approximate specific

• Aluminum 0.9 kJ/(kg C) • Glass 0.84 • Steel 0.45 • Wood 1.7 • Water (liquid) 4.2 • Ice 2.1 • Air (const V) 1

Persans 24 Latent heat of phase change • When the phase of material changes, it takes up or gives off heat but the temperature does not change. The energy goes into breaking or making bonds between atoms. • The energy relationship is expressed using the latent heat of melting (or condensation, or vaporization, or freezing) !Q = Lm

Persans 25 Persans 26 Example: Energy required to heat and boil a cup of water • Estimate the amount of energy that is required to heat a cup of water (0.2 kg) from room temperature to boiling.

• Estimate the amount of energy that is required to convert that cup of water to vapor.

Persans 27 Example continued

To raise the temperature of 0.2 kg of water from 20°C to 100°C we use the approximate formula: J !Q = cm!T = 4190 •.2kg •80K = 67000J = 67kJ kgK To vaporize the same amount: kJ !Q = L m = 2256 ".2kg = 451kJ v v kg The total energy is thus:

!QT = 518kJ

Persans 28 Comparison of thermal and gravitational potential energy Let's estimate temperature change due to conversion of gravitational potential energy to heat energy from dropping a student from from the top of a 9 story building: m m !KE = mgh " 70kg # 4 #9 floors#10 = 25kJ (6 kcal). floor s2 If all of this energy were converted to heat the temperature of the water would increase by m J 10 #36m 360 !Q mgh gh 2 kg !T " = = = s = = 0.086K. cm cm c J J water 4200 4200 kgK kgK Persans 29 The flow of heat

Persans 30 The flow of heat • Heat energy flows from warmer volume to colder volumes. • Convection: energy transfer by collective motion of a macroscopic volume of the fluid. • Conduction: energy transfer by transfer of vibrational kinetic energy on the atomic scale. • Radiative: energy transfer by emission of electromagnetic radiation from a hot material.

Persans 31 Heat Conduction • Heat transfer by conduction can be represented using a simple and sensible relation: For a slab of thickness d and area A

dQ ! TH #TC " ! dT " H = = kA$ % = kA$ % dt & xH # xC ' & dx ' A H= rate of heat flow ! W " k = specific $ % & m • K ' A = area; dx TH (TC ) is the hot (cold) temperature Persans 32 Thermal conductivity of common materials

Material ρ (W/m.K) BTU/(ft 0F hr)

Still Air 0.026 0.011

Aluminum 235 140

Glass 1.0 0.50

Concrete 0.7 0.35

Wood 0.1 0.05

Persans 33 Solving a differential equation for temperature of an object

• We can use what Q = cmTH we know about dQ dT # T !T $ heat flow and = !cm H = kA H C internal energy to % & dt dt ' xH ! xC ( deduce the dTC temperature of an TC is a constant, so = 0 object as a function dt of time. d (TH !TC ) kA ! = dt (TH !TC ) "xcm final final kA !ln T !T = t ( H C )initial "xcm initial kA ! t "xcm (TH !TC )= e Persans 34 Exercise ______• The temperature difference between two sides of a 1 cm thick sheet of foam board is 0 T1 C. The rate at which heat passes through it is 400 W per square meter. – What would the heat flow be if the thickness were increased to 2 cm?

– What would the heat flow be if the temperature difference were doubled?

Persans 35 Convection • “Still air” is a very good insulator. The problem is that air can flow and its density depends on temperature. (1/ρ~V/nR~T). – Cold air is denser than hot air and flows downward. – Good insulators frequently just use air, but trap it so in can’t convect. • Heat can then be carried around by the flow of macroscopic volumes of air whose motion is caused by temperature differences.

Persans 36 Radiative heat transfer • All objects radiate heat, but hot objects radiate a lot more of it per unit area. $Q emit = !" AT 4 $t obj ! = Stephan-Boltzman constant=5.7%10#8 W/m2. " = emissivity (~1 for black surface, ~0.05 for shiny metal) • All objects absorb heat from their environment as well.

#Q absorb = !" AT 4 #t env

Persans 37 Heat energy and work

Persans 38 Work by a system

• If we change the volume and there is a net pressure difference between inside and outside a cylinder, there is work done.

V f W = ! pdV Vi For many cases we have to do the integral because pressure varies with volume and temp.

Persans 39 The p-V diagram p

• isothermal • adiabatic • constant volume work • constant pressure

V

Persans 40 First Law of Thermodynamics

• The change in internal energy (cmΔT) of a material is the difference between the heat added to the material system and the work done by the system.

!Eint = Q +W Q = heat added to the material W = work done on the system (W is negative if the system expands against an external pressure) Persans 41 Using Thermodynamic Laws: The Ideal Gas

Equation of state for an ideal gas:

pV = nRT = NkBT n = number of moles of gas N = number of molecules !23 kB =1.38"10 J / K

Persans 42 Ideal gas: Adiabatic process • In an adiabatic process, no heat energy enters or leaves the system.

#Q = 0 so dEint = dW = " pdV But the internal energy only depends on the temperature dEint = nCV dT so pdV = "nCV dT We also know d( pV ) = d(nRT ) so Vdp + pdV = nRdT from which Vdp = nCpdT. dp dV C = "! ; ! = p p V CV p = AV "!

Persans 43 Ideal gas: Isothermal process No change in internal energy: Q +W = 0 nRT p = V

Constant volume Work done by system must be zero.

dEint = dQ + dW = dQ All the heat added to a system is stored as internal energy.

Persans 44 Work in special cases: Ideal gas Constant pressure:

V f V f Wp = " pdV = p " dV = p(Vf !Vi ). Vi Vi Constant Temperature:

NkBT dV #Vf $ p = % WT = NkBT " = NkBT ln& ' V V ( Vi )

Persans 45 Some special processes

Adiabatic process: No transfer of heat is made between the system and the outside world. (!Q = 0) " !Eadiabatic = -W

Constant volume process: No work is done on/by the system.

W = 0; !Ecv = !Q

Free expansion: An adiabatic process in which no work

is done on or by the system. !E fe = 0.

Cyclical process: A process in which the system returns to

a specific state or position on a p-V diagram. !Ecyclic = 0. Work done by the system must equal energy added.

Persans 46 A Heat/Work Conversion Engine

• By applying pressure, allowing a piston to slide, and adding or subtracting heat, we can convert heat energy into work, or mechanical, energy.

Persans 47 Heat engines move heat between two reservoirs and produce or use work

W ! = (any engine) QH Persans 48 The Carnot Cycle • The most efficient heat engine cycle is the Carnot cycle, consisting of two isothermal processes and two adiabatic processes. The Carnot cycle is the most efficient heat engine cycle allowed by physical laws.

QH " QL TH !CARNOT = =1" QH TL

Persans 49 Persans 50