THERMALPHYSICSNOTES

PHYSICS B4B

BAKERSFIELD COLLEGERick Darke (Instructor)

THERMALPHYSICSNOTES

THERMODYNAMICS TERMS

thermodynamics - that branch of physics which dealswith heat and temperature (also called thermal phys-ics)

system - a definite quantity of matter enclosed byboundaries (real or imaginary)

open system - a system, into or out of which massmay be transferred

closed system - a system for which there is no trans-fer of mass across the boundaries

THERMODYNAMICS TERMS

temperature - an index of the average random trans-lational kinetic energy of particles in a system; a rela-tive measure of hotness or coolness

heat (energy) - the energy exchanged between ob-jects because of a difference of temperature; a mea-sure of the random kinetic energy of molecules in asubstance

thermal contact - a condition in which heat may beexchanged between two objects

THERMODYNAMICS TERMS

thermal equilibrium - the condition in which there isno net heat exchange between objects in thermalcontact

thermally isolated system - a system for which thereis no transfer of heat energy across its boundaries

completely isolated system - a system for whichthere is no transfer of mass or heat energy across itsboundaries (thermally isolated and closed)

HOW TO MAKE A THERMOMETER

STEP 1: Obtain a thermometric sub-stance with some temperature responseyou believe to be a linear function oftemperature.

example: liquid mercury. The volume ofthe mercury is a function of its tempera-ture, and in a capillary of fixed diameter,the volume changes in mercury will beobservable as height changes in themercury column.

HOW TO MAKE A THERMOMETER

STEP 2: Create a temperature scale by defining twofixed-pointtemperatures.

example: Define thetemperature of a wa-ter-ice equilibriumsystem at 1.0 atm tobe 0°C. Define thetemperature of a wa-ter-steam equilibriumsystem at 1.0 atm tobe 100°C.

100°C

0°C

HOW TO MAKE A THERMOMETER

STEP 3: Create a graduated scale bydividing the interval between the fixed-point temperatures linearly into a num-ber of divisions. This graduated scalecan then be extrapolated in both direc-tions from the two fixed-point tempera-tures.

100°C80°C60°C40°C20°C

0°C

P(atm) PHASE DIAGRAM OF H2O

218 vapori-

fusion zation curve

liquid

curve

1.00

(water)

solid triple(ice) point

.006 sublimation gas

curve (steam)

-273.15 0 .01 T(°C) 100 374

DEFINITION OF THE KELVIN

1 kelvin (S.I. unit of temperature) = 1/273.16 of thethermodynamic temperature of the triple-point of H2O(P = 0.61 kPa and T = 0.01°C)

note: You must use kelvin temperatures in all expres-sions in which the (absolute) temperature T is involved,but you may use either kelvins or degrees celsius inexpressions in which temperature difference ΔT is in-volved.

examples: P = σeAT4 must use kelvinsQ = mcΔT may use kelvins or °C

TEMPERATURE SCALES

scale absolute? conversions

Kelvin yes TK = TC + 273.2

Rankine yes TR = TF + 451.2

Celsius no TC = (TF - 32)(5/9)

Fahrenheit no TF = 1.8TC + 32.0

DON HERBERT (MR. WIZARD)

Donald Jeffry Herbert (7-10-1917to 6-12-2007) was the creator andhost of the shows "Watch Mr.Wizard" (1951-65, 1971-72) and"Mr. Wizard's World" (1983-90),educational television programsfor children devoted to scienceand technology. He also pro-duced many short video pro-grams about science andauthored several popular books about science forchildren.

ABSOLUTE ZERO EXPERIMENT DATA

bath T (°C) P (psia)

dry ice + alcohol -64.8 9.6

room air 37.0 14.7

boiling water 100.0 17.1

ice water 0.0 13.0

ABSOLUTE ZERO EXPERIMENT GRAPHP (psia)

16

14

12

10

8

6

4

2

-300 -250 -200 -150 -100 -50 0 50 100

T (°C)

ZEROTH LAW OF THERMODYNAMICS

If two systems A and B are in thermal equilibrium witha third system C, then they will be in thermal equilib-rium with each other if placed in thermal contact. Twoobjects in thermal equilibrium with each other are atthe same temperature.

and

system A

nonetheatflow

system C

system B

nonetheatflow

system C

system A

nonetheatflow

system B

LINEAR EXPANSION COEFFICIENTS

material ααααα at 20°C (°C-1)

aluminum 24 x 10-6

brass 19 x 10-6

copper 17 x 10-6

concrete 12 x 10-6

steel 11 x 10-6

glass (ordinary) 9.0 x 10-6

glass (pyrex) 3.2 x 10-6

invar (Ni-Fe alloy) 0.9 x 10-6

VOLUME EXPANSION COEFFICIENTS

material βββββ at 20°C (°C-1)

air 37 x 10-4

gasoline 9.6 x 10-4

glycerine 4.9 x 10-4

mercury 1.8 x 10-4

acetone 1.5 x 10-4

ethanol 1.1 x 10-4

PHASE CHANGE TERMS

freezing condensing

ice watersteam

melting boiling

sublimating

resublimating

THERMAL EXPANSION PROBLEM

A copper sphere has a diameter of 2.000 cm and is atroom temperature (20°C). An aluminum plate has acircular cut-out with a diameter of 1.995 cm (also atroom temperature). At what (common) temperaturewould the copper sphere just barely be able to passthrough the hole in the aluminum plate?

THERMAL EXPANSION PROBLEM

A steel rod of circular cross-section and a diameter of5.0 cm spans the 2.5-meter gap between two concretefixtures. At 20°C the beam is not compressed and justtouches each of the fixtures (exerting no force on them).What force is exerted on the fixtures when the tem-perature of the steel beam rises to 80°C, and thedistance between the fixtures remains the same dur-ing the warming? The coefficient of expansion of steelis 1.1x10-5 °C-1, and the Young's modulus of steel is2.0x10-6 Pa.

IDEAL GAS LAW PROBLEM

How many moles of carbon diox-ide would there be in a 3.5-cm3

CO2 cartridge at room tempera-ture (20°C) if the gauge pressureof the gas in the cartridge is 500psi? How many molecules of car-bon dioxide would there be in thecartridge?

IDEAL GAS LAW PROBLEM

An aerosol can contains a gaswhose gauge pressure is 2.0 atmat 22°C. Suppose that the canwill rupture when the gauge pres-sure of the gas inside rises to3.5 atm. If the can were tossedinto a fire, at what temperaturewill the can rupture?

IDEAL GAS LAW PROBLEM

66.0 ft3 of air at atmospheric pres-sure and at 22°C is to be placedinto a 10.0-liter scuba tank. Justafter this transfer, it is found thatthe tank's gauge shows a pres-sure of 3000 psig. What tempera-ture is the air immediately afterthe tank is filled?

IDEAL GAS LAW PROBLEM

A bubble is released from thebottom of a fresh-water lake, 15.0meters below the surface. Thegas in the bubble is 4°C whenreleased and warms to 20°C bythe time it reaches the surface. Ifthe bubble had a volume of 2.0cm3 at release, what will be itsvolume when it reaches the sur-face?

HEAT ENERGY UNITS

calorie: the amount of heat energy required to raisethe temperature of 1 gram of water from 14.5°C to15.5°C (also called the "15-degree calorie" or the "littlecalorie")

Calorie: 1000 calories (also called the kilocalorie [kcal]or the "big calorie")

Btu (British thermal unit): the amount of heat energyrequired to raise the temperature of 1 pound of waterfrom 63°F to 64°F

MECHANICAL EQUIVALENT OF HEAT

mechanical equivalent of heat (MEH): the conver-sion factor between mechanical energy and heat en-ergy:

1 cal = 4.186 J1 Cal = 4186 J1 Btu = 1055 J

JOULE'S EXPERIMENT

In 1845 British physicist JamesJoule presented the paper "Onthe Mechanical Equivalent ofHeat" to the British Associationmeeting in Cambridge. In thiswork he reported the results ofhis best-known experiment, inwhich he estimated the mechani-cal equivalent of heat to be 819ft·lbf/Btu (4.41 J/cal). In 1850,Joule obtained a refined measurement of 773 ft·lbf/Btu(4.16 J/cal).

JOULE'S APPARATUS

Joule's apparatus employed a fall-ing weight, in which gravity doesthe mechanical work in spinninga paddle-wheel in an insulatedbarrel of water. The temperatureof the water is increased throughthe viscous dissipation of me-chanical energy which is con-verted into heat energy.

SPECIFIC HEATS

material c (J/kg.°C) c (cal/g.°C)

aluminum 900 0.22brass 380 0.09copper 387 0.09iron 448 0.11lead 128 0.03glass 837 0.20ice 2090 0.50water 4186 1.00steam 2010 0.48ethanol 2400 0.58

LATENT HEATS (FUSION)

material M.P. (°C) Lf (J/kg) Lf (cal/g)

H2O 0 333,000 80aluminum 660 397,000 95copper 1083 134,000 32lead 327 24,500 5.8ethanol -114 104,000 25sulfur 119 38,100 9.1helium -270 5,230 1.25

LATENT HEATS (VAPORIZATION)

material B.P. (°C) Lv (J/kg) Lv (cal/g)

H2O 100 2,260,000 540aluminum 2450 11,400,000 2700copper 1187 5,060,000 1207lead 1750 870,000 208ethanol 78 854,000 204sulfur 445 326,000 78helium -269 20,900 5.0

HEAT ENERGY PROBLEM

At Vernal Falls in Yosemite Na-tional Park, California, water in theMerced River plummets 97meters from the rim of the falls toa pool below. What is the tem-perature increase of the waterafter dropping this distance? Hint:You need to consider two energyconversions.

HEAT ENERGY PROBLEM

Compute the amount of heat en-ergy required to convert 10 gramsof ice originally at -20°C to steamat 150°C?

CALORIMETRY PROBLEM

A styrofoam cup contains 100grams of water at 60°C. A 200-gram piece of ice at -40°C isplaced in the cup, and the sys-tem is allowed to come to ther-mal equilibrium. Assuming thecup does not take part in any heatsharing, describe thermal equi-librium reached by the system.

THERMODYNAMIC PROCESSES

A process is a continu-ous change in the state ofa material. If the materialis a gas, it is usual to con-sider P as a function of V(with temperature sup-pressed) in graphing theprocess (PV-diagram). Anarrow is used to make thetime-progression of theprocess obvious.

P

P1

P2

V1 V2 V

FIRST LAW OF THERMODYNAMICS

The internal energy change (ΔE12) of a system takenthrough a process from state 1 to state 2 is accountedfor by the heat energy (Q12) transferred to the systemand the work (W12) done on the system by the envi-ronment (conservation of energy).

ΔE12 = Q12 + W12

or in differential form:

dE = dQ + dW = dQ + PdV

ISOBARIC PROCESS

A process throughoutwhich the system pressureremains constant (P = con-stant or dP = 0).

Q12 = (ν/2 + 1)P(V2 - V1)W12 = - P(V2 - V1)ΔE12 = (ν/2)P(V2 - V1)

P

P

V1 V2 V

ISOCHORIC PROCESS

A process throughoutwhich the system volumeremains constant (V = con-stant or dV = 0).

Q12 = (ν/2)V(P2 - P1)W12 = 0ΔE12 = (ν/2)V(P2 - P1)

P

P1

P2

V V

ISOTHERMAL PROCESS

A process throughoutwhich the system tem-perature remains constant(T = constant or dT = 0).

Q12 = PVln(P2 / P1)W12 = - PVln(P2 / P1)ΔE12 = 0

P

P1

P2

V1 V2 V

ADIABATIC PROCESS

A process throughoutwhich the system does notabsorb or expel heat en-ergy (Q12 = 0 or dQ = 0).

Q12 = 0W12 = (ν/2)(P2V2 - P1V1)ΔE12 = (ν/2)(P2V2 - P1V1)

P

P1

P2

V1 V2 V

PROCESSES FOR IDEAL GASES

process Q W ΔΔΔΔΔE

(ν/2)Δ(PV)general (ν/2)Δ(PV)+SPdV -SPdV

isobaric (ν/2+1)PΔV -PΔV (ν/2)PΔV(dP = 0) nCPΔT -nRΔT nCVΔT

isochoric (ν/2)VΔP 0 (ν/2)VΔP(dV = 0) nCVΔT 0 nCVΔT

isothermal PVln(P2/P1) -PVln(P2/P1) 0(dT = 0) nRTln(P2/P1) -nRTln(P2/P1) 0

adiabatic 0 (ν/2)Δ(PV) (ν/2)Δ(PV)(dQ = 0) 0 nCVΔT nCVΔT

THERMAL CONDUCTIVITY

The rate PPPPP at which heat is conducted through a slab ofmaterial (W or Btu/h) with cross-sectional area A andthickness L whose faces are maintained at temperaturesTH and TC is given by:

P = kA(TH - TC)/L

where k is the thermal conduc-tivity of the material of the slab(measured in W/m.°C or Btu/h.ft.°F).

L

A

TH TC

THERMAL CONDUCTIVITIES

substance k (W/m.oC) substance k (W/m.oC)

silver 427 water 0.6copper 397 rubber 0.2gold 314 hydrogen 0.17aluminum 238 helium 0.14iron 80 asbestos 0.08ice 2 wood 0.008concrete 0.8 oxygen 0.024glass 0.8 air 0.023

R-VALUES

substance R-valueand thickness (ft2.°F.h/Btu)

drywall (0.5" thick) 0.45glass pane (1/8" thick) 0.89insulating glass (1/4") 1.54concrete block (filled cores) 1.93brick (4" thick) 4.00styrofoam (1" thick) 5.00fiber glass (3.5" thick) 10.9fiber glass (6" thick) 18.8

THERMAL CONDUCTIVITY PROBLEM

A pane of glass is 1/8" thickand has dimensions 1.5 mby 2.5 m. It separates twoheat reservoirs whose tem-peratures are 10°C and22°C. What is the rate ofheat transport through thepane?

insulator90

Cu Al15

°Cinsulator

°C

THERMAL CONDUCTIVITY PROBLEM

A copper and an alumi-num cylinder are joinedend-to-end as shown.Each has a diameter of4.0 cm and a length of25.0 cm. One end of thecopper cylinder is main-tained at 90°C and one end of the aluminum cylinderis maintained at 15°C . What is the junction tempera-ture?

THERMAL RADIATION

The rate PPPPP of radiant energy loss (watts) from thesurface of an object with surface area A and surfacetemperature T is given by:

P = σAeT4

where σσσσσ is the Stefan-Boltzmann constant (5.67x10-8

W/m2.K4), and e is the emissivity of the surface of theobject (dimensionless parameter). This is called theStefan-Boltzmann Law.

THERMAL RADIATION PROBLEM

The surface of a cube with sides oflength 0.05 m has an emissivity of0.85. If the surface of the cube ismaintained at a temperature of550°C, what is the radiant energy fluxthrough the surface of the cube?

MOLAR SPECIFIC HEATS (J/mol.K)

gas CP CV γγγγγ = CP/Cv

He 20.8 12.5 1.67Ar 20.8 12.5 1.67

H2 28.8 20.4 1.41N2 29.1 20.8 1.40O2 29.4 21.1 1.40

CO2 37.0 28.5 1.30H20 35.4 27.0 1.30CH4 35.5 27.1 1.31

SECOND LAW OF THERMODYNAMICS

It is impossible to construct a heat engine that, oper-ating in a cycle, produces no effect other than theinput of energy by heat from a reservoir and the per-formance of an equal amount of work (Kelvin-Planckformulation).

It is impossible to construct a cyclical machine whosesole effect is to transfer energy continuously by heatfrom one object to another at a higher temperaturewithout the input of energy by work (Clausius state-ment).

ENTROPY

Entropy (S) is a quantity describing the disorder of asystem. It has dimensions of heat energy per tem-perature (cal/K or J/K). It is equal to the Boltzmannconstant (kB) times the natural log of the total numberof ways (W) in which the system can arrange itselfconsistent with energy being conserved.

S = kBlnW

ENTROPY

Entropy is an extensive physical quantity. The totalentropy of a system is equal to the sum of the entro-pies of each of the system's components.

n

Stotal = Σ Sii=1

ENTROPY CHANGE

Calculating the entropy change ΔS for a system isusually a more tractable problem than calculating theentropy itself. The change in entropy of a system takenthrough a process from state 1 to state 2 is most oftenderivable from

2

ΔS12 = S2 - S1 = S dQ/T 1

ENTROPY CHANGE

warming: In changing the temperature of a materialfrom T1 to T2 without changing its phase, use:

ΔS12 = mcln(T2/T1) or nCln(T2/T1)

example: Find the change in entropy of a 250-grampiece of copper heated from 20oC to 180oC. Take thespecific heat of copper to be .092 cal/g.oC.

ENTROPY CHANGE

phase change: In changing the phase of a materialwithout changing its temperature, use:

ΔSΔphase = + mL/T

example: Find the change in entropy of a 1.5-kg massof water (initially at 0oC) when it is fully frozen, produc-ing ice at 0oC. Take the latent heat of fusion for H2Oto be 2.26 x 106 J/K.

ENTROPY CHANGE

free expansion: In allowing a gas to free-expand frominitial volume V1 to final volume V2, use:

ΔS12 = nRln(V2/V1)

example: Find the change in entropy when an idealgas at 1.0 atm and 20oC occupying a chamber whosevolume is 0.5 m3 is allowed to free-expand into anevacuated chamber whose (extra) volume is 2.5 m3.

ENTROPY CHANGE

mixing of gases: In mixing nA moles of gas A, initiallyoccupying volume VA, with nB moles of gas B, initiallyoccupying volume VB:

ΔS12 = nARln([VA + VB]/VA) + nBRln([VA + VB]/VB)

example: A chamber is divided into two compartmentsseparated by a partition. Compartment A (H2 gas): V= .75 m3, P = 1.0 atm, and T = 30oC. Compartment B(CO2 gas): V = .25 m3, P = 1.0 atm, and T = 30oC. Thepartition is removed and the gases mix. Find the changein entropy of the system.

ENTROPY PROBLEM

A 60-gram piece of ice at 0oC isplaced in a container with 200grams of water at 50oC. Find thetemperature of this system afterthermal equilibrium is attained.Use this result to find ΔS for: (1)the water; (2) the ice; and (3) theentire system during this process.Assume that the container doesnot transfer any heat energy.