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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 temp- erature 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.
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