Quantity of Heat.Pdf

ChapterChapter 17.17. QuantityQuantity ofof HeatHeat

AA PowerPointPowerPoint PresentationPresentation byby PaulPaul E.E. Tippens,Tippens, ProfessorProfessor ofof PhysicsPhysics SouthernSouthern PolytechnicPolytechnic StateState UniversityUniversity

© 2007 Photo © Vol. 05 Photodisk/Getty FOUNDRY: It requires about 289 Joules of heat to melt one gram of steel. In this chapter, we will define the quantity of heat to raise the temperature and to change the phase of a substance. Objectives:Objectives: AfterAfter finishingfinishing thisthis unit,unit, youyou shouldshould bebe ableable to:to: • Define the quantity of heat in terms of the calorie, the kilocalorie, the joule, and the Btu.

• Write and apply formulas for specific heat capacity and solve for gains and losses of heat.

• Write and apply formulas for calculating the latent heats of fusion and vaporization of various materials. HeatHeat DefinedDefined asas EnergyEnergy

HeatHeat isis notnot somethingsomething anan objectobject has,has, butbut ratherrather energyenergy thatthat itit absorbsabsorbs oror givesgives up.up. TheThe heatheat lostlost byby thethe hothot coalscoals isis equalequal toto thatthat gainedgained byby thethe water.water.

Cool Thermal Equilibrium water

Hot coals UnitsUnits ofof HeatHeat

One calorie (1 cal) is the quantity of heat required to raise the temperature of 1 g of water by 1 C0.

ExampleExample 10 calories of heat will raise the temperature of 10 g of water by 10 C0. UnitsUnits ofof HeatHeat (Cont.)(Cont.)

One kilocalorie (1 kcal) is the quantity of heat required to raise the temperature of 1 kg of water by 1 C0.

ExampleExample 10 kilocalories of heat will raise the temperature of 10 kg of water by 10 C0. UnitsUnits ofof HeatHeat (Cont.)(Cont.)

One British Thermal Unit (1 Btu) is the quantity of heat required to raise the temperature of 1 lb of water by 1 F0.

ExampleExample 10 Btu of heat will raise the temperature of 10 lb of water by 10 F0. TheThe BtuBtu isis anan OutdatedOutdated UnitUnit

The British Thermal Unit (1 Btu) is discouraged, but unfortunately remains in wide-spread use today. If it is to be used, we must recognize that the pound unit is actually a unit of mass, not weight.

When working with the Btu, we must recall that the pound-mass is not a variable quantity that depends on gravity -- 1 lb one reason that the use of the Btu is discouraged! 1 lb (1/32) slug TheThe SISI UnitUnit ofof HeatHeat

SinceSince heatheat isis energy,energy, thethe joulejoule isis thethe preferredpreferred unit.unit. Then,Then, mechanicalmechanical energyenergy andand heatheat areare measuredmeasured inin thethe samesame fundamentalfundamental unit.unit.

ComparisonsComparisons ofof HeatHeat Units:Units:

11 calcal == 4.1864.186 JJ 11 BtuBtu == 778778 ftft lblb

11 kcalkcal == 41864186 JJ 11 BtuBtu == 252252 calcal

11 BtuBtu == 10551055 JJ TemperatureTemperature andand QuantityQuantity ofof HeatHeat

The effect of heat on temp- 200C 220C erature depends on the quantity of matter heated.

The same quantity of heat is applied to each mass of 600 g water in the figure. 200C 300C The larger mass experiences a smaller increase in temperature. 200 g HeatHeat CapacityCapacity

The heat capacity of a substance is the heat required to raise the temperature a unit degree.

Lead Glass Al Copper Iron

1000C 1000C 1000C 1000C 1000C

37 s 52 s 60 s 83 s 90 s

Heat capacities based on time to heat from zero to 1000C. Which has the greatest heat capacity? HeatHeat CapacityCapacity (Continued)(Continued)

AllAll atat 10010000CC placedplaced onon ParaffinParaffin SlabSlab Lead Glass Al Copper Iron

Iron and copper balls melt all the way through; others have lesser heat capacities. SpecificSpecific HeatHeat CapacityCapacity TheThe specificspecific heatheat capacitycapacity ofof aa materialmaterial isis thethe quantityquantity ofof heatheat neneedededed toto raiseraise thethe temperaturetemperature ofof aa unitunit massmass throughthrough aa unitunit degree.degree. Q cQmct ;  mt

Water:Water: cc == 1.01.0 cal/gcal/g CC00 oror 11 Btu/lbBtu/lb FF00 oror 41864186 J/kgJ/kg KK

Copper:Copper: cc == 0.0940.094 cal/gcal/g CC00 oror 390390 J/kgJ/kg KK ComparisonComparison ofof HeatHeat Units:Units: HowHow muchmuch heatheat isis neededneeded toto raiseraise 11--kgkg ofof waterwater fromfrom 000 toto 1001000C?C? The mass of one kg of water is:

1 kg = 1000 g = 0.454 lbm

Qmct 1 lbm = 454 g 1 kg For water: c = 1.0 cal/g C0 or 1 Btu/lb F0 or 4186 J/kg K

TheThe heatheat requiredrequired toto dodo thisthis jobjob is:is: 10,000 cal 10 kcal 39.7 Btu 41, 860 J ProblemProblem SolvingSolving ProcedureProcedure

1. Read problem carefully and draw a rough sketch. 2. Make a list of all given quantities 3. Determine what is to be found. 4. Recall applicable law or formula and constants. Q cQmct ;  mt Water:Water: c c = = 1.0 1.0 cal/g cal/g C C00 or or 1 1 Btu/lb Btu/lb F F00 oror 41864186 J/kgJ/kg KK 5. Determine what was to be found. ExampleExample 1:1: AA 500500--gg coppercopper coffeecoffee mugmug isis filledfilled withwith 200200--gg ofof coffee.coffee. HowHow muchmuch heatheat waswas requiredrequired toto heatheat cupcup andand coffeecoffee fromfrom 2020 toto 96960CC?? 1.1. DrawDraw sketchsketch ofof problem.problem 2.2. ListList givengiven information.information.

MugMug massmass mmm == 0.5000.500 kgkg

CoffeeCoffee massmass mmc == 0.2000.200 kgkg 0 InitialInitial temperaturetemperature ofof coffeecoffee andand mug:mug: tt0 == 2020 CC 0 FinalFinal temperaturetemperature ofof coffeecoffee andand mug:mug: ttf == 9696 CC 3.3. ListList whatwhat isis toto bebe found:found: Total heat to raise temperature of coffee (water) and mug to 960C. ExampleExample 1(Cont.):1(Cont.): HowHow muchmuch heatheat neededneeded toto heatheat cupcup andand coffeecoffee fromfrom 2020 toto 96960CC?? mmm == 0.20.2 kgkg;; mmw == 0.50.5 kgkg.. 4.4. RecallRecall applicableapplicable formulaformula oror law:law: Heat Gain or Loss: Q = mc t 5.5. DecideDecide thatthat TOTALTOTAL heatheat isis thatthat requiredrequired toto raiseraise temperaturetemperature ofof mugmug andand waterwater (coffee).(coffee). WriteWrite equation.equation.

QQT == mmmc cm tt ++ mmwc cw tt

0 6.6. LookLook upup specificspecific Copper:Copper: ccm == 390390 J/kgJ/kg CC 0 heatsheats inin tables:tables: CoffeeCoffee (water):(water): ccw == 41864186 J/kgJ/kg CC ExampleExample 1(Cont.):1(Cont.): HowHow muchmuch heatheat neededneeded toto heatheat cupcup andand coffeecoffee fromfrom 2020 toto 96960CC?? mmc == 0.20.2 kgkg;; mmw == 0.50.5 kgkg.. 7.7. SubstituteSubstitute infoinfo andand solvesolve problem:problem:

0 Copper:Copper: ccm == 390390 J/kgJ/kg CC 0 CoffeeCoffee (water):(water): ccw == 41864186 J/kgJ/kg CC

QT =mmc m t+ mwc w t

Water: (0.20 kg)(4186 J/kgC0)(76 C0) t=t= 996600CC -- 202000CC Cup: (0.50 kg)(390 J/kgC0)(76 C0) == 7676 CC00

Q = 78.4 kJ QT = 63,600 J + 14,800 J QTT = 78.4 kJ AA WordWord AboutAbout UnitsUnits

TheThe substitutedsubstituted unitsunits mustmust bebe consistentconsistent withwith thosethose ofof thethe chosenchosen valuevalue ofof specificspecific heatheat capacity.capacity.

0 0 For example: Water cw = 4186 J/kg C or 1 cal/g C

QQ == mmwc cw tt

IfIfTheThe youyou units units useuse 41861 41861forfor cal/gcal/g QQ J/kgJ/kg,, Cm,m,C00 CforandandCfor00 for c,for c, t t c,c, thenthenmustmust QQ bebe mustmust consistentconsistent bebe inin calories,joules,calories,joules, withwith andand mandmandthosethose mustmust mm must must basedbased bebe inin bebe kilograms.ononkilograms. inin thethe grams.grams. valuevalue ofof thethe constantconstant c.c. ConservationConservation ofof EnergyEnergy Whenever there is a transfer of heat within a system, the heat lost by the warmer bodies must equal the heat gained by the cooler bodies:

(Heat(Heat Losses)Losses) == (Heat(Heat Gained)Gained)

Cool Thermal Equilibrium water Hot iron 0 ExampleExample 2:2: AA handfulhandful ofof coppercopper 10 shotshot isis heatedheated toto 90900CC andand thenthen water droppeddropped intointo 8080 gg ofof waterwater inin anan 0 insulatedinsulated cupcup atat 10100CC.. IfIf thethe 90 shot 0 equilibriumequilibrium temperaturetemperature isis 18180CC,, Insulator whatwhat waswas thethe massmass ofof thethe copper?copper?

0 0 0 te = 18 C cw = 4186 J/kg C ; cs = 390 J/kg C 0 0 mw = 80 g; tw= 10 C; ts = 90 C Heat lost by shot = heat gained by water

0 0 0 0 msc s(90 C - 18 C) = mwcw(18 C - 10 C)

Note: Temperature differences are [High - Low] to insure absolute values (+) lost and gained. ExampleExample 2:2: (Cont.)(Cont.)

ms = ? 100 180C water 900 shot Insulator 80 g of Water Heat lost by shot = heat gained by water

0 0 0 0 msc s(90 C - 18 C) = mwcw(18 C - 10 C)

0 0 0 0 ms(390 J/kgC )(72 C ) = (0.080 kg)(4186 J/kgC )(8 C ) 2679 J m 0.0954 kg mms == 95.495.4 gg s 28,080 J/kg s ChangeChange ofof PhasePhase

WhenWhen aa changechange ofof phasephase occurs,occurs, therethere isis onlyonly aa changechange inin potentialpotential energyenergy ofof thethe molecules.molecules. TheThe temperaturetemperature isis constantconstant duringduring thethe change.change.

Liquid Vaporization Solid Gas fusion

Q = mLf Q = mLv

Terms:Terms: Fusion,Fusion, vaporization,vaporization, condensation,condensation, latentlatent heats,heats, evaporation,evaporation, freezingfreezing point,point, meltingmelting point.point. ChangeChange ofof PhasePhase

The latent heat of fusion (Lf ) of a substance is the heat per unit mass required to change the Q Lf  substance from the solid to the liquid phase of m its melting temperature. For Water: L = 80 cal/g = 333,000 J/kg For Water: Lf f = 80 cal/g = 333,000 J/kg

The latent heat of vaporization (Lv ) of a substance is the heat per unit mass required Q Lv  to change the substance from a liquid to a m vapor at its boiling temperature.

For Water: L = 540 cal/g = 2,256,000 J/kg For Water: Lvv = 540 cal/g = 2,256,000 J/kg MeltingMelting aa CubeCube ofof CopperCopper TheThe heatheat QQ requiredrequired toto meltmelt aa substancesubstance atat itsits meltingmelting temperaturetemperature What Q cancan bebe foundfound ifif thethe massmass andand latentlatent 2 kg to melt heatheat ofof fusionfusion areare known.known. copper?

Q = mL Q = mLvv Lf = 134 kJ/kg Example:Example: ToTo completelycompletely meltmelt 22 kgkg ofof coppercopper atat 104010400C,C, wewe need:need:

QQ == mLmLf == (2(2 kg)(134,000kg)(134,000 J/kg)J/kg) QQ == 268268 kJkJ ExampleExample 3:3: HowHow muchmuch heatheat isis neededneeded toto convertconvert 1010 gg ofof iceice atat --20200CC toto steamsteam atat 1001000CC??

First, let’s review the process graphically as shown: temperature t ice steam 540 cal/g 1000C 1 cal/gC0 steam steam only 80 cal/g and 00C water water ice and 0 cice = 0.5 cal/gConly -200C ice water Q ExampleExample 33 (Cont.):(Cont.): StepStep oneone isis QQ1 toto convertconvert 1010 gg ofof iceice atat --20200CC toto iceice atat 000CC (no(no waterwater yet).yet).

0 Q1 to raise ice to 0 C: Q1 = mct -200C 00C t 0 0 1000C Q1 = (10 g)(0.5 cal/gC )[0 - (-20 C)] 0 0 Q1 = (10 g)(0.5 cal/gC )(20 C ) Q = 100 cal Q11 = 100 cal 00C 0 cice = 0.5 cal/gC -200C ice Q ExampleExample 33 (Cont.):(Cont.): StepStep twotwo isis QQ2 toto convertconvert 1010 gg ofof iceice atat 000CC toto waterwater atat 000CC..

0 Q2 to melt 10 g of ice at 0 C: Q2 = mLf Melting t Q = (10 g)(80 cal/g) = 800 cal 1000C 2 Q = 800 cal Q22 = 800 cal

80 cal/g Add this to Q1 = 100 cal: 00C 900 cal used to this point. ice and -200C water Q ExampleExample 33 (Cont.):(Cont.): StepStep threethree isis QQ3 toto changechange 1010 gg ofof waterwater atat 000CC toto waterwater atat 1001000CC..

0 0 Q3 to raise water at 0 C to 100 C. 0 0 0 0 C to 100 C Q3 = mct; cw= 1 cal/gC

t 0 0 0 Q3 = (10 g)(1 cal/gC )(100 C - 0 C) 1000C QQ3 == 10001000 calcal 1 cal/gC0 3

Total = Q1 + Q2 + Q3 = 100 +900 + 1000 00C water = 1900 cal only -200C Q ExampleExample 33 (Cont.):(Cont.): StepStep fourfour isis QQ4 toto convertconvert 0 1010 gg ofof waterwater toto steamsteam atat 100100 CC?? ((QQ4 == mLmLv) )

0 Q4 to convert all water at 100 C 0 vaporization to steam at 100 C. (Q = mLv )

Q4 = (10 g)(540 cal/g) = 5400 cal 1000C

800 cal 5400 cal Total Heat: 1000 100 cal 7300 cal cal steam 7300 cal 00C ice and water and -200C ice water only water Q ExampleExample 4:4: HowHow manymany gramsgrams ofof iceice atat 000CC mustmust bebe mixedmixed withwith fourfour gramsgrams ofof steamsteam inin orderorder toto produceproduce waterwater atat 60600CC?? mi = ? 0 IceIce mustmust meltmelt andand thenthen riserise toto 6060 C.C. ice 0 SteamSteam mustmust condensecondense andand dropdrop toto 6060 C.C. steam Total Heat Gained = Total Heat Lost 4 g 0 miLf + micw t= msLv + mscw t te = 60 C

Note:Note: AllAll losseslosses andand gainsgains areare absoluteabsolute valuesvalues (positive).(positive).

00 00 0 TotalTotal Gained:Gained: mmi(80 i(80 cal/g)cal/g) ++ mmi(1 i(1 cal/gCcal/gC)(60)(60 CC )-0C) TotalLost: Lost:(4 g)(540 (4 g)(540 cal/g) cal/g)+ (4 g)(1 + (4 cal/gC g)(1 cal/gC0)(1000 )(40C0 -6 C000)C) ExampleExample 44 (Continued)(Continued)

0 0 Total Gained: mi(80 cal/g) + mi(1 cal/gC )(60 C ) Total Lost: (4 g)(540 cal/g) + (4 g)(1 cal/gC0)(40 C0)

mi = ? Total Heat Gained = Total Heat Lost 4 g 80mi + 60mi = 2160 g +160 g

0 2320 g te = 60 C mi  mmi == 16.616.6 gg 140 i ExampleExample 5:5: FiftyFifty gramsgrams ofof iceice ice water areare mixedmixed withwith 200200 gg ofof waterwater 0 initiallyinitially atat 70700CC.. FindFind thethe 00C 700C equilibriumequilibrium temperaturetemperature ofof thethe 50 g 200 g mixture.mixture.

Ice melts and rises to te te = ? Water drops from 70 to te.

0 Heat Gained: miL f + mic w t ; t = te -0C

0 0 Gain = (50 g)(80 cal/g) + (50 g)(1 cal/gC )(te -0C)

Gain = 4000 cal + (50 cal/g)te ExampleExample 55 (Cont.):(Cont.):

GainGain = 4000 cal + (50 cal/g)te 00C 700C 50 g 200 g Heat Lost = mwcw t 0 t = 70 C -te [high - low] te = ?

0 0 Lost = (200 g)(1 cal/gC )(70 C- te ) 0 Lost = 14,000 cal - (200 cal/C ) te

Heat Gained Must Equal the Heat Lost:

0 4000 cal + (50 cal/g)te = 14,000 cal - (200 cal/C ) te ExampleExample 55 (Cont.):(Cont.): Heat Gained Must Equal the Heat Lost:

0 4000 cal + (50 cal/g)te = 14,000 cal - (200 cal/C ) te

0 Simplifying, we have: (250 cal/C ) te = 10,000 cal

10, 000 cal 0 te 0 40 C 00C 700C 250 cal/C 50 g 200 g

0 t = 40 0C te = ? tee = 40 C SummarySummary ofof HeatHeat UnitsUnits One calorie (1 cal) is the quantity of heat required to raise the temperature of 1 g of water by 1 C0. One kilocalorie (1 kcal) is the quantity of heat required to raise the temperature of 1 kg of water by 1 C0.

One British thermal unit (Btu) is the quantity of heat required to raise the temperature of 1 lb of water by 1 F0. Summary:Summary: ChangeChange ofof PhasePhase

The latent heat of vaporization (Lv ) of a substance is the heat per unit mass required Q Lv  to change the substance from a liquid to a m vapor at its boiling temperature.

For Water: L = 540 cal/g = 2,256,000 J/kg For Water: Lvv = 540 cal/g = 2,256,000 J/kg Summary:Summary: SpecificSpecific HeatHeat CapacityCapacity

TheThe specificspecific heatheat capacitycapacity ofof aa materialmaterial isis thethe quantityquantity ofof heatheat toto raiseraise thethe temperaturetemperature ofof aa unitunit massmass throughthrough aa unitunit degree.degree.

Q cQmct ;  mt Summary:Summary: ConservationConservation ofof EnergyEnergy

Whenever there is a transfer of heat within a system, the heat lost by the warmer bodies must equal the heat gained by the cooler bodies:

(Heat(Heat Losses)Losses) == (Heat(Heat Gained)Gained)

SummarySummary ofof Formulas:Formulas: Q cQmct ;  mt

(Heat(Heat Losses)Losses) == (Heat(Heat Gained)Gained)

Q LQmL; f m f Q LQmL; vvm CONCLUSION:CONCLUSION: ChapterChapter 1717 QuantityQuantity ofof HeatHeat