PHYSICS B4B

BAKERSFIELD COLLEGE

Rick Darke (Instructor)

THERMALPHYSICSNOTES

THERMALPHYSICSNOTES

THERMODYNAMICS TERMS

thermodynamics - that branch of physics which deals with heat and tem-perature (also called thermal physics)

system - a definite quantity of matter enclosed by boundaries (real orimaginary)

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

closed system - a system for which there is no transfer of mass across theboundaries

temperature - an index of the average random translational kinetic energyof particles in a system; a relative measure of hotness or coolness

heat (energy) - the energy exchanged between objects because of a differ-ence of temperature; a measure of the random kinetic energy of moleculesin a substance

THERMODYNAMICS TERMS

thermal contact - a condition in which heat may be exchanged between twoobjects

thermal equilibrium - the condition in which there is no net heat exchangebetween objects in thermal contact

thermally isolated system - a system for which there is no transfer of heatenergy across its boundaries

completely isolated system - a system for which there is no transfer ofmass or heat energy across its boundaries (thermally isolated and closed)

HOW TO MAKE A THERMOMETER

STEP 1: Select a thermometric substance (a substancewith some measurable property that is temperature de-pendent) with a temperature response you believe to bea linear function of temperature.

example: liquid-in-glass thermometer

details: Liquid mercury is the thermometric substance,and the volume of a fixed amount of liquid mercury is themeasurable property that is known to be temperaturedependent. In a capillary and reservoir of fixed diam-eters, the volume changes in the mercury with tempera-ture will be observable as height changes in the mercurycolumn in the capillary.

HOW TO MAKE A THERMOMETER

STEP 2: Create a tempera-ture scale by defining twofixed-point temperatures.

example: Define the tempera-ture of a water-ice equilibriumsystem at 1.0 atm to have atemperature of 0 °C. Definethe temperature of a water-steam equilibrium system at1.0 atm to have a temperatureof 100 °C. Note that the abovetwo fixed-point temperaturesare defined, not measured.

ice / water (1 atm) water / steam (1 atm)

100 °C

0 °C

HOW TO MAKE A THERMOMETER

STEP 3: Create a graduated scale by dividing the intervalbetween the fixed-point temperatures linearly into a num-ber of divisions. This graduated scale can then be ex-trapolated in both directions from the two fixed-point tem-peratures.

100 °C80 °C60 °C40 °C20 °C0 °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 the thermodynamic tempera-ture of the triple-point of H2O (P = 0.61 kPa and T = 0.01 °C)

note: You must use kelvin temperatures in all expressions in which the(absolute) temperature T is involved, but you may use either kelvins ordegrees celsius in expressions 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 (born Donald HerbertKemske) [1917-2007] was the creator and hostof the shows "Watch Mr. Wizard" (1951-65,1971-72) and "Mr. Wizard's World" (1983-90),educational television programs for childrendevoted to science and technology. He alsoproduced many short video programs aboutscience and authored several popular booksabout science for children.

ABSOLUTE ZERO EXPERIMENT SETUP

Four pressures of a constant volume of air were recorded for four knowntemperature environments into which the contained air was immersed.

ice + water boiling water dry ice + alcohol room airT = 0.0 °C T = 100.0 °C T = -64.8 °C T = 37.0 °C

13.0psia

17.1psia

9.6psia

14.7psia

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 GRAPH

P (psia)

16

14

12

10

8

6

4

2

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

T (°C)

linear regression equation:P = 12.78 psia + (.04556 psia/°C)Tr = .9965

regression-interpretedvalue of absolute zero:T(P = 0 psia) = -281 °C

ZEROTH LAW OF THERMODYNAMICS

If two systems A and B are in thermal equilibrium with a third system C, thenthey will be in thermal equilibrium with each other if placed in thermal con-tact. Two objects in thermal equilibrium with each other are at the sametemperature.

and

system A

nonetheatflow

system C

system B

nonetheatflow

system C

system A

nonetheatflow

system B

LINEAR EXPANSION COEFFICIENTS

material α α α α α at 20 °C (°C-1)aluminum 24x10-6

brass 19x10-6

copper 17x10-6

concrete 12x10-6

steel 11x10-6

glass (ordinary) 9.0x10-6

glass (pyrex) 3.2x10-6

invar (Ni-Fe alloy) 0.9x10-6

VOLUME EXPANSION COEFFICIENTS

material βββββ at 20 °C (°C-1)air 37x10-4

gasoline 9.6x10-4

mercury 1.8x10-4

ethanol 1.1x10-4

PHASE CHANGE TERMS

freezing condensing

ice watersteam

melting boiling

sublimating

resublimating

PRACTICE PROBLEM: THERMAL EXPANSION

A solid copper sphere has a diameter of 2.000cm and is at room temperature (20 °C). Analuminum plate has a circular cutout with adiameter of 1.995 cm (also at room tempera-ture). At what common temperature would thecopper sphere just barely be able to passthrough the hole in the aluminum plate? Thecoefficient of expansion of copper is 17x10-6

°C-1, and the coefficient of expansion of alumi-num is 24x10-6 °C-1.

Cu

Al

PRACTICE PROBLEM: THERMAL EXPANSION

A steel rod of circular cross-section and diam-eter 5.0 cm spans a 2.5-meter gap betweentwo concrete fixtures. At 20 °C the rod is notcompressed and just touches each of the fix-tures (exerting no force on them). What forceis exerted on the fixtures when the tempera-ture of the rod rises to 80 °C, but the distancebetween the fixtures does not change? Thecoefficient of expansion of steel is 11x10-6 °C-

1, and the Young's modulus of steel is 2.0x1011

Pa.

concrete

concrete

stee

l

PRACTICE PROBLEM: IDEAL GAS LAW

How many moles of carbon dioxide would therebe in a 3.5-cm3 CO2 cartridge at room tem-perature (20 °C) if the gauge pressure of thegas in the cartridge is 500 psi? How manymolecules of carbon dioxide would there be inthe cartridge?

PRACTICE PROBLEM: IDEAL GAS LAW

An aerosol can contains a gas whose gaugepressure is 2.0 atm at 22 °C. Suppose that thecan will rupture when the gauge pressure ofthe gas inside rises to 3.5 atm. If the can weretossed into a fire, at what temperature will thecan rupture?

PRACTICE PROBLEM: IDEAL GAS LAW

66.0 ft3 of air at atmospheric pressure and at22 °C is to be placed into a 10.0-liter scubatank. Just after this transfer, it is found thatthe tank's gauge shows a pressure of 3000psig. What temperature is the air immediatelyafter the tank is filled?

PRACTICE PROBLEM: IDEAL GAS LAW

A bubble is released from the bottom of afresh-water lake, 15.0 meters below the sur-face. The gas in the bubble is 4 °C when re-leased and warms to 20 °C by the time itreaches the surface. If the bubble had a vol-ume of 2.0 cm3 at release, what will be itsvolume when it reaches the surface?

HEAT ENERGY: UNITS & CONVERSIONS

calorie: the amount of heat energy required to raise the temperature of 1gram of water from 14.5 °C to 15.5 °C (also called the "15-degree calorie"or the "little calorie")

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

Btu (British thermal unit): the amount of heat energy required to raise thetemperature of 1 pound of water from 63 °F to 64 °F

mechanical equivalent of heat (MEH): the conversion factor betweenmechanical energy and heat energy:

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

JOULE'S EXPERIMENT

In 1845 British physicist James Joule presentedthe paper "On the Mechanical Equivalent ofHeat" to the British Association meeting inCambridge. In this work he reported the re-sults of his best-known experiment, in whichhe estimated the mechanical equivalent of heatto be 819 ft·lbf/Btu (4.41 J/cal). In 1850, Jouleobtained a refined measurement of 773 ft·lbf/Btu (4.16 J/cal).

JOULE'S APPARATUS

Joule's apparatus employed a falling weight, inwhich gravity does the mechanical work inspinning a paddle-wheel in an insulated barrelof water. The temperature of the water is in-creased through the viscous dissipation of me-chanical energy which is converted into heatenergy.

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

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 25helium -270 5,230 1.25

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 204helium -269 20,900 5.0

PRACTICE PROBLEM: HEAT ENERGY

At Vernal Falls in Yosemite National Park,California, water in the Merced River plum-mets 97 meters from the rim of the falls to apool below. What is the temperature increaseof the water after dropping this distance? Hint:You need to consider two energy conversions.

PRACTICE PROBLEM: HEAT ENERGY

Compute the amount of heat energy requiredto convert 10 grams of ice originally at -20 °Cto steam at 150 °C?

PRACTICE PROBLEM: CALORIMETRY

A styrofoam cup contains 100 grams of waterat 50 °C. A 40-gram piece of ice at -20 °C isplaced in the cup, and the system is allowedto come to thermal equilibrium. Assuming thecup does not take part in any heat sharing,describe thermal equilibrium reached by thesystem.

PRACTICE PROBLEM: CALORIMETRY

A styrofoam cup contains 100 grams of waterat 60 °C. A 200-gram piece of ice at -40 °C isplaced in the cup, and the system is allowedto come to thermal equilibrium. Assuming thecup does not take part in any heat sharing,describe thermal equilibrium reached by thesystem.

PRACTICE PROBLEM: CALORIMETRY

A styrofoam cup contains 100 grams of waterat 20 °C. A 200-gram piece of ice at -60 °C isplaced in the cup, and the system is allowedto come to thermal equilibrium. Assuming thecup does not take part in any heat sharing,describe thermal equilibrium reached by thesystem.

FIRST LAW OF THERMODYNAMICS

The increase in internal energy (ΔE12) of asystem taken through a process from state 1to state 2 is accounted for by the amount ofheat energy (Q12) transferred (added) to thesystem during the process and the amount ofwork (W12) done on the system by the envi-ronment during the process.

ΔE12 = Q12 + W12 = Q12 - S PdV

The differential expression for the above is:

dE = dQ + dW = dQ - PdV

W12 ΔE12

Q12

PRACTICE PROBLEM: FIRST LAW

Calculate the increase in internal energy of 1.0gram of H2O when it is taken from water at100 °C to steam at 100 °C (assume that theprocess occurs at atmospheric pressure). H2O(g)

H2O(l)

THERMODYNAMIC PROCESS

A thermodynamic process is a continu-ous change in the state of a material.If the material is a gas, it is usual todescribe the process graphically on aPV-diagram (where P is shown as afucntion of V, and where the state vari-able temperature is suppressed). Sographically, a process is a path on thePV-diagram. An arrow is used to makethe time-progression of the process ob-vious.

P

P1

P2

V1 V2 V

MOLAR SPECIFIC HEATS

The molar specific heats (CV and CP) of a gas are the constants used inthe expressions Q = nCVΔT and Q = nCPΔT for computing the heat energyneeded to raise the temperature of n moles of a gas ΔT degrees. CV = (ν/2)R is the molar specific heat at constant volume, and CP = (ν/2 + 1)R is themolar specific heat at constant pressure. The parameter ν is the effectivenumber of degrees of freedom of internal energy sharing for the gas.

gas ννννν CP (J/mol.K) CV (J/mol.K) γγγγγ = CP/Cv

He 3.01 20.8 12.5 1.67Ar 3.01 20.8 12.5 1.67H2 4.91 28.8 20.4 1.41N2 5.01 29.1 20.8 1.40O2 5.08 29.4 21.1 1.40

CO2 6.86 37.0 28.5 1.30H20 6.50 35.4 27.0 1.30

GENERAL PROCESS

In any general processthat an ideal gasis taken through the three thermody-namic quantities Q, W, and ΔE canalways be given by:

Q12 = (ν/2)Δ(PV) + S PdVW12 = -S PdVΔE12 = (ν/2)Δ(PV)

or, using nRT for PV:

Q12 = n(ν/2)RΔT + S PdVW12 = -S PdVΔE12 = n(ν/2)RΔT

where: Δ(PV) = P2V2 - P1V1and ΔT = T2 - T1

P

P1

P2

V1 V2 V

ISOBARIC PROCESS

An isobaric process is a processthroughout which the system pressureremains constant. (P = constant or dP =0 throughout the process).

Q12 = (ν/2 + 1)PΔV = n(ν/2 + 1)RΔTW12 = - PΔV = - nRΔTΔE12 = (ν/2)PΔV = n(ν/2)RΔT

where: ΔV = V2 - V1and ΔT = T2 - T1

P

P

V1 V2 V

PRACTICE PROBLEM: ISOBARIC PROCESS

A weather balloon at sea level is filled with600 m3 of helium gas at 15.0 °C and 1.00 atmapressure. The helium inside the balloon slowlyabsorbs heat energy from the warmer sur-rounding air. The balloon's volume increases,but the pressure of the helium inside remainsconstant. If the temperature of the helium in-creases to 25.0 °C find: (1) W = the work doneon the helium by the environment; (2) ΔE = thechange in internal energy of the helium; and(3) Q = the heat energy absorbed by the he-lium. The effective ν for helium at these tem-peratures is 3.01.

ISOCHORIC PROCESS

An isochoric process is a processthroughout which the system volume re-mains constant. (V = constant or dV = 0throughout the process).

Q12 = (ν/2)VΔP = n(ν/2)RΔTW12 = 0ΔE12 = (ν/2)VΔP = n(ν/2)RΔT

where: ΔP = P2 - P1and ΔT = T2 - T1

P

P1

P2

V V

PRACTICE PROBLEM: ISOCHORIC PROCESS

A pressure tank has a volume of 30.0 m3. Thistank is filled with air to a gauge pressure of5.00 atm early in the day when the tempera-ture is 20.0 °C. In the late afternoon it is foundthat the ambient temperature (and the air withinthe tank) has risen to 44.0 °C. Use this infor-mation to find: (1) W = the work done on theair by the environment; (2) ΔE = the change ininternal energy of the air; and (3) Q = the heatenergy absorbed by the air. The effective νfor air at these temperatures is 5.03.

ISOTHERMAL PROCESS

An isothermal process is a processthroughout which the system tempera-ture remains constant. (T = constant ordT = 0 throughout the process).

Q12 = PVln(V2 / V1) = nRTln(V2 / V1)W12 = -PVln(V2 / V1) = -nRTln(V2 / V1)ΔE12 = 0

note: P1/P2 can be substituted for V2/V1 in the above expressions. PV in theabove expressions is the product of anyP-V combination along the isotherm.

P

P1

P2

V1 V2 V

PRACTICE PROBLEM: ISOTHERMAL PROCESS

During the upstroke of the handle of a bicyclepump, air at a temperature of 20 °C and 1.00atma pressure enters a 0.75-L volume of thecylinder. A slow downstroke compresses theair to a volume of 0.15 L. Assuming that thedownstroke is slow enough that heat is con-ducted through the cylinder wall so that thetemperature of the gas in the cylinder remainsconstant during the process, find: (1) W = thework done on the air by the environment; (2)ΔE = the change in internal energy of the air;and (3) Q = the heat energy absorbed by theair.

ADIABATIC PROCESS

An adiabatic process is a processthroughout which no heat energy isexchanged between the system and itssurroundings. (PVγ

= constant or dQ = 0throughout the process). γ = (ν + 2)/ν.

Q12 = 0W12 = (ν/2)Δ(PV) = n(ν/2)RΔTΔE12 = (ν/2)Δ(PV) = n(ν/2)RΔT

where: Δ(PV) = P2V2 - P1V1and ΔT = T2 - T1

P

P1

P2

V1 V2 V

PRACTICE PROBLEM: ADIABATIC PROCESS

During the upstroke of the handle of a bicyclepump, air at a temperature of 20 °C and 1.00atma pressure enters a 0.75-L volume of thecylinder. A quick downstroke compresses theair to a volume of 0.15 L. Assuming that thedownstroke is rapid enough that there is notime for heat energy to be conducted throughthe cylinder wall, find: (1) W = the work doneon the air by the environment; (2) ΔE = thechange in internal energy of the air; and (3) Q= the heat energy absorbed by the air.

IDEAL GAS PROCESSES MATRIX

process Q12 W12 ΔΔΔΔΔE12

general(ν/2)Δ(PV) + S PdV (ν/2)Δ(PV)(ν/2)nRΔT + S PdV

-S

PdV(ν/2)nRΔT

isobaric (ν/2 + 1)PΔV -PΔV (ν/2)PΔV(constant P) (ν/2 + 1)nRΔT -nRΔT (ν/2)nRΔT

isochoric (ν/2)VΔP (ν/2)VΔP(constant V) (ν/2)nRΔT

0(ν/2)nRΔT

isothermal PVln(V2 / V1) -PVln(V2 / V1)(constant T) nRTln(V2 / V1) -nRTln(V2 / V1)

0

adiabatic (ν/2)Δ(PV) (ν/2)Δ(PV)(constant PVγ)

0(ν/2)nRΔT (ν/2)nRΔT

note: ΔP = P2 - P1, ΔV = V2 - V1, ΔT = T2 - T1, and Δ(PV) = P2V2 - P1V1

THERMAL CONDUCTIVITY

The rate PPPPP at which heat is conducted througha slab of material (in W or Btu/h) with cross-sectional area A and thickness L whose facesare maintained at temperatures TH and TC isgiven by:

P = kA(TH - TC)/L

where k is the thermal conductivity of thematerial of the slab (measured in W/m.°C orBtu/h.ft.°F).

L

A

TH TC

THERMAL CONDUCTIVITIES

substance k (W/m.°C) substance k (W/m.°C)

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

PRACTICE PROBLEM: THERMAL CONDUCTIVITY

A pane of window glass is 1/8" thick and hasdimensions 1.5 m by 2.5 m. It separates twoheat reservoirs whose temperatures are 10 °Cand 22 °C. What is the rate of heat transportthrough the pane?

PRACTICE PROBLEM: THERMAL CONDUCTIVITY

A copper and an aluminum cylinder are joinedend-to-end as shown. Each has a diameter of4.0 cm and a length of 25.0 cm. One end ofthe copper cylinder is maintained at 90 °C andone end of the aluminum cylinder is maintainedat 15 °C . The thermal conductivity of copperis 397 W/m.°C and the thermal conductivity ofaluminum is 238 W/m.°C. What is the steadystate junction temperature?

insulator

90Cu Al

15°C

insulator

°C

THERMAL RADIATION

The rate PPPPP of radiant energy loss (watts) fromthe surface of an object with surface area Aand surface temperature T is given by:

P = σAeT4

where σσσσσ is the Stefan-Boltzmann constant(5.67x10-8 W/m2.K4), and e is the emissivity ofthe surface of the object (dimensionless pa-rameter). This is called the Stefan-BoltzmannLaw.

A T

PRACTICE PROBLEM: THERMAL RADIATION

The surface of a cube with sides of length0.05 m has an emissivity of 0.85. If the surfaceof the cube is maintained at a temperature of550 °C, estimate the radiant energy flux throughthe surface of the cube?

MOLAR SPECIFIC HEATS

The molar specific heats (CV and CP) of a gas are the constants used inthe expressions Q = nCVΔT and Q = nCPΔT for computing the heat energyneeded to raise the temperature of n moles of a gas ΔT degrees. CV = (ν/2)R is the molar specific heat at constant volume, and CP = (ν/2 + 1)R is themolar specific heat at constant pressure. The parameter ν is the effectivenumber of degrees of freedom of internal energy sharing for the gas.

gas ννννν CP (J/mol.K) CV (J/mol.K) γγγγγ = CP/Cv

He 3.01 20.8 12.5 1.67Ar 3.01 20.8 12.5 1.67H2 4.91 28.8 20.4 1.41N2 5.01 29.1 20.8 1.40O2 5.08 29.4 21.1 1.40

CO2 6.86 37.0 28.5 1.30H20 6.50 35.4 27.0 1.30

PRACTICE PROBLEM: MOLAR SPECIFIC HEAT

A weather balloon is filled with 600 m3 of he-lium gas at a temperature of 15.0 °C. The gasinside the balloon slowly absorbs heat energyfrom the warmer surrounding air in such a waythat the balloon's volume increases, but thepressure of the helium remains the same asthat of the surrounding air (1.0 atm). How muchheat energy would the helium need to absorbfor its temperature to increase by 1.0 °C? Themolar specific heat at constant pressure ofhelium is 20.8 J/mol.K, and the molecular ofweight of helium is 4.00 g/mol.

PRACTICE PROBLEM: MOLAR SPECIFIC HEAT

A 600-m3 pressure tank is filled with heliumgas at a temperature of 15.0 °C and pressureof 1.0 atm (absolute). The helium inside thetank slowly absorbs heat energy through thetank from the warmer surrounding air in sucha way that the helium's pressure increases,but the volume of the tank (and helium) re-mains the same. How much heat energy wouldthe helium need to absorb for its temperatureto increase by 1.0 °C? The molar specific heatat constant volume of helium is 12.5 J/mol.K.

SECOND LAW OF THERMODYNAMICS

Kelvin-Planck formulation: It is impossible to construct a heat engine that,operating in a cycle, produces no effect other than the input of energy byheat from a reservoir and the performance of an equal amount of work.

Clausius statement: It is impossible to construct a cyclical machine whosesole effect is to transfer energy continuously by heat from one object toanother at a higher temperature without the input of energy by work.

Lord Kelvin Max Planck Rudolf Clausius

ENTROPY

Entropy (S) is a quantity describing the disorder of a system. It has dimen-sions of heat energy per temperature (cal/K or J/K). It is equal to theBoltzmann constant (kB) times the natural log of the total number of ways(W) in which the system can arrange itself consistent with energy beingconserved.

S = kBlnW

Entropy is an extensive physical quantity. The total entropy of a system isequal to the sum of the entropies of each of the system's components.

n

Stotal = Σ Sii=1

ENTROPY CHANGE

Calculating the entropy change ΔS for a system is usually a more tractableproblem than calculating the entropy itself. The change in entropy of asystem taken through a process from state 1 to state 2 is most oftenderivable from:

2

ΔS12 = S2 - S1 = S

dQ/T1

example: The entropy change of m grams of a substance with specific heatc heated from T1 to T2 would be obtained as:

2 2

ΔS12 = S

dQ/T = S

mcdT/T = mcln(T2/T1)1 1

ENTROPY CHANGE: WARMING / COOLING

In changing the temperature of a material from T1 to T2 without changing itsphase, the change in entropy is given by:

ΔS12 = mcln(T2/T1) or nCln(T2/T1)

example: Find the change in entropy of a 28.3-gram gold krugerrand when it is heated from20 °C to 180 °C. The specific heat of gold is.0308 cal/g.°C.

ENTROPY CHANGE: PHASE CHANGE

In changing the phase of a material without changing its temperature, thechange in entropy is given by:

ΔS = + mL/T

example: Find the change in entropy of 2.5 kgof molten aluminum becoming a solid ingot atits melting point (660 °C). The latent heat offusion of aluminum is 397,000 J/kg.

ENTROPY CHANGE: FREE EXPANSION

In allowing a gas to free-expand from initial volume V1 to final volume V2,the entropy change is given by:

ΔS12 = nRln(V2/V1)

example: Find the change in entropy whenhelium gas at 1.0 atm and 20 °C, occupying achamber whose volume is 0.5 m3, is allowedto free-expand into an adjacent evacuatedchamber whose (extra) volume is 2.5 m3.

ENTROPY CHANGE: MIXING

In mixing nA moles of gas A, initially occupying volume VA, with nB moles ofgas B, initially occupying volume VB:

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

example: A chamber is divided into two com-partments separated by a partition. Compart-ment A (H2 gas): VA = 75 L, PA = 1.0 atm, andTA = 0 °C. Compartment B (CO2 gas): VB = 25L, PB = 1.0 atm, and TB = 0 °C. The partition isremoved and the gases mix. Find the changein entropy of the system.

PRACTICE PROBLEM: ENTROPY CHANGE

A 60-gram piece of ice at 0 °C is placed in acontainer with 200 grams of water at 50 °C.Find the temperature of this system after ther-mal equilibrium is attained. Use this result tofind ΔS for: (1) the water; (2) the ice; and (3)the entire system for this process. Assumethat the container does not transfer any heatenergy.

KINETIC THEORY ASSUMPTIONS

1 Gases are composed of a large number of similar particles thatobey Newton's Laws of Motion.

2 Gas particle velocity vectors are randomized in magnitude and di-rection at any time.

3 Particles are much smaller than the distance between particles.Most of the volume of a gas is therefore empty space.

4 The only particle-particle and particle-wall forces are contact forces.Particle-particle and particle-wall collisions are perfectly elastic.

5 The average kinetic energy of a collection of gas particles is di-rectly proportional to the absolute temperature of the gas (and nothingelse).

SOLUTIONS TO PRACTICE PROBLEMS

PROBLEM: FIRST LAW OF THERMODYNAMICSn = m/M = (1.0 g)/(18.0 g/mol) = .0556 molV1 = m/ρ = (.001 kg)/(103 kg/m3) = .000001 m3

V2 = nRT/P = (.0556 mol)(8.31 J/mol.K)(373 K)/(101,300 Pa) = .00171 m3

Q = +mLv = (.001 kg)(2.26x106 J/kg) = 2260 JW = -(101,300 Pa)(.00171 m3 - .000001 m3) = -173 JΔE = Q + W = 2260 J - 173 J = 2087 J

PROBLEM: ISOBARIC PROCESSn = PV/RT = (101,300 Pa)(600 m3)/(8.31 J/mol.K)/(288 K) = 25,396 molΔE = (ν/2)nR(T2 - T1) = (3.01/2)(25,396)(8.31 J/mol.K)(10 K) = 3.18 MJW = -nR(T2 - T1) = -(25,396)(8.31 J/mol.K)(10 K) = -2.11 MJQ = ΔE - W = 3.18 MJ - (-2.11 MJ) = 5.29 MJ

PROBLEM: ISOCHORIC PROCESSn = PV/RT = (506,500 Pa)(30 m3)/(8.31 J/mol.K)/(293 K) = 6241 molΔE = (ν/2)nR(T2 - T1) = (5.03/2)(6241 mol)(8.31 J/mol.K)(24 K) = 3.13 MJW = 0 MJQ = ΔE = 3.13 MJ

SOLUTIONS TO PRACTICE PROBLEMS

PROBLEM: ISOTHERMAL PROCESSΔE = 0 JW = -PVln(V2/V1) = -(101,300 Pa)(.00075 m3)ln(.15 L/.75 L) = 122 JQ = -W = -122 J

PROBLEM: ADIABATIC PROCESSP2 = P1(V1/V2)

γ = (101,300 Pa)(.75 L/.15 L)1.40 = 964,200 Pa

ΔE = (ν/2)(P2V2 - P1V1)ΔE = (5.03/2)[(964,396 Pa)(.00015 m3) - (101,300 Pa)(.00075 m3)] = 173 JW = ΔE = 173 JQ = 0 J

PROBLEM: THERMAL CONDUCTIVITYL = (.125 ")(2.54 cm/")/(100 cm/m) = .00318 m and A = (1.5 m)(2.5 m) = 3.75 m2

P = kA(TH - TC)/L = (.8 W/m.°C)(3.75 m2)(22 °C - 10 °C)/(.00318 m) = 11,320 W

PROBLEM: THERMAL CONDUCTIVITYPCu = PAl must hold for steady state conditionskAlA(TJ - TC)/LAl = kCuA(TH - TJ)/LCuSince LAl = LCu, kAl(TJ - TC) = kCu(TH - TJ) and TJ = (kAlTH + kCuTC)/(kAl + kCu)= [(238 W/m.°C)(15 °C) + (397 W/m.°C)(90 °C)]/(238 W/m.°C + 397 W/m.°C) = 61.9 °C

SOLUTIONS TO PRACTICE PROBLEMS

PROBLEM: THERMAL RADIATIONA = 6(.05 m)2 = .0150 m2

P = (5.67x10-8 W/m2.K4)(.0150 m2)(.85)(823 K)4 = 332 W

PROBLEM: MOLAR SPECIFIC HEAT CPQ = nCPΔT = (PV/RT)CPΔT = PVCPΔT/(RT)= (101,300 Pa)(600 m3)(20.8 J/mol.K)(1 K)/(8.31 J/mol.K)/(298 K) = 511 kJ

PROBLEM: MOLAR SPECIFIC HEAT CVQ = nCVΔT = (PV/RT)CVΔT = PVCVΔT/(RT)= (101,300 Pa)(600 m3)(12.5 J/mol.K)(1 K)/(8.31 J/mol.K)/(298 K) = 307 kJ

PROBLEM: ΔS FOR WARMING / COOLINGΔS = mcln(T2/T1) = (28.3 g)(.0308 cal/g.°C)ln[(453 K)/(293 K)] = +.380 cal/K

PROBLEM: ΔS FOR PHASE CHANGEΔS = -mLf/T = -(2.5 kg)(397,000 J/kg)/(933 K) = -1064 J/K

SOLUTIONS TO PRACTICE PROBLEMS

PROBLEM: ΔS FOR FREE EXPANSIONΔS = nRln(V2/V1) = (PV/T)ln(V2/V1)= (101,300 Pa)(.5 m3)ln[(3.0 m3/.5 m3)]/(293 K) = 310 J/K

PROBLEM: ΔS FOR MIXINGnA = (75 L)/(22.4 L/mol) = 3.35 molnB = (25 L)/(22.4 L/mol) = 1.12 molΔS = (8.31 J/mol.K)[(3.35 mol)ln[(100 L)/(75 L)] + (1.12 mol)ln[(100 L)/(25 L)]] = 20.9 J/K

PROBLEM: ΔS FOR COMPOUND SYSTEMQlost = Qgained(200 g)(1.0 cal/g.°C)(50 °C - T) = (60 g)(80 cal/g) + (60 g)(1.0 cal/g.°C)(T)Solve above for equilibrium temperature T and get T = 20.0 °CΔSwater = (200 g)(1.0 cal/g.°C)ln[(293 K)/(323 K)] = -19.5 cal/KΔSice = (60 g)(80 cal/g)/(273 K) + (60 g)(1.0 cal/g.°C)ln[(293 K)/(273 K)] = +21.8 cal/KΔSsystem = -19.5 cal/K + 21.8 cal/K = +2.3 cal/K