Calorimetry Burning of a Match - PowerPoint Presentation

Calorimetry

Burning of a Match Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 293 Energy released to the surrounding as heat Surroundings System (Reactants) D (PE) Potential energy (Products)

Conservation of Energy in a Chemical Reaction Surroundings System Surroundings System Energy Before reaction After reaction In 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 41 Endothermic Reaction Reactant + Energy Product

Conservation of Energy in a Chemical Reaction Surroundings System Surroundings System Energy Before reaction After reaction In 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 41 Exothermic Reaction Reactant Product + Energy

Direction of Heat Flow Surroundings ENDOthermic q sys > 0 EXOthermic q sys < 0 System Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 207 System H 2 O( s ) + heat  H 2 O( l ) melting H 2 O( l )  H 2 O( s ) + heat freezing

Caloric Values Food joules/grams calories/gram Calories/gram Protein 17 000 4000 4 Fat 38 000 9000 9 Carbohydrates 17 000 4000 4 Smoot, Smith, Price, Chemistry A Modern Course , 1990, page 51 1000 calories = 1 Calorie "science" "food" 1calories = 4.184 joules

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 Metal

A Coffee Cup Calorimeter Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 302 Thermometer Styrofoam cover Styrofoam cups Stirrer

Bomb Calorimeter thermometer stirrer full of water ignition wire steel “bomb” sample

1997 Encyclopedia Britanica, Inc. oxygen supply stirrer thermometer magnifying eyepiece air space crucible steel bomb sample ignition coil bucket heater water ignition wires insulating jacket

A Bomb Calorimeter

Causes of Change - Calorimetry Outline Keys http://www.unit5.org/chemistry/Matter.html

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

Heating Curves Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time Melting - PE  Solid - KE  Liquid - KE  Boiling - PE  Gas - KE 

Heating Curves Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time Melting - PE  Solid - KE  Liquid - KE  Boiling - PE  Gas - KE 

Heating Curves Temperature Change change in KE (molecular motion) depends on heat capacity Heat Capacityenergy required to raise the temp of 1 gram of a substance by 1°C “Volcano” clip -water has a very high heat capacity Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Heating Curves Phase Change change in PE (molecular arrangement) temp remains constant Heat 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

Heating Curves Heat of Vaporization ( Hvap )energy required to boil 1 gram of a substance at its b.p.usually larger than Hfus…why? EX: sweating, steam burns, the drinking bird Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Phase Diagrams Show the phases of a substance at different temps and pressures. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Humor A 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” “It’s just a phase you’re going through”.

A  B warm ice B  C melt ice (solid  liquid) C  D warm waterD  E boil water (liquid  gas) E  D condense steam (gas  liquid)E  F superheat steam Heating Curve for Water(Phase Diagram) 140 120 100 80 60 40 20 0 -20 -40 -60 -80-100 Temperature (o C) Heat BP MP A B C D E F Heat = m x C fus C f = 333 J/g Heat = m x C vap C v = 2256 J/g Heat = m x D T x C p, liquid C p = 4.184 J/g o C Heat = m x D T x C p, solid C p (ice) = 2.077 J/g o C Heat = m x D T x C p, gas C p (steam) = 1.87 J/g o C

Calculating Energy Changes - Heating Curve for Water Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time D H = mol x D H fus D H = mol x D H vap Heat = mass x D t x C p, liquid Heat = mass x D t x C p, gas Heat = mass x D t x C p, solid

Equal Masses of Hot and Cold Water Thin metal wall Insulated box Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

Water Molecules in Hot and Cold Water Hot water Cold Water 90 o C 10 oC Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

Water Molecules in the same temperature water Water (50 o C) Water(50 oC)Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

Heat Transfer Al Al m = 20 g T = 40 o C SYSTEM Surroundings m = 20 g T = 20 o C 20 g (40 o C) 20 g (20oC) 30oC Block “A” Block “B” Final Temperature Assume NO heat energy is “lost” to the surroundings from the system. What will be the final temperature of the system ? a) 60 o C b) 30 o C c) 20 o C d) ?

Heat Transfer Al Al m = 20 g T = 40 o C SYSTEM Surroundings m = 10 g T = 20 o C 20 g (40 o C) 20 g (20oC) 30.0oC Block “A” Block “B” Final Temperature Assume NO heat energy is “lost” to the surroundings from the system. 20 g (40oC) 10 g (20oC) 33.3 o C What will be the final temperature of the system ? a) 60 o C b) 30 o C c) 20 o C d) ? ?

Heat Transfer Al Al m = 20 g T = 20 o C SYSTEM Surroundings m = 10 g T = 40 o C 20 g (40 o C) 20 g (20oC) 30.0oC Block “A” Block “B” Final Temperature Assume NO heat energy is “lost” to the surroundings from the system. 20 g (40oC) 10 g (20oC) 33.3 o C 20 g (20 o C) 10 g (40 o C) 26.7 o C

Heat Transfer m = 75 g T = 25 o C SYSTEM Surroundings m = 30 g T = 100 oC 20 g (40o C)20 g (20 oC) 30.0oC Block “A” Block “B” Final Temperature 20 g (40oC) 10 g (20oC) 33.3 oC 20 g (20 o C)10 g (40oC) 26.7 o C Ag H 2 O Real Final Temperature = 26.7 o C Why? We’ve been assuming ALL materials transfer heat equally well.

Specific Heat Water and silver do not transfer heat equally well. Water has a specific heat C p = 4.184 J/goC Silver has a specific heat Cp = 0.235 J/goCWhat 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!

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

c p = Specific Heat q = Heat lost or gained  T = Temperature change OR m = Mass Calculations involving Specific Heat

Substance Specific 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) 0.129 Specific Heats of Some Common Substances at 298.15 K Table of Specific Heats

The 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.

The 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 #2 Molar Heat of Vaporization

Latent Heat – Sample Problem Problem : The molar heat of fusion of water is 6.009 kJ/mol. How much energy is needed to convert 60 grams of ice at 0  C to liquid water at 0C? Mass of ice Molar Mass of water Heat offusion

Heat of Reaction The 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.

“loses” heat Calorimetry m = 75 g T = 25 o C SYSTEM Surroundings m = 30 g T = 100 o C Ag H 2 O T final = 26.7 o C

Calorimetry m = 75 g T = 25 o C SYSTEM Surroundings m = 30 g T = 100 oC Ag H 2 O

1 Calorie = 1000 calories “food” = “science” Candy bar 300 Calories = 300,000 calories English Metric = _______ Joules 1 calorie - amount of heat needed to raise 1 gram of water 1 oC 1 calorie = 4.184 Joules 1 BTU (British Thermal Unit) – amount of heat needed to raise 1 pound of water 1 o F.

C p (ice) = 2.077 J/g oC It takes 2.077 Joules to raise 1 gram ice 1oC. X Joules to raise 10 gram ice 1 oC.(10 g)(2.077 J/g o C) = 20.77 Joules X Joules to raise 10 gram ice 10oC. (10oC)(10 g)(2.077 J/g o C) = 207.7 Joules Heat = (specific heat) (mass) (change in temperature) q = Cp . m . D T Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time D H = mol x D H fus D H = mol x D H vap Heat = mass x D t x C p, liquid Heat = mass x D t x C p, gas Heat = mass x D t x C p, solid

Heat = (specific heat) (mass) (change in temperature) q = Cp . m . DT Given T i = -30 o C Tf = -20oC q = 207.7 Joules Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time D H = mol x D H fus D H = mol x D H vap Heat = mass x D t x C p, liquid Heat = mass x D t x C p, gas Heat = mass x D t x C p, solid

240 g of water (initially at 20 o C) are mixed with an unknown mass of iron (initially at 500 oC). When thermal equilibrium is reached, the system has a temperature of 42oC. Find the mass of the iron. Calorimetry Problems 2 question #5 Fe T = 500 oC mass = ? grams T = 20 o C mass = 240 g LOSE heat = GAIN heat - - [(C p , Fe) (mass) (DT)] = (Cp ,H2O ) (mass) (DT) - [(0.4495 J/goC) (X g) (42oC - 500o C)] = (4.184 J/goC) (240 g) (42oC - 20oC)] Drop Units: - [(0.4495) (X) (-458)] = (4.184) (240 g) (22) 205.9 X = 22091 X = 107.3 g Fe

A 97 g sample of gold at 785 o C is dropped into 323 g of water, which has an initial temperature of 15o C. 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 #8 Au T = 785oC mass = 97 g T = 15 o C mass = 323 g LOSE heat = GAIN heat - - [(C p , Au ) (mass) ( DT)] = (Cp, H2O) (mass) (DT) - [(0.129 J/goC) (97 g) (T f - 785oC)] = (4.184 J/goC) (323 g) (Tf - 15oC)] Drop Units: - [(12.5) (T f - 785 o C)] = (1.35 x 10 3 ) (T f - 15 o C)] -12.5 T f + 9.82 x 10 3 = 1.35 x 10 3 T f - 2.02 x 10 4 3 x 10 4 = 1.36 x 10 3 T f T f = 22.1 o C

If 59 g of water at 13 o C are mixed with 87 g of water at 72 oC, find the final temperature of the system. Calorimetry Problems 2 question #9 T = 13 o C mass = 59 g LOSE heat = GAIN heat - - [(C p,H2O) (mass) ( DT)] = (Cp,H2O ) (mass) (DT) - [(4.184 J/goC) (87 g) (T f - 72oC)] = (4.184 J/goC) (59 g) (Tf - 13oC) Drop Units: - [(364.0) (T f - 72oC)] = (246.8) (T f - 13oC) -364 Tf + 26208 = 246.8 Tf - 3208 29416 = 610.8 Tf Tf = 48.2 o C T = 72 o C mass = 87 g

A 38 g sample of ice at -11 o C is placed into 214 g of water at 56 oC. Find the system's final temperature. Calorimetry Problems 2 question #10 ice T = -11oC mass = 38 g T = 56 o C mass = 214 g LOSE heat = GAIN heat - - [(C p , H 2O(l)) (mass) (DT)] = (Cp, H2O(s)) (mass) ( DT) + (Cf) (mass) + (Cp,H 2O(l)) (mass) (DT) - [(4.184 J/goC)(214 g)(Tf - 56 oC)] = (2.077 J/goC)(38 g)(11oC) + (333 J/g)(38 g) + (4.184 J/goC)(38 g)(Tf - 0 o C) - [(895) (T f - 56 o C)] = 868 + 12654 + (159) (T f )] - 895 T f + 50141 = 868 + 12654 + 159 T f - 895 T f + 50141 = 13522 + 159 T f T f = 34.7 o C 36619 = 1054 T f Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time D H = mol x D H fus D H = mol x D H vap Heat = mass x D t x C p, liquid Heat = mass x D t x C p, gas Heat = mass x D t x C p, solid A B C D warm ice melt ice warm water water cools D B A C

25 g of 116 o C steam are bubbled into 0.2384 kg of water at 8 o C. Find the final temperature of the system. Calorimetry Problems 2 question #11 - [(C p , H 2 O) (mass) (D T)] + (-Cv,H2O) (mass) + (Cp ,H2O) (mass) (DT) = [(C p,H2O) (mass) ( DT)] - [ - 748 + -56400 + 104.5Tf - 10460] = 997Tf - 7980 - [qA + qB + qC] = qD q A = [(Cp,H2O) (mass) ( DT)] qA = [(1.87 J/goC) (25 g) (100o - 116 oC)] qA = - 748 J q B = (C v , H 2 O ) (mass) q B = (-2256 J/g) (25 g) q B = - 56400 J q C = [(C p , H 2 O ) (mass) ( D T)] q C = [(4.184 J/g o C) (25 g) (T f - 100 o C)] q C = 104.5T f - 10460 q D = (4.184 J/g o C) (238.4 g) (T f - 8 o C) q D = 997Tf - 7980 - [q A + qB + qC] = qD748 + 56400 - 104.5Tf + 10460 = 997Tf - 7980 67598 - 104.5Tf = 997Tf - 7979 75577 = 1102Tf 1102 1102 T f = 68.6 o C Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time D H = mol x D H fus D H = 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

25 g of 116 o C steam are bubbled into 0.2384 kg of water at 8 o C. Find the final temperature of the system. Calorimetry Problems 2 question #11 - [(C p , H 2 O) (mass) (D T) + (-Cv,H2O) (mass) + (Cp ,H2O) (mass) (DT)] = (C p,H2O) (mass) ( DT) - [ - 748 + -56400 + 104.5Tf - 10460] = 997Tf - 7980 - [qA + qB + qC] = qD - [(C p,H2O) (mass) (DT) + - [(1.87 J/g oC) (25 g) (100o - 116oC) + - [ - 748 J (C v , H 2 O ) (mass) + (-2256 J/g) (25 g) + + - 56400 J (C p , H 2 O ) (mass) ( D T)] (4.184 J/g o C) (25 g) (T f - 100 o C)] + 104.5T f - 10460 ] = (4.184 J/g o C) (238.4 g) (T f - 8 o C) = 997T f - 7980 748 + 56400 - 104.5T f + 10460 = 997T f - 7980 67598 - 104.5T f = 997T f - 7979 75577 = 1102T f 1102 1102 T f = 68.6 o C Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time D H = mol x D H fus D H = mol x D H vap Heat = 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 - 7980

A 322 g sample of lead (specific heat = 0.138 J/g o C) is placed into 264 g of water at 25 oC. If the system's final temperature is 46oC, what was the initial temperature of the lead? Calorimetry Problems 2 question #12 Pb T = ? oC mass = 322 g T i = 25 oC mass = 264 g LOSE heat = GAIN heat - - [(C p ,Pb) (mass) (DT)] = (Cp, H2O) (mass) ( DT) - [(0.138 J/goC) (322 g) (46oC - Ti)] = (4.184 J/g oC) (264 g) (46oC- 25oC)] Drop Units: - [(44.44) (46 o C - T i )] = (1104.6) (21 o C)] - 2044 + 44.44 T i = 23197 44.44 T i = 25241 T i = 568 o C Pb T f = 46 o C

A sample of ice at –12 o C is placed into 68 g of water at 85 o C. If the final temperature of the system is 24 oC, what was the mass of the ice? Calorimetry Problems 2 question #13 H2O T = -12o C mass = ? g T i = 85 oC mass = 68 g GAIN heat = - LOSE heat [ q A + qB + qC ] = - [(Cp,H 2O) (mass) (D T)] 458.2 m = - 17339 m = 37.8 g ice T f = 24 o C q A = [(C p , H 2 O ) (mass) ( D T)] q C = [(C p , H 2 O ) (mass) ( D T)] q B = (C f , H 2 O ) (mass) q A = [(2.077 J/g o C) (mass) (12 o C)] q B = (333 J/g) (mass) q C = [(4.184 J/g o C) (mass) (24oC)] [ q A + qB + q C ] = - [(4.184 J/goC) (68 g) (-61oC)]24.9 m 333 m100.3 m 458.2 mqTotal = qA + q B + qC 458.2 458.2

Endothermic Reaction Energy + Reactants  Products + D H Endothermic Reaction progress Energy Reactants Products Activation Energy

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

Calorimetry Problems 2 Keys Calorimetry 2   Specific Heat Values Calorimetry 2   Specific Heat Values http://www.unit5.org/chemistry/Matter.html

Heat Energy Problems Keys a b c   Heat Problems ( key )          Heat Energy of Water Problems (Calorimetry) Specific Heat Problems Heat Energy Problems Heat Energy Problems   Heat Problems ( key )          Heat Energy of Water Problems (Calorimetry) Specific Heat Problems http://www.unit5.org/chemistry/Matter.html

Enthalpy Diagram H 2 O(g) H 2 O(l) H 2 (g) + ½ O 2 (g) 44 kJ Exothermic +44 kJ Endothermic D H = +242 kJ Endothermic - 242 kJ Exothermic - 286 kJ Endothermic D H = -286 kJ Exothermic Energy H 2 (g) + 1/2O 2 (g)  H 2 O(g) + 242 kJ D H = -242 kJ Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 211

Hess’s Law Calculate the enthalpy of formation of carbon dioxide from its elements. C(g) + 2O(g)  CO 2(g) Use the following data: 2O(g)  O2(g) DH = - 250 kJC(s)  C(g) D H = +720 kJCO2(g)  C(s) + O 2(g) DH = +390 kJ Smith, Smoot, Himes, pg 141 2O(g)  O 2(g) DH = - 250 kJ C(g) + 2O(g)  CO2(g) DH = -1360 kJ C(g)  C(s) DH = - 720 kJ C(s) + O2(g)  CO2(g) D H = - 390 kJ

In 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 position of ball final position of ball 10 yard pass 5 yard penalty 5 yard net gain