MAE110AHomework3:Solutions

10/20/2017

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𝑇 = 140°C

3.10:ForH2O,determinethespecifiedpropertyattheindicatedstate.LocatethestateonasketchoftheT-vdiagram.Givena) 𝑇 = 140°C, 𝑣 = 0.5m3 kgb) 𝑝 = 30MPa, 𝑇 = 100°Cc) 𝑝 = 10MPa, 𝑇 = 485°Cd) 𝑇 = 80°C, 𝑥 = 0.75BasicEquations𝑣8 = 𝑣9 + 𝑥 𝑣; − 𝑣9 Analysisa) FromTableA-3at𝑇 = 140°C:𝑣9 = 1.0435×10>? m3 kg , 𝑣; = 1.673m3 kg . 𝑣9 > 𝑣 >

𝑣;,thereforethestateisinthetwo-phaseliquid-vaporregionasshowninthediagrambelow.Thepressureisthesaturationpressureat140°CFromTableA-3:𝑝 = 3.613bar

b) Pressureishigherthanthecriticalpressureasshowninthediagrambelow(criticalproperties

pulledfromtableA-3).Therefore,thestateisinthecompressedliquidregion.FromTableA-5:𝑣 = 1.0290m3 kg

𝑝K = 220.9bar𝑝 = 30MPa = 300bar

𝑇K = 374.14°C

𝑇 = 100°C

Determinea) p,inbar,andlocateon𝑇-𝑣diagramb) 𝑣,in m3 kg⁄ , andlocateon𝑇 − 𝑣diagramc) 𝑣,in m3 kg⁄ , andlocateon𝑇 − 𝑣diagramd) 𝑝,inbar,and𝑣,in m3 kg⁄ , andlocateon𝑇 − 𝑣diagram

𝑝 = 𝑝sat,140°C = 3.613bar

2

c) Thegiventemperatureisgreaterthanthesaturationtemperatureat100bar(saturationpropertiesfromtableA-3).Therefore,thestateisinthesuperheatedvaporregion.InterpolatingfromTableA-4:𝑦 = 𝑦S + 𝑥 − 𝑥S

TU>TV8U>8V

⇒ 𝑣 = 𝑣XYSZ + 𝑇 − 𝑇XYSZ𝑣[\SZ − 𝑣XYSZ𝑇[\SZ − 𝑇XYSZ

= 0.03160m? kg + 485K − 480K0.03394m? kg– 0.03160m? kg

520K– 480K

= 0.03189m? kg d) Thegivendatatellusthestateissaturatedwaterwithacertainquality,asshowninthe

diagrambelow.FromTableA-2:𝑝 = 0.4739barUsingequation3.2anddatafromTableA-2⇒ 𝑣 = 𝑣9 + 𝑥 𝑣; − 𝑣9 = 1.0291×10>? m? kg + 0.75 3.407m? kg − 1.0291×10>? m? kg = 2.556m? kg

𝑇 = 485°C

𝑇sat,1SSbar = 311.1°C

𝑝 = 10MPa = 100bar

𝑇K = 374.14°C

𝑇 = 80°C𝑥 = 0.75

𝑝 = 𝑝sat,80°C = 0.4739bar

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3.50:Refrigerant22undergoesaconstantpressureprocesswithinapiston-cylinderassemblyfromsaturatedvaporat4bartoafinaltemperatureof30°C.Kineticandpotentialenergyeffectsarenegligible.Fortherefrigerant,showtheprocessonap-vdiagram.Evaluatetheworkandtheheattransfer,eachinkJperkgofrefrigerant.GivenConversionsModel𝑝_,sat = 𝑝\,sat = 4bar1bar = 100kPaNegligiblepotentialandkineticenergyeffects𝑇\ = 30°C1kJ = 1kPa ∙ m3RefrigerantistheclosedsystemConstantpressureprocessVolumechangeistheonlyworkdoneDetermine𝑤, specificwork, in kJ kg𝑞, specificheattransfer, in kJ kg𝑝-𝑣and𝑇-𝑣diagramsfortheprocessBasicEquations

𝑊 = 𝑝𝑑∀\

_

Δ𝐾𝐸 + Δ𝑃𝐸 + Δ𝑈 = 𝑄pq + 𝑊pq − 𝑄rst − 𝑊rst

𝑞 =𝑄𝑚,𝑤 =

𝑊𝑚

Analysis

𝑊 = 𝑝𝑑∀\

_⇒ 𝑤 = 𝑝𝑑𝑣

\

_= 𝑝 𝑣\ − 𝑣_

FromTableA-8:𝑣_ = 𝑣;,4bar = 0.0581m? kgFromTableA-9:𝑣\ = 0.06872m? kg

𝑤 = 4bar100kPa1bar

0.06872m? kg − 0.0581m? kg

= 4.248 kJ kgΔ𝐾𝐸 + Δ𝑃𝐸 + Δ𝑈 = 𝑄pq + 𝑊pq − 𝑄rst − 𝑊rst⇒ Δ𝑈 = mΔ𝑢 = 𝑄pq − 𝑊rst ⇒ Δ𝑢 = 𝑞pq − 𝑤rst ⇒ 𝑢\ − 𝑢_ = 𝑞pq − 𝑤rst ⇒ 𝑞pq = 𝑢\ − 𝑢_ + 𝑤rstFromTableA-8:𝑢_ = 𝑢;,4bar = 224.24 kJ kgFromTableA-9:𝑢\ = 245.73 kJ kg𝑞pq = 235.73 kJ kg − 224.24 kJ kg + 4.248 kJ kg= 15.738 kJ kgDiscussionTheproblemdescribestheworkingfluid,refrigerant22,initiallyasasaturatedvaporundergoingaconstantpressureprocesstoafinaltemperature.Thep-vdiagramindicatesthatthesecondstateiswithinthesuperheatedvaporregion.Withknownstatesthepropertyvaluescanbeeasilydeterminedfromthetables.Afterthepropertyvaluesareevaluatedtheproblembecomesidenticaltothosepreviouslysolvedinhomework2.

𝑞pq

𝑤rst

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3.63:Aclosed,rigidtankfilledwithwater,initiallyat20bar,aqualityof80%,andavolumeof0.5m3,iscooleduntilthepressureis4bar.ShowtheprocessofthewateronasketchoftheT-vdiagramandevaluatetheheattransfer,inkJ.GivenConversionsModel𝑝_ = 20bar1bar = 100kPaNegligiblepotentialandkineticenergyeffects𝑥_ = 0.81kJ = 1kPa ∙ m3Wateristheclosedsystem𝑉 = 0.5m3Constantvolumeprocess,𝑣_ = 𝑣\𝑝\ = 4barOnlyenergytransferisviaheatDetermineQ,heattransfer,inkJ𝑇-𝑣diagramfortheprocess𝑝-𝑣diagramfortheprocessBasicEquationsΔ𝐾𝐸 + Δ𝑃𝐸 + Δ𝑈 = 𝑄pq + 𝑊pq − 𝑄rst − 𝑊rst𝑣 = 𝑣9 + 𝑥 𝑣; − 𝑣9

𝑣 =∀𝑚

AnalysisΔ𝐾𝐸 + Δ𝑃𝐸 + Δ𝑈 = 𝑄pq + 𝑊pq − 𝑄rst − 𝑊rstAssumeheattransferisnetintothesystem.⇒ Δ𝑈 = −𝑄rst⇒ 𝑄rst = 𝑚 𝑢_ − 𝑢\ FromTableA-3at20bar:𝑣_ = 𝑣9_ + 𝑥_ 𝑣;_ − 𝑣9_ = 1.1764×10>? m? kg + 0.8 0.09963m? kg − 1.1764×10>? m? kg = 0.07994m? kg𝑢_ = 𝑢9_ + 𝑥_ 𝑢;_ − 𝑢9_ = 906.44kJ/kg + 0.8 2600.3kJ/kg − 906.44kJ/kg = 2261.53kJ/kgFromTableA-3at4bar:

𝑣\ = 𝑣9\ + 𝑥 𝑣;\ − 𝑣9\ ⇒ 𝑥\ =𝑣\ − 𝑣9\𝑣;\ − 𝑣9\

=0.07994m? kg − 1.0836×10>? m? kg0.4625m? kg − 1.0836×10>? m? kg

= 0.1709𝑢\ = 𝑢9\ + 𝑥\ 𝑢;\ − 𝑢9\ = 604.31kJ/kg + 0.1709 2553.6kJ/kg − 604.31kJ/kg = 937.44kJ/kg

𝑣 =∀𝑚⇒ 𝑚 =

∀𝑣=

0.5m3

0.07994m? kg

= 6.255kg𝑄rst = 6.255kg 2261.53kJ/kg − 937.44kJ/kg = 8282.18kJ

water

𝑄rst

T

v

1

2

𝑝_ =20bar

𝑝\ =4bar

𝑓\

𝑓_

𝑔\

𝑔_𝑥_ = 0.8

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DiscussionAsinthepreviousproblem,thisproblemreliesheavilypropertyevaluationsviathetables.Inthiscase,thequalityisneededtodeterminethepropertiesofthewaterateachsetpoint.Thequalityfortheinitialstateisgiven,butthequalityforthefinalstatemustbedeterminedbyusingthefactthattheprocessisconstantvolume.Oncethenecessarypropertiesareevaluated,thisproblemboilsdowntojustthefirstlaw.

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H3.2:Africtionlesspiston-cylinderassemblycontains3.5kgofH2Oinitiallyat𝑇_ = 130°Cand∀_=1.1m3.Initially,thepistonrestsonstops.Themassofthepistonandeffectoftheatmosphericpressureabovearesuchthatapressureofthewaterof400kPa(abs)isrequiredtomovethepiston,andifitmovesitdoessowithnoacceleration.Thewaterisheateduntilthetemperaturereaches𝑇\ = 240°C.Assumeaquasiequillibriumprocesswithnegligiblekineticandpotentialenergyeffects.GivenConversionsModel𝑇_ = 130°C1bar = 100kPaNegligiblepotentialandkineticenergyeffects𝑚 = 3.5kg1kJ = 1kPa ∙ m3Wateristheclosedsystem∀_= 1.1m3Constantpressureprocess,𝑝_ = 𝑝\𝑇\ = 240°C𝑝\ = 400kPaDetermineW,work,inkJQ,heattransfer,inkJBasicEquations

𝑊 = 𝑝𝑑∀\

_

Δ𝐾𝐸 + Δ𝑃𝐸 + Δ𝑈 = 𝑄pq + 𝑊pq − 𝑄rst − 𝑊rst𝑣 = 𝑣9 + 𝑥 𝑣; − 𝑣9

𝑣 =∀𝑚

Analysis

𝑣 =∀𝑚⇒ 𝑣_ =

∀_𝑚⇒ 𝑣_ =

1.1m3

3.5kg= 0.314 m3 kg

FromTableA-2at𝑇_ = 130°C:𝑣9 < 𝑣_ < 𝑣; ⇒ 1.0697×10>? m3 kg < 0.314m3 kg < 0.6685m3 kg ∴ saturatedliquid-vapor𝑣 = 𝑣9 + 𝑥 𝑣; − 𝑣9 ⇒ 𝑣_ = 𝑣9_ + 𝑥 𝑣;_ − 𝑣9_

⇒ 𝑥 =𝑣_ − 𝑣9_𝑣;_ − 𝑣9_

=0.314 m3 kg − 1.0697×10>? m3 kg0.6685m3 kg − 1.0697×10>? m3 kg

= 0.469

𝑢_ = 𝑢9_ + 𝑥 𝑢;_ − 𝑢9_ ⇒ 546.02kJ/kg + 0.469 2539.9kJ/kg − 546.02kJ/kg = 1481.15kJ/kgTheinitialstateofthisproblemisdefinedby𝑇_and𝑣_tobewithinthesaturationzone.Thep-vandT-vdiagrams,shownabove,cannowbedrawn.Notetheinitialpressureislessthan400kPa.

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Thevolumedoesnotbegintochangeuntilthepressurereaches400kPa,atwhichpointthepressureremainsconstanttothefinalstate.Thisallowsfortheworkoutofthesystemtobecalculated:

𝑊 = 𝑝𝑑∀\

_= 𝑝 ∀\ − ∀_

FromInterpolationofTableA-4:

𝑣\ = 𝑣S + 𝑝 − 𝑝S𝑣_ − 𝑣S𝑝_ − 𝑝S

= 0.781m3 kg + 4.0bar − 3.0bar0.4646m3 kg − 0.781m3 kg

5.0bar − 3.0bar

= 0.6228m3 kg

𝑣 =∀𝑚⇒ 𝑣\ =

∀\𝑚⇒ ∀\= 𝑣\𝑚 = 0.6228m3 kg 3.5kg

= 2.18m3𝑊 = 400kPa 2.18m3 − 1.1m3 = 432kJ(out)FromInterpolationofTableA-4:

𝑢\ = 𝑢S + 𝑝 − 𝑝S𝑢_ − 𝑢S𝑝_ − 𝑝S

= 2713.1kJ/kg + 4.0bar − 3.0bar2707.6 kJ kg − 2713.1 kJ kg

5.0bar − 3.0bar

= 2710.3 kJ kgΔ𝐾𝐸 + Δ𝑃𝐸 + Δ𝑈 = 𝑄pq + 𝑊pq − 𝑄rst − 𝑊rst ⇒ Δ𝑈 = 𝑄pq − 𝑊rst ⇒ 𝑄pq = 𝑚 𝑢\ − 𝑢_ + 𝑊rst𝑄pq = 3.5kg 2710.3 kJ kg − 1481.15kJ/kg + 432kJ= 4734.0kJ

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3.104:Determinethespecificvolume,inm3 kg,ofRefrigerant134aat16bar,100°Cusingvariousmethods.GivenConversionsModel𝑝 = 16bar1bar = 100kPaRefrigerantistheclosedsystem𝑇 = 100°CDeterminev,specificvolume,inm3 kga) UsingTableA-12b) UsingFigureA-1c) UsingtheidealgaslawBasicEquations𝑝𝑣 = 𝑅𝑇

𝑍 =𝑝𝑣𝑅𝑇

𝑝� =𝑝𝑝K

𝑇� =𝑇𝑇K

Analysisa) FromTableA-12usingthegivenpandT:𝑣 = 0.01601m3 kgb) FromTableA-1:𝑝K = 40.7bar

𝑝� =𝑝𝑝K=

16bar40.7bar

= 0.393

𝑇� =𝑇𝑇K=373K374K

= 1.0

FromFigureA-1:𝑍 ≅ 0.86

𝑍 =𝑝𝑣𝑅𝑇

⇒ 𝑣 =𝑍𝑅𝑇𝑝

=0.86 8.314J/mol∙K

102.03g/mol 1000g/kg 373K

16bar 100×10? Pa bar

=0.01634m3 kgc)

𝑝𝑣 = 𝑅𝑇 ⇒ 𝑣 =𝑅𝑇𝑝=

8.314J/mol∙K102.03g/mol 1000g/kg 373K

16bar 100×10? Pa bar

=0.0190m3 kgDiscussionThefirsttwovaluesareonly2%different,withthetablevaluebeingslightlylower.Whilethevaluecalculatedwiththeidealgaslawis19%greaterthanthetablevalue.Thisillustratesfallibilityoftheidealgaslaw,onlyundercertainconditionsisthelawconsideredaccurate.

R134ap=16barT=100°C

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3.116:Twokgofoxygenfillsthecylinderofapiston-cylinderassembly.Theinitialvolumeandpressureare2m3and1bar,respectively.Heattransfertotheoxygenoccursatconstantpressureuntilthevolumeisdoubled.GivenConversionsModel𝑝_ = 𝑝\ = 1bar1bar = 100kPaOxygenistheclosedsystem𝑘 = 1.351kJ = 1kPa ∙ m3Negligiblekineticandpotentialenergyeffects∀_= 2m3Pressureisconstant∀\= 2∀_= 4m3Specificheatratioisconstant,𝑐�, 𝑐�constantm=2kg OxygenisanidealgasDetermineQ,heattransfer,inkJBasicEquations

𝑊 = 𝑝𝑑∀\

_

Δ𝐾𝐸 + Δ𝑃𝐸 + Δ𝑈 = 𝑄pq + 𝑊pq − 𝑄rst − 𝑊rst𝑝𝑣 = 𝑅𝑇

𝑐� =𝑅

𝑘 − 1 𝑒𝑞3.47𝑏

𝑑𝑢 = 𝑐�𝑑𝑇Analysis

𝑊 = 𝑝𝑑∀\

_⇒ 𝑊 = 𝑝 ∀\ − ∀_ = 1bar 100 kPa bar 4m3 − 2m3 = 200kJ

forconstant𝑐�: 𝑑𝑢\

_= 𝑐� 𝑑𝑇

\

_⇒ 𝑢\ − 𝑢_ = 𝑐� 𝑇\ − 𝑇_

Δ𝐾𝐸 + Δ𝑃𝐸 + Δ𝑈 = 𝑄pq + 𝑊pq − 𝑄rst − 𝑊rst ⇒ Δ𝑈 = 𝑄pq − 𝑊rst⇒ 𝑄pq = 𝑚 𝑢\ − 𝑢_ + 𝑝 ∀\ − ∀_ = 𝑚𝑐� 𝑇\ − 𝑇_ + 𝑝 ∀\ − ∀_

𝑝𝑣 = 𝑅𝑇 ⇒ 𝑇_ =𝑝𝑣_𝑅

=1bar 100×10? Pa bar 2m3

2kg8.314J/mol∙K32.0g/mol 1000g/kg

= 384.9K

𝑝_𝑅=𝑝\𝑅∴𝑇_𝑣_=𝑇\𝑣\⇒ 𝑇\ = 𝑇_

𝑣\𝑣_= 384.9K

4m3

2kg2m3

2kg

= 769.8K

𝑐� =𝑅

𝑘 − 1=

8.314J/mol∙K32.0g/mol1.35 − 1

= 0.742kJ/kg∙K

𝑄pq = 2kg 0.742kJ/kg∙K 769.8K − 384.9K + 1bar 100 kPa bar 4m3 − 2m3 = 770kJ

𝑄pq

Oxygen

𝑊rst

10

DiscussionWithanidealgas,propertyevaluationissimplified.Theequationofstaterelatesp,v,T.Withfurtherassumptionofconstantspecificheat,𝑑𝑢 = 𝑐�𝑑𝑇canbeintegratedtoobtain𝑢\ − 𝑢_ =𝑐� 𝑇\ − 𝑇_ andpropertytablesarenotneeded.

11

H3.3:Apiston-cylinderdevicecontainsairinitiallyat𝑝_ = 90kPa,𝑇_ = 20°C,and∀_= 0.2𝐿.Theairisthencompressedinaquasiequillibriumpolytropicprocesswithn=1.25toavolumeone-seventhoftheoriginalvolume.Assumeidealgasbehaviorandnegligiblekineticandpotentialenergyeffects.GivenConversionsModel𝑝_ = 90kPa 1kJ = 1kPa ∙ m3Airistheclosedsystem∀_= 0.2L 1L =0.001m3Negligiblekineticandpotentialenergyeffects𝑇_ = 20°CAirisanidealgas𝑛 = 1.25Polytropicprocess

∀\=∀_7=0.27L

Determinea) 𝑝\,finalpressure,inkPab) 𝑇\,finaltemperature,in°Cc) 𝑊,work,inkJ[magnitudeanddirection]d) 𝑄,heattransfer,inkJ[magnitudeanddirection]BasicEquations

𝑊 = 𝑝𝑑∀\

_

Δ𝐾𝐸 + Δ𝑃𝐸 + Δ𝑈 = 𝑄pq + 𝑊pq − 𝑄rst − 𝑊rst𝑝∀q= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑝∀= 𝑚𝑅𝑇Analysisa)

𝑝∀q= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ⇒ 𝑝_∀_q= 𝑝\∀\q⇒ 𝑝\ =𝑝_∀_q

∀\q=

90×10?Pa 0.2L ∙ 0.001m3 L_.\[

0.27 L ∙ 0.001m3 L

_.\[

= 1.025MPab)

𝑝∀q= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ⇒ 𝑝_∀_q= 𝑝\∀\q⇒𝑝\𝑝_=

∀_∀\

q

𝑝∀= 𝑚𝑅𝑇 ⇒𝑝_∀_𝑇_

=𝑝\∀\𝑇\

⇒𝑇\𝑇_=𝑝\∀\𝑝_∀_

=∀_∀\

q ∀\∀_

=∀_∀\

q>_⇒𝑇\𝑇_=

∀_∀\

q>_

∀_= 0.2L𝑝_ = 90kPa𝑇_ = 20°C

𝑣\ =𝑣_7 𝑣_

𝑄rst

𝑊pq

12

𝑇\𝑇_=

∀_∀\

q>_⇒ 𝑇\ = 𝑇_

∀_∀\

q>_= 20°C

0.2L ∙ 0.001m3 L0.27 L ∙ 0.001m3 L

_.\[>_

= 32.5°Cc)

𝑊 = 𝑝𝑑∀\

_= 𝑝\∀\ − 𝑝_∀_

1 − 𝑛=

1025kPa 0.27 L ∙ 0.001m3 L − 90kPa 0.2L ∙ 0.001m3 L

1 − 1.25

= −0.0451kJ(in)Sincecompression𝑊pq = 𝑊 = 0.0451kJd) Workisintothesystem,soIwillassumeheattransferisoutofthesystem.Δ𝐾𝐸 + Δ𝑃𝐸 + Δ𝑈 = 𝑄pq + 𝑊pq − 𝑄rst − 𝑊rst ⇒ Δ𝑈 = 𝑊pq − 𝑄rst ⇒ 𝑄rst = 𝑊pq − 𝑚 𝑢\ − 𝑢_ InterpolatingfromTableA-22using𝑇_ = 20°C = 293K:

𝑢_ = 𝑢\�S� + 𝑇_ − 𝑇\�S�𝑢\�[� − 𝑢\�S�𝑇\�[� − 𝑇\�S�

= 206.91kJ/kg + 293K − 290K210.49kJ/kg − 206.91kJ/kg

295K − 290K

= 209.06kJ/kgInterpolatingfromTableA-22using𝑇\ = 32.5°C = 305.5K:

𝑢\ = 𝑢?S[� + 𝑇\ − 𝑇?S[�𝑢?_S� − 𝑢?S[�𝑇?_S� − 𝑇?S[�

= 217.67kJ/kg + 305.5K − 305K221.25kJ/kg − 217.67kJ/kg

310K − 305K

= 218.03kJ/kg

𝑝∀= 𝑚𝑅𝑇 ⇒ 𝑚 =𝑝_∀_𝑅𝑇_

=90×10?Pa 0.2L ∙ 0.001m3 L8.314J/mol∙K29.0g/mol 1000g/kg 293K

= 2.14×10>Xkg

𝑄rst = 0.0451kJ − 2.14×10>Xkg 218.03kJ/kg − 209.06kJ/kg = 0.0432kJ(out)AssumptionofheattranTypeequationhere.sferoutofthesystemisconfirmedbythepositivevalue.DiscussionSincespecificheatsarenotconstanttheidealgastablesareusedtofindu(T).

13

3.88:Inaheat-treatingprocess,a1-kgmetalpart,initiallyat1075K,isquenchedinaclosedtankcontaining100kgofwater,initiallyat295K.Thereisanegligibleheattransferbetweenthecontentsofthetankandtheirsurroundings.Modelingthemetalpartandthewaterasincompressiblewithconstantspecificheats0.5kJ/kg•Kand4.4kJ/kg•K,respectively.GivenConversionsModel𝑚��t�� = 1kg 1kJ = 1kPa ∙ m3Water+metalistheclosedsystem𝑚��t�� = 100kg1L =0.001m3Negligiblekineticandpotentialenergyeffects𝑇_,��t�� = 1075KAdiabaticandnowork𝑇_,��t�� = 295KConstantspecificheats𝑐��t�� = 0.5kJ/kg∙K Waterandmetalbehaveasincompressiblesubstances𝑐��t�� = 4.4kJ/kg ∙ KDetermineTeq,theequilibriumtemperature,inKBasicEquationsΔ𝐾𝐸 + Δ𝑃𝐸 + Δ𝑈 = 𝑄pq + 𝑊pq − 𝑄rst − 𝑊rstAnalysisΔ𝐾𝐸 + Δ𝑃𝐸 + Δ𝑈 = 𝑄pq + 𝑊pq − 𝑄rst − 𝑊rst ⇒ Δ𝑈��t�� + Δ𝑈��t�� = 0⇒ 𝑚��t��𝑐��t�� 𝑇�  − 𝑇_,��t�� + 𝑚��t��𝑐��t�� 𝑇�  − 𝑇_,��t�� = 0

⇒ 𝑇�  =𝑚��t��𝑐��t��𝑇_,��t�� + 𝑚��t��𝑐��t��𝑇_,��t��

𝑚��t��𝑐��t�� + 𝑚��t��𝑐��t��

=1kg 0.5kJ/kg∙K 1075K + 100kg 4.4kJ/kg∙K 295K

1kg 0.5kJ/kg∙K + 100kg 4.4kJ/kg∙K

= 295.9KDiscussionPhysically,thereisheattransferfromthemetaltothewater.However,sincethatoccursinsidethedefinedsystem,itisnotconsideredintheanalysis.