ENERGY SOURCES RENEWABLE energy sources are those sources of
8.1 energy sources - St Leonard's College · 8.1 energy sources 1 a Specific energy is the energy...
7
ANSWERS TO TEST YOURSELF QUESTIONS 8 1 PHYSICS FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2015 Answers to test yourself questions Topic 8 8.1 Energy sources 1a Specific energy is the energy that can be extracted from a unit mass of a fuel while energy density is the energy that can be extracted from a unit volume of fuel. b The available energy is mgh. The energy density is the energy per unit volume that can be obtained and so E mgh V mgh m gh D = = = = × × = × - ρ ρ 10 98 75 74 10 3 5 3 . . Jm 2a In 1 s the energy is 500 MJ. b In one year the energy is 5.0 × 10 8 × 365 × 24 × 60 × 60=1.6 × 10 16 J. 3a The overall efficiency is 0.80 × 0.40 × 0.12 × 0.65 = 0.025. b 4a The energy produced by burning the fuel is 10 7 × 30 × 10 6 = 3.0 × 10 14 J. Of this 70% is converted to useful energy and so the power output is P = × × × × = × 0 30 30 10 24 60 60 1 04 10 14 9 . . . W. b The rate at which energy is discarded is P discard = × × × × = × 0 70 30 10 24 60 60 2 43 10 14 9 . . . W. c Use Δ Δ Δ m t c P discard θ = to get Δ Δ Δ m t P c discard = = × × = × θ 2 43 10 4200 5 12 10 9 5 . . kg s -1 . 5 In time t the energy used by the engine is 20 × 10 3 × t.The energy available is 0.40 × 34 × 10 6 J and so 20 × 10 3 × t = 0.40 × 34 × 10 6 ⇒ t =680 s. The distance travelled is then x = vt = 9.0 × 680 = 6.1 km. 6 The useful energy produced each day is E = Pt =1.0×10 9 ×24×60×60=8.64×10 13 J. The energy produced by burning the coal is therefore 0 40 8 64 10 2 16 10 13 14 . . . = × ⇒ = × E E c c J. So the mass of coal to be burned is m = × × = × 2 16 10 30 10 72 10 14 6 6 . . kg per day. 20 48 28 100 80 32 4 1.5 2.5
Transcript of 8.1 energy sources - St Leonard's College · 8.1 energy sources 1 a Specific energy is the energy...
ANSWERS TO TEST YOURSELF QUESTIONS 8 1physics for the iB Diploma © camBriDge University press 2015
answers to test yourself questionstopic 88.1 energy sources
1 a Specificenergyistheenergythatcanbeextractedfromaunitmassofafuelwhileenergydensityistheenergythatcanbeextractedfromaunitvolumeoffuel.
b Theavailableenergyismgh.Theenergydensityistheenergyperunitvolumethatcanbeobtainedand
soEmgh
V
mghm ghD = = = = × × = × −
ρ
ρ 10 9 8 75 7 4 103 5 3. . Jm
2 a In1stheenergyis500MJ. b Inoneyeartheenergyis5.0×108×365×24×60×60=1.6×1016J.
3 a Theoverallefficiencyis0.80×0.40×0.12×0.65=0.025.
b
4 a Theenergyproducedbyburningthefuelis107× 30× 106=3.0× 1014J.Ofthis70%isconvertedtouseful
energyandsothepoweroutputisP =× ×× ×
= ×0 30 3 0 10
24 60 601 04 10
149. .
. W.
b TherateatwhichenergyisdiscardedisPdiscard =× ×× ×
= ×0 70 3 0 10
24 60 602 43 10
149. .
. W.
c Use∆∆
∆m
tc Pdiscardθ = toget
∆∆ ∆m
t
P
cdiscard= =
××
= ×θ
2 43 10
4200 51 2 10
95.
. kg s−−1.
5 Intimettheenergyusedbytheengineis20× 103× t.Theenergyavailableis0.40× 34× 106Jandso20× 103× t =0.40× 34× 106⇒ t =680s.Thedistancetravelledisthenx=vt=9.0× 680=6.1km.
6 TheusefulenergyproducedeachdayisE=Pt=1.0×109×24×60×60=8.64×1013J.Theenergyproduced
byburningthecoalistherefore0 408 64 10
2 16 1013
14..
.=×
⇒ = ×E
Ec
c J.Sothemassofcoaltobeburnedis
m =××
= ×2 16 10
30 107 2 10
14
66.
. kgperday.
20 48 28
100
80
32
4
1.5
2.5
2 physics for the iB Diploma © camBriDge University press 2015ANSWERS TO TEST YOURSELF QUESTIONS 8
7 a ThefissionableisotopeofuraniumisU–235.ThisisfoundinverysmallconcentrationsinuraniumorewhichismostlyU–238.EnrichmentmeansincreasingtheconcentrationofU–235inasampleofuranium.
b Themoderatoristhepartofthenuclearreactorwhereneutronsreleasedfromthefissionreactionsslowdownasaresultofcollisionswiththeatomsofthemoderator.Thetemperatureofthemoderatorcanbekeptconstantwithacoolingsystemthatremovestheexcessthermalenergygeneratedinthemoderator.
c Criticalmassreferstotheleastmassofuraniumthatmustbepresentfornuclearfissionreactionstobesustained.Ifthemassofuraniumistoosmall(i.e.belowthecriticalmass)theneutronsmayescapewithoutcausingfissionreactions.
8 a Thereactionis 92235
01
54140
3894
012U+ n Xe Sr n→ + + .Themassdifferenceis
δ = − × +
−
( . . ) .
(
235 043992 92 0 000549 1 008665
1339 921636 54 0 000549 93 915360 38 0 000. . ) ( . .− × + − × 5549 2 1 008665
0 198331
) . )
.
+ ×( )= u
AndsotheenergyreleasedisE=0.198331× 931.5c2MeVc-2=185MeV. b WithNreactionspersecondthepoweroutputisN× 185MeVs-1.Inotherwords,N× 185× 106× 1.6× 10-19 =
200× 106⇒N =6.8× 1018s-1.
9 a Onekilogramofuraniumcorrespondsto1000
2354 26= . molesandso4.26× 6.02× 1023=2.57× 1024
nuclei.Eachnucleusproduces200MeVandsotheenergyproducedby1kg(theenergydensity)is2.57× 1024× 200× 106× 1.6× 10-19≈8× 1013Jkg-1.
b8 10
30 102 7 10
13
66×
×= ×. kg
10 a Theenergythatmustbeproducedin1 sisE =×
= ×500 10
0 401 25 10
69
.. J.Hencethenumberoffission
reactionspersecondis1 25 10
200 10 1 6 103 9 10
9
6 1919.
..
×× × ×
= ×−.
b Thenumberofnucleirequiredtofissionpersecondis3.9× 1019whichcorrespondsto3 9 10
6 02 106 5 10
19
235.
..
××
= × − mol,i.e.amassof6.5× 10-5× 235× 10-3 =1.5× 10-5kgs-1.
11
a i fuelrodsarepipesinwhichthefuel(i.e.uranium–235)iskept. ii controlrodsarerodsthatcanabsorbneutrons.Theseareloweredinorraisedoutofthemoderatorsothat
therateofreactionsiscontrolled.Therodsarelowerediniftherateistoohigh–therodsabsorbneutronssothattheseneutronsdonotcauseadditionalreactions.Theyraisedoutiftherateistoosmallleavingtheneutronstocausefurtherreactions.
iii Themoderatoristhepartofthenuclearreactorwhereneutronsreleasedfromthefissionreactionsslowdownasaresultofcollisionswiththeatomsofthemoderator.Thetemperatureofthemoderatorcanbekeptconstantwithacoolingsystemthatremovestheexcessthermalenergygeneratedinthemoderator.
pump
fuelrod
fuelrod
control rods(can be raised or lowered)
control rods(can be raised or lowered)
coolant(pressurisedwater)
water
water
steam
steamsteamgenerator
steamgenerator
coolant (carbondioxide gas)
solidmoderator(graphite)
gas-cooled reactorpressurised water reactor (PWR)
pump
ANSWERS TO TEST YOURSELF QUESTIONS 8 3physics for the iB Diploma © camBriDge University press 2015
b Itiskineticenergyoftheneutronsproducedwhichisconvertedintothermalenergyinthemoderatorastheneutronscollidewiththemoderatoratoms.
12 Asolarpanelreceivessolarradiationincidentonitandusesittoheatupwater,i.e.itconvertssolarenergyintothermalenergy.Thephotovoltaiccellconvertsthesolarincidentonittoelectricalenergy.
13 a Thepowerthatmustbesuppliedis3.0kWandthisequals700×A×0.70×0.50.Hence700×A×0.70×0.50=3.0×103⇒A =12m2.
b
14 Thepowerthatisprovidedis0.65×240×Aandthismustequalmc
t
∆∆
θ=
× ×× ×
300 4200 35
12 60 60.
Hence0 65 700300 4200 35
12 60 606 5 2. .× × =
× ×× ×
⇒ =A A m .
15 Thepowerincidentonthepanelis600×4.0×0.60=1440W.Theenergyneededtowarmthewateris
mc∆θ=150×4200×30=1.89×107Jandso1440 1 89 101 89 10
14401 31 10 37
74× = × ⇒ =
×= × =t t.
.. .s 66 hr.
16 a FromthegraphthisisaboutT = 338K b P =IA =400×2 =800W
c Theusefulpoweristhe320 Wthatisextracted.Theefficiencyisthus320
8000 40= . .
17 Thepowersuppliedatthegivenspeedis(readingfromthegraph)100 kW.Hencetheenergysuppliedin100 hrsis100×103×1000×60×60=3.6×1011J.
18 Welookatthepowerformulaforwindmills,P Av=1
23ρ todeduce:
a i theareawillincreasebyafactorof4ifthelengthisdoubledandsothepowergoesupbyafactorof4, ii thepowerwillincreasebyafactorof23=8 iii thecombinedeffectis4×8 =32 b Notallthekineticenergyofthewindcanbeextractedbecausenotallthewindisstoppedbythewindmill(as
theformulahasassumed).Inaddition,therewillbefrictionallossesastheturbinesturnaswellaslossesduetoturbulence.
19 Assumptionsinclude: i nolossesofenergyduetofrictionalforcesastheturbinesturn ii noturbulenceintheair iii thatalltheairstopsattheturbinessothatthespeedoftheairbehindtheturbinesiszero(whichis
impossible).
25 kW 30 kW 30 kW
85 kW
4 physics for the iB Diploma © camBriDge University press 2015ANSWERS TO TEST YOURSELF QUESTIONS 8
20 Thepowerinthewindbeforehittingtheturbineis1
2 1 13ρAv andrightafterpassingthroughtheturbineis
1
2 2 23ρ Av sotheextractedpoweris
1
2
1
2
1
21 5 1 2 8 0 1 8 31 1
32 2
3 2 3ρ ρ πAv Av− = × × × − ×. ( . . . .00 2 03 ) .≈ kW.
21 FromP Av=1
23ρ weget25 10
1
21 2 9 0 57 163 3 2× = × × × ⇒ =. . .A A m andsoπR R2 57 16 4 3= ⇒ =. . m.The
assumptionsmadearetheusualones(seequestion19) i nolossesofenergyduetofrictionalforcesastheturbinesturn ii noturbulenceintheair,and iii thatalltheairstopsattheturbinessothatthespeedoftheairbehindtheturbinesiszero(whichisimpossible).
22 Thepotentialenergyofamass∆mofwateris∆mgh andsothepowerdevelopedistherateofchangeofthis
energyi.e.∆∆m
tgh = × × = × ≈ ×500 9 8 40 1 96 10 2 0 105 5. . . W.
23 Thepotentialenergyofamass∆mofwateris∆mghandsothepowerdevelopedistherateofchangeofthis
energyi.e.∆∆
∆∆
m
tgh
V
tgh Qgh= =ρ ρ .
24 Theamountofelectricalenergygeneratedwillalwaysbelessthantheenergyrequiredtoraisethewaterbacktoitsoriginalheight.Thisisbecausetheelectricalenergygeneratedislessthanwhattheoreticallycouldbeprovidedbythewater(becauseofvariouslosses).Sothisclaimcannotbecorrect.
25 a Coalpowerplant:chemicalenergyofcoal→thermalenergy→kineticenergyofsteam→kineticenergyofturbine→electricalenergy.
b hydroelectricpowerplant:potentialenergyofwater→kineticenergyofwater→kineticenergyofturbine→electricalenergy.
c windturbine:kineticenergyofwind→kineticenergyofturbine→electricalenergy. d nuclearpowerplant:nuclearenergyoffuel→kineticenergyofneutrons→thermalenergyinmoderator→
kineticenergyofsteam→kineticenergyofturbine→electricalenergy.
chapter 8.2 thermal energy transfer
26 a Energyhastobeconservedsowhateverenergyentersthejunctionhastoleavethejunctionandsotheratesofenergytransferarethesame.
b ThetemperaturedifferencesarenotthesamebecauseXandYhavedifferentthermalconductivity.
27 Thereis;inordertosendthewarmairthatcollectshigherupintheroomdownwards.
28 PowerisproportionaltoT 4andsotheratiois900
3003 81
44
= =
29 a AblackbodyisanybodyatabsolutetemperatureTwhoseradiatedpowerperunitareaisgivenbyσT4.Ablackbodyappearsblackwhenitstemperatureisverylow.Itabsorbsalltheradiationincidentonitandreflectsnone.
b Apieceofcharcoalisagoodapproximationtoablackbodyasistheopeningofasoftdrinkcan.
c Itincreasesby 273 100
273 501 8
4++
≈ .
30 a Thewavelengthatthepeakofthegraphisdeterminedbytemperatureandsincethewavelengthisthesamesoisthetemperature.
b Theratiooftheintensitiesatthepeakisabout1 1
1 90 6
.
..≈ .
31 Wehavethate AT P TP
e Aσ
σ4 4= ⇒ = i.e.T =
×× × × ×
=−
1 35 10
0 800 5 67 10 5 00 10278
9
8 64
.
. . .K.
ANSWERS TO TEST YOURSELF QUESTIONS 8 5physics for the iB Diploma © camBriDge University press 2015
32 a WemusthavethatσATd
Td
42
1 1∝ ⇒ ∝ .
b Usingourknowledgeofpropagationofuncertainties,wededucethat∆ ∆T
T
d
d=1
2andso
∆T
T= × =1
21 0 0 005. % . .Hence,∆T=0.005T=0.005×288=1.4K.
33 a Intensityisthepowerreceivedperunitareafromasourceofradiation. b P=e σAT4.P=0.90×5.67×10-8×1.60×(273+37)4=754W.Assuminguniformradiationinalldirections,the
intensityisthenIP
d= = = −
4
754
4 5 02 42 2
2
π π( . ). W m .
34 a Imagineaspherecenteredatthesourceofradiusd.ThepowerPradiatedbythesourceisdistributedoverthe
surfaceareaAofthisimaginarysphere.Thepowerperunitareai.e.theintensityisthusIP
A
P
d= =
4 2π.
b Wehaveassumedthattheradiationisuniforminalldirections.
35 a Thepeakwavelengthisapproximatelyλ0=0.65×10-5mandsofromWien’slaw:λ0T=2.9×10-3Kmwefind
T =××
=−
−
2 9 10
0 65 10450
3
5
.
.K.
b Thecurvewouldbesimilarinshapebuttallerandthepeakwouldbeshiftedtotheleft.
36 a Albedoistheratioofthereflectedintensitytotheincidentintensityonasurface. b Thealbedoofaplanetdependsonfactorssuchascloudcoverintheatmosphere,amountoficeonthesurface,
amountofwateronthesurfaceandcolorandnatureofthesoil.
37 a Theearthreceivesradiantenergyfromthesunandinturnradiatesitself.Theradiatedenergyisintheinfraredregionoftheelectromagneticspectrum.Gasesintheatmosphereabsorbpartofthisradiatedenergyandreradiateitinalldirections.Someofthisradiationreturnstotheearthsurfacewarmingitfurther.
b Themaingreenhousegasesarewatervapour,carbondioxideandmethane.Seepage336forsources.
38 a TheenergyflowdiagramissimilartothatinFig.8.10onpage334. b Thereflectedintensityis350-250=100Wm-2andsothealbedois
100
3500 29= . .
c Ithastobeequaltothatabsorbedi.e.250Wm-2.
d Usee T I TI
e Aσ
σ4 4= ⇒ = i.e.(assumingablackbody)T =
×=−
250
5 67 10258
84
.K.Youmustbecarefulwith
thesecalculationsintheexam.Youmustbesureastowhetherthequestionwantsyoutoassumeablackbodyornot.Strictlyspeaking,inamodelwithoutanatmospheretheearthsurfacecannotbetakentobeablackbody–ifitwerenoradiationwouldbereflected!
39 a TheintensityradiatedbythesurfaceisI 1andafractiontofthisescapessoweknowthatI3=tI1.Theintensity
ofradiationenteringthesurfaceis( )14 1 2− + +α αS
I I .Thismustequaltheintensityofradiationleavingthe
surfacewhichisI 1:( )14 1 2 1− + + =α αS
I I I .Theintensityofradiationenteringtheatmosphereis(1-α)I 1
andthatleavingitis2I 2+I 3.Hence2I 2+I 3=(1-α)I 1.Sowehavetosolvetheequations(weusedI3=tI1)
( )14 1 2 1− + + =α αS
I I I
2 12 1 1I tI I+ = −( )α
Thesesimplifyto
( ) ( )
( )( )
14
1
2 11
2
2 1
2 1 2
− + = −
= − − ⇒ =− −
α α
α α
SI I
I t I It
II1
6 physics for the iB Diploma © camBriDge University press 2015ANSWERS TO TEST YOURSELF QUESTIONS 8
Hence
( )( )
( )
( ) ( )(
14
1
21
14
1
1 1
1
− +− −
= −
− = − −
α α α
α α
S tI I
SI
11
2
14
1
22
11
1
1
1
− −
− =− +
=− +
−
α
α α
αα
tI
S tI
It
S
)
( )( )
( )44
Therefore,It
t
S2
1
1
1
4=
− −− +
×−α
αα( )
andIt
t
S3
2
1
1
4=
− +×
−α
α( )asrequired.
b TheintensityenteringisS
4.Theintensityleavingisα S
I I4 3 2+ + .Theintensityleavingsimplifiesto
αα
α αα
α αS t
t
S t
t
S S
4
2
1
1
4
1
1
1
4 4+
− +×
−+
− −− +
×−
= +( ) ( )
(( ) ( )
(
12
11
1
1
41
−− +
+ −− −− +
= + −
αα
α αα
α
t
t
t
tS αα α
α
α α
)
( )
1
1
41
4
− +− +
= + −( )
=
t
tS
S
c TheintensityradiatedbythesurfaceisIt
S1
2
11
4=
− +−
αα( ) andmustequalσ T 4,whereTisthesurface
temperature.Hence
2
11
4
2 14 1
2 1 0 3
4
4
− +− =
=−
− +
=−
αα σ
α
σα
t
ST
t
S
T
t
( )
( )
( . 00 350
5 67 10 2881 0 30
0 556
8 4
)
..
.
×× ×
− +
≡
−
t
d i LetanintensityIbeincidentontheatmosphereofemissivitye.AnamounteIwillbeabsorbedandreradiated,anamountαIwillbereflectedandanamounttIwillbetransmitted.ByenergyconservationwehavethatI =eI + αI + tIfromwhiche = 1 - α - tasrequired.
ii WeequateIt
t
S2
1
1
1
4=
− −− +
×−α
αα( )
toe σ T 4whereTistheatmospheretemperaturetoget
(recallthate = 1 - α - t)
e Tt
t
S
Tt
S
T
σ αα
α
σα
α
4
4
1
1
1
41
1
1
4
=− −− +
×−
=− +
×−
=
( )
( )
11
1 0 30 0 556
1 0 30 350
5 67 10242
84
− +×
− ××
=
−. .
( . )
.T K
ANSWERS TO TEST YOURSELF QUESTIONS 8 7physics for the iB Diploma © camBriDge University press 2015
40 Radiationisamainmechanismtoboththeatmosphereandtospace.Inadditionthereisconductiontotheatmosphere,aswellasconvection.
41 a Drysub–tropicallandhasahighalbedo,around0.4whereasawarmoceanhasanalbedooflessthan0.2. b Radiationandconvectioncurrentsarethemainmechanisms. c Replacingdrylandbywaterreducesthealbedooftheregion.Reducingthealbedomeansthatlessradiation
isreflectedandmoreisabsorbedandsoanincreaseintemperaturemightbeexpected.Theincreaseintemperaturemightinvolveadditionalevaporationandsomorerain.
42 Therateofevaporationfromwaterdependsonthetemperatureofthewaterandthetemperatureofthesurroundingair.Thesearebothhigherinthecaseofthetropicaloceanwaterandevaporationwillbemoresignificantinthatcase.
43 Thereismoreevaporationinregionbimplyingthatitisbothwarmerandwet.Thefactthatregionbiswarmerisfurthersupportedbythesomewhatgreaterconductionwhichwouldbeexpectedifthedifferenceintemperaturebetweentheatmosphereandthelandwerelarger.
44 Wehavethattheoriginalaveragealbedooftheareawas0.6×0.10+0.4×0.3=0.18andthenewoneis0.7×0.10+0.3×0.3=0.16forareductioninalbedoof0.02.Hencetheexpectedtemperaturechangeisestimatedtobe2°C.
