Calorimetry

Calorimetry1Conservation of Energy in a Chemical ReactionSurroundings

SystemSurroundings

SystemEnergyBeforereactionAfterreactionIn this example, the energy of the reactants and products increases, while the energy of the surroundings decreases.

In every case, however, the total energy does not change.Myers, Oldham, Tocci, Chemistry, 2004, page 41Endothermic Reaction

Reactant + Energy Product3Conservation of Energy in a Chemical ReactionSurroundings

SystemSurroundings

SystemEnergyBeforereactionAfterreactionIn this example, the energy of the reactants and products decreases, while the energy of the surroundings increases.

In every case, however, the total energy does not change.Myers, Oldham, Tocci, Chemistry, 2004, page 41Exothermic Reaction

Reactant Product + Energy4Direction of Heat FlowSurroundingsENDOthermicqsys > 0EXOthermicqsys < 0SystemKotz, Purcell, Chemistry & Chemical Reactivity 1991, page 207

SystemH2O(s) + heat H2O(l)meltingH2O(l) H2O(s) + heat freezing5Caloric ValuesFood joules/grams calories/gram Calories/gramProtein 17 000 4000 4

Fat 38 000 9000 9

Carbohydrates 17 000 4000 4Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 511000 calories = 1 Calorie"science" "food"1calories = 4.184 joules6

Typical apparatus used in this activity include a boiler (such as large glass beaker), a heat source (Bunsen burner or hot plate), a stand or tripod for the boiler, a calorimeter, thermometers, samples (typically samples of copper, aluminum, zinc, tin, or lead), tongs (or forceps or string) to handle samples, and a balance.

Experimental Determination of Specific Heat of a Metal7A Coffee Cup CalorimeterZumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 302

ThermometerStyrofoamcoverStyrofoamcupsStirrer8Bomb Calorimeterthermometerstirrerfull of waterignition wiresteel bombsample

91997 Encyclopedia Britanica, Inc.

oxygen supplystirrerthermometermagnifyingeyepieceair spacecruciblesteel bombsampleignition coilbucketheaterwaterignitionwiresinsulatingjacket

10A Bomb Calorimeter

11Causes of Change - CalorimetryOutlineKeys

http://www.unit5.org/chemistry/Matter.html12Heating Curves

Melting - PE Solid - KE Liquid - KE Boiling - PE Gas - KE Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

13Heating CurvesTemperature (oC)40200-20-40-60-80-1001201008060140TimeMelting - PE Solid - KE Liquid - KE Boiling - PE Gas - KE 14Heating CurvesTemperature (oC)40200-20-40-60-80-1001201008060140TimeMelting - PE Solid - KE Liquid - KE Boiling - PE Gas - KE 15Heating CurvesTemperature Changechange in KE (molecular motion) depends on heat capacityHeat Capacityenergy required to raise the temp of 1 gram of a substance by 1CVolcano clip -water has a very high heat capacityCourtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

16Heating CurvesPhase Changechange in PE (molecular arrangement)temp remains constantHeat of Fusion (Hfus)energy required to melt 1 gram of a substance at its m.p.Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

17Heating CurvesHeat of Vaporization (Hvap)energy required to boil 1 gram of a substance at its b.p.usually larger than Hfuswhy?EX: sweating, steam burns, the drinking bird

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

18Phase DiagramsShow the phases of a substance at different temps and pressures.

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

19HumorA small piece of ice which lived in a test tube fell in love with a Bunsen burner.

Bunsen! My flame! I melt whenever I see you said the ice.

The Bunsen burner replied Its just a phase youre going through.20A B warm iceB C melt ice (solid liquid)C D warm waterD E boil water (liquid gas)E D condense steam (gas liquid)E F superheat steamHeating Curve for Water(Phase Diagram) 140

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-100Temperature (oC)HeatBPMPABCDEFHeat = m x CfusCf = 333 J/gHeat = m x CvapCv = 2256 J/gHeat = m x DT x Cp, liquidCp = 4.184 J/goCHeat = m x DT x Cp, solidCp (ice) = 2.077 J/goCHeat = m x DT x Cp, gasCp (steam) = 1.87 J/goC21Calculating Energy Changes - Heating Curve for WaterTemperature (oC)40200-20-40-60-80-1001201008060140TimeDH = mol x DHfusDH = mol x DHvapHeat = mass x Dt x Cp, liquidHeat = mass x Dt x Cp, gasHeat = mass x Dt x Cp, solid22Equal Masses of Hot and Cold Water

Thin metal wallInsulated boxZumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 29123Water Molecules in Hot and Cold Water

Hot water Cold Water90 oC 10 oCZumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 29124Water Molecules in the same temperature water

Water(50 oC)Water(50 oC)Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 29125Heat TransferAlAlm = 20 gT = 40oCSYSTEMSurroundingsm = 20 gT = 20oC20 g (40oC)20 g (20oC)30oCBlock ABlock BFinalTemperatureAssume NO heat energy is lost to the surroundings from the system.

What will be the final temperature of the system ?

a) 60oC b) 30oC c) 20oC d) ?26Heat TransferAlAlm = 20 gT = 40oCSYSTEMSurroundingsm = 10 gT = 20oC20 g (40oC)20 g (20oC)30.0oCBlock ABlock BFinalTemperatureAssume NO heat energy is lost to the surroundings from the system.20 g (40oC)10 g (20oC)33.3oC

What will be the final temperature of the system ?

a) 60oC b) 30oC c) 20oC d) ??27Heat TransferAlAlm = 20 gT = 20oCSYSTEMSurroundingsm = 10 gT = 40oC20 g (40oC)20 g (20oC)30.0oCBlock ABlock BFinalTemperatureAssume NO heat energy is lost to the surroundings from the system.20 g (40oC)10 g (20oC)33.3oC

20 g (20oC)10 g (40oC)26.7oC28Heat Transferm = 75 gT = 25oCSYSTEMSurroundingsm = 30 gT = 100oC20 g (40oC)20 g (20oC)30.0oCBlock ABlock BFinalTemperature20 g (40oC)10 g (20oC)33.3oC

20 g (20oC)10 g (40oC)26.7oCAgH2OReal Final Temperature = 26.7oCWhy?Weve been assuming ALL materialstransfer heat equally well.29Specific HeatWater and silver do not transfer heat equally well. Water has a specific heat Cp = 4.184 J/goC Silver has a specific heat Cp = 0.235 J/goC

What does that mean? It requires 4.184 Joules of energy to heat 1 gram of water 1oC and only 0.235 Joules of energy to heat 1 gram of silver 1oC.

Law of Conservation of Energy In our situation (silver is hot and water is cold) this means water heats up slowly and requires a lot of energy whereas silver will cool off quickly and not release much energy.

Lets look at the math!

30

The amount of heat required to raise the temperature of one gram of substance by one degree Celsius.Specific Heat31

cp = Specific Heatq = Heat lost or gainedT = Temperature changeOR

m = MassCalculations involving Specific Heat32 SubstanceSpecific heat J/(g.K) Water (l)4.18 Water (s)2.06 Water (g)1.87 Ammonia (g)2.09 Benzene (l)1.74 Ethanol (l)2.44 Ethanol (g)1.42 Aluminum (s)0.897 Calcium (s)0.647 Carbon, graphite (s)0.709 Copper (s)0.385 Gold (s)0.129 Iron (s)0.449 Mercury (l)0.140 Lead (s)0.129Specific Heats of Some CommonSubstances at 298.15 KTable of Specific Heats33The energy that must be absorbed in order to convert one mole of solid to liquid at its melting point.Latent Heat of Phase ChangeMolar Heat of Fusion

The energy that must be removed in order to convert one mole of liquid to solid at its freezing point.34The energy that must be absorbed in order to convert one mole of liquid to gas at its boiling point.The energy that must be removed in order to convert one mole of gas to liquid at its condensation point.Latent Heat of Phase Change #2Molar Heat of Vaporization35Latent Heat Sample ProblemProblem: The molar heat of fusion of water is 6.009 kJ/mol. How much energy is needed to convert 60 grams of ice at 0C to liquid water at 0C?

Massof iceMolarMass ofwaterHeatoffusion36Heat of ReactionThe amount of heat released or absorbed during a chemical reaction.Endothermic: Exothermic: Reactions in which energy is absorbed as the reaction proceeds.Reactions in which energy is released as the reaction proceeds.37loses heatCalorimetry

m = 75 gT = 25oCSYSTEMSurroundingsm = 30 gT = 100oCAgH2OTfinal = 26.7oC38Calorimetry

m = 75 gT = 25oCSYSTEMSurroundingsm = 30 gT = 100oCAgH2O391 Calorie = 1000 caloriesfood = scienceCandy bar300 Calories = 300,000 caloriesEnglishMetric = _______Joules1 calorie - amount of heat needed to raise 1 gram of water 1oC1 calorie = 4.184 Joules1 BTU (British Thermal Unit) amount of heat needed to raise 1 pound of water 1oF.40Cp(ice) = 2.077 J/g oCIt takes 2.077 Joules to raise 1 gram ice 1oC.X Joules to raise 10 gram ice 1oC.(10 g)(2.077 J/g oC) = 20.77 JoulesX Joules to raise 10 gram ice 10oC.(10oC)(10 g)(2.077 J/g oC) = 207.7 JoulesHeat = (specific heat) (mass) (change in temperature)q = Cp . m . DTTemperature (oC)40200-20-40-60-80-1001201008060140TimeDH = mol x DHfusDH = mol x DHvapHeat = mass x Dt x Cp, liquidHeat = mass x Dt x Cp, gasHeat = mass x Dt x Cp, solid41Heat = (specific heat) (mass) (change in temperature)q = Cp . m . DT

GivenTi = -30oCTf = -20oCq = 207.7 JoulesTemperature (oC)40200-20-40-60-80-1001201008060140TimeDH = mol x DHfusDH = mol x DHvapHeat = mass x Dt x Cp, liquidHeat = mass x Dt x Cp, gasHeat = mass x Dt x Cp, solid42240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC). When thermal equilibrium is reached, the system has a temperature of 42oC. Find the mass of the iron.Calorimetry Problems 2 question #5FeT = 500oCmass = ? gramsT = 20oCmass = 240 gLOSE heat = GAIN heat-- [(Cp,Fe) (mass) (DT)] = (Cp,H2O) (mass) (DT)- [(0.4495 J/goC) (X g) (42oC - 500oC)] = (4.184 J/goC) (240 g) (42oC - 20oC)] Drop Units: - [(0.4495) (X) (-458)] = (4.184) (240 g) (22)205.9 X = 22091X = 107.3 g Fe43A 97 g sample of gold at 785oC is dropped into 323 g of water, which has an initial temperature of 15oC. If gold has a specific heat of 0.129 J/goC, what is the final temperature of the mixture? Assume that the gold experiences no change in state of matter.Calorimetry Problems 2 question #8AuT = 785oCmass = 97 gT = 15oCmass = 323 gLOSE heat = GAIN heat-- [(Cp,Au) (mass) (DT)] = (Cp,H2O) (mass) (DT)- [(0.129 J/goC) (97 g) (Tf - 785oC)] = (4.184 J/goC) (323 g) (Tf - 15oC)] Drop Units: - [(12.5) (Tf - 785oC)] = (1.35 x 103) (Tf - 15oC)] -12.5 Tf + 9.82 x 103 = 1.35 x 103 Tf - 2.02 x 1043 x 104 = 1.36 x 103 TfTf = 22.1oC44If 59 g of water at 13oC are mixed with 87 g of water at 72oC, find the final temperature of the system. Calorimetry Problems 2 question #9T = 13oCmass = 59 gLOSE heat = GAIN heat-- [(Cp,H2O) (mass) (DT)] = (Cp,H2O) (mass) (DT)- [(4.184 J/goC) (87 g) (Tf - 72oC)] = (4.184 J/goC) (59 g) (Tf - 13oC)Drop Units: - [(364.0) (Tf - 72oC)] = (246.8) (Tf - 13oC)-364 Tf + 26208 = 246.8 Tf - 320829416 = 610.8 TfTf = 48.2oCT = 72oCmass = 87 g45A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC. Find the system's final temperature.Calorimetry Problems 2 question #10iceT = -11oCmass = 38 gT = 56oCmass = 214 gLOSE heat = GAIN heat-- [(Cp,H2O(l)) (mass) (DT)] = (Cp,H2O(s)) (mass) (DT) + (Cf) (mass) + (Cp,H2O(l)) (mass) (DT) - [(4.184 J/goC)(214 g)(Tf - 56oC)] = (2.077 J/goC)(38 g)(11oC) + (333 J/g)(38 g) + (4.184 J/goC)(38 g)(Tf - 0oC) - [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)] - 895 Tf + 50141 = 868 + 12654 + 159 Tf- 895 Tf + 50141 = 13522 + 159 TfTf = 34.7oC 36619 = 1054 TfTemperature (oC)40200-20-40-60-80-1001201008060140TimeDH = mol x DHfusDH = mol x DHvapHeat = mass x Dt x Cp, liquidHeat = mass x Dt x Cp, gasHeat = mass x Dt x Cp, solidABCDwarm icemelt icewarm waterwater coolsDBAC4625 g of 116oC steam are bubbled into 0.2384 kg of water at 8oC. Find the final temperature of the system.Calorimetry Problems 2 question #11- [(Cp,H2O) (mass) (DT)] + (-Cv,H2O) (mass) + (Cp,H2O) (mass) (DT) = [(Cp,H2O) (mass) (DT)] - [ - 748 + -56400 + 104.5Tf - 10460] = 997Tf - 7980 - [qA + qB + qC] = qDqA = [(Cp,H2O) (mass) (DT)] qA = [(1.87 J/goC) (25 g) (100o - 116oC)] qA = - 748 J qB = (Cv,H2O) (mass)qB = (-2256 J/g) (25 g) qB = - 56400 J qC = [(Cp,H2O) (mass) (DT)] qC = [(4.184 J/goC) (25 g) (Tf - 100oC)] qC = 104.5Tf - 10460qD = (4.184 J/goC) (238.4 g) (Tf - 8oC)qD = 997Tf - 7980- [qA + qB + qC] = qD748 + 56400 - 104.5Tf + 10460 = 997Tf - 7980 67598 - 104.5Tf = 997Tf - 7979 75577 = 1102Tf11021102Tf = 68.6oCTemperature (oC)40200-20-40-60-80-1001201008060140TimeDH = mol x DHfusDH = mol x DHvapHeat = mass x Dt x Cp, liquidHeat = mass x Dt x Cp, gasHeat = mass x Dt x Cp, solidABCD(1000 g = 1 kg)238.4 g4725 g of 116oC steam are bubbled into 0.2384 kg of water at 8oC. Find the final temperature of the system.Calorimetry Problems 2 question #11- [(Cp,H2O) (mass) (DT) + (-Cv,H2O) (mass) + (Cp,H2O) (mass) (DT)] = (Cp,H2O) (mass) (DT)- [ - 748 + -56400 + 104.5Tf - 10460] = 997Tf - 7980 - [qA + qB + qC] = qD- [(Cp,H2O) (mass) (DT) + - [(1.87 J/goC) (25 g) (100o - 116oC) + - [ - 748 J (Cv,H2O) (mass) +(-2256 J/g) (25 g) + + - 56400 J (Cp,H2O) (mass) (DT)] (4.184 J/goC) (25 g) (Tf - 100oC)] + 104.5Tf - 10460 ]= (4.184 J/goC) (238.4 g) (Tf - 8oC)= 997Tf - 7980748 + 56400 - 104.5Tf + 10460 = 997Tf - 7980 67598 - 104.5Tf = 997Tf - 7979 75577 = 1102Tf11021102Tf = 68.6oCTemperature (oC)40200-20-40-60-80-1001201008060140TimeDH = mol x DHfusDH = mol x DHvapHeat = mass x Dt x Cp, liquidHeat = mass x Dt x Cp, gasHeat = mass x Dt x Cp, solidABCD(1000 g = 1 kg)238.4 g= 997Tf - 798048A 322 g sample of lead (specific heat = 0.138 J/goC) is placed into 264 g of water at 25oC.If the system's final temperature is 46oC, what was the initial temperature of the lead?Calorimetry Problems 2 question #12PbT = ? oCmass = 322 gTi = 25oCmass = 264 gLOSE heat = GAIN heat-- [(Cp,Pb) (mass) (DT)] = (Cp,H2O) (mass) (DT)- [(0.138 J/goC) (322 g) (46oC - Ti)] = (4.184 J/goC) (264 g) (46oC- 25oC)] Drop Units: - [(44.44) (46oC - Ti)] = (1104.6) (21oC)] - 2044 + 44.44 Ti = 2319744.44 Ti = 25241Ti = 568oCPbTf = 46oC49A sample of ice at 12oC is placed into 68 g of water at 85oC. If the final temperature of the system is 24oC, what was the mass of the ice?Calorimetry Problems 2 question #13H2OT = -12oCmass = ? gTi = 85oCmass = 68 gGAIN heat = - LOSE heat [ qA + qB + qC ] = - [(Cp,H2O) (mass) (DT)]458.2 m = - 17339 m = 37.8 giceTf = 24oCqA = [(Cp,H2O) (mass) (DT)] qC = [(Cp,H2O) (mass) (DT)] qB = (Cf,H2O) (mass)qA = [(2.077 J/goC) (mass) (12oC)] qB = (333 J/g) (mass)qC = [(4.184 J/goC) (mass) (24oC)] [ qA + qB + qC ] = - [(4.184 J/goC) (68 g) (-61oC)]24.9 m333 m100.3 m458.2 mqTotal = qA + qB + qC

458.2458.250Endothermic ReactionEnergy + Reactants Products+DH EndothermicReaction progressEnergyReactantsProductsActivation Energy51Calorimetry Problems 1Keys

Calorimetry 1Calorimetry 1http://www.unit5.org/chemistry/Matter.html52Calorimetry Problems 2Keys

Calorimetry 2Specific Heat ValuesCalorimetry 2Specific Heat Valueshttp://www.unit5.org/chemistry/Matter.html53Heat Energy ProblemsKeys

a b c

Heat Problems (key) Heat Energy of Water Problems (Calorimetry) Specific Heat Problems Heat Energy ProblemsHeat Energy Problems Heat Problems (key) Heat Energy of Water Problems (Calorimetry) Specific Heat Problems http://www.unit5.org/chemistry/Matter.html54Enthalpy DiagramH2O(g)H2O(l)

H2(g) + O2(g)44 kJExothermic+44 kJEndothermicDH = +242 kJEndothermic-242 kJExothermic-286 kJEndothermicDH = -286 kJExothermicEnergyH2(g) + 1/2O2(g) H2O(g) + 242 kJ DH = -242 kJKotz, Purcell, Chemistry & Chemical Reactivity 1991, page 21155Hesss LawCalculate the enthalpy of formation of carbon dioxide from its elements.C(g) + 2O(g) CO2(g)Use the following data:2O(g) O2(g)DH = - 250 kJC(s) C(g)DH = +720 kJCO2(g) C(s) + O2(g)DH = +390 kJSmith, Smoot, Himes, pg 1412O(g) O2(g)DH = - 250 kJC(g) + 2O(g) CO2(g)DH = -1360 kJ

C(g) C(s)DH = - 720 kJC(s) + O2(g) CO2(g) DH = - 390 kJ56In football, as in Hess's law, only the initial and final conditions matter.A team that gains 10 yards on a pass play but has a five-yard penalty,has the same net gain as the team that gained only 5 yards.initial positionof ballfinal positionof ball10 yard pass5 yard penalty5 yard net gain

57pg 353 Holt Chemistry Myers, Oldham, Tocci , 2004 (Holt Rinehart)