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Calorimetry Burning of a Match Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 293 Energy released to the surrounding as heat SurroundingsSystem.

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Presentation on theme: "Calorimetry Burning of a Match Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 293 Energy released to the surrounding as heat SurroundingsSystem."— Presentation transcript:

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2 Calorimetry

3 Burning of a Match Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 293 Energy released to the surrounding as heat SurroundingsSystem (Reactants)  (PE) Potential energy (Products)

4 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

5 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

6 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

7 Caloric Values Food joules/grams calories/gram Calories/gram Protein Fat Carbohydrates Smoot, Smith, Price, Chemistry A Modern Course, 1990, page calories = 1 Calorie "science" "food" 1calories = joules

8 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

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

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

11 1997 Encyclopedia Britanica, Inc. oxygen supply stirrer thermometer magnifying eyepiece air space crucible steel bombsampleignition coil bucket heater water ignition wires insulating jacket

12 A Bomb Calorimeter

13 Causes of Change - Calorimetry Outline KeysKeys

14 Heating Curves Melting - PE  Solid - KE  Liquid - KE  Boiling - PE  Gas - KE  Courtesy Christy Johannesson

15 Heating Curves Temperature ( o C) Time Melting - PE  Solid - KE  Liquid - KE  Boiling - PE  Gas - KE 

16 Heating Curves Temperature ( o C) Time Melting - PE  Solid - KE  Liquid - KE  Boiling - PE  Gas - KE 

17 Heating Curves Temperature Change –change in KE (molecular motion) –depends on heat capacity Heat Capacity –energy 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

18 Heating Curves Phase Change –change in PE (molecular arrangement) –temp remains constant Heat of Fusion (  H fus ) –energy required to melt 1 gram of a substance at its m.p. Courtesy Christy Johannesson

19 Heating Curves Heat of Vaporization (  H vap ) –energy required to boil 1 gram of a substance at its b.p. –usually larger than  H fus …why? EX: sweating, steam burns, the drinking bird Courtesy Christy Johannesson

20 Phase Diagrams Show the phases of a substance at different temps and pressures. Courtesy Christy Johannesson

21 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”.

22 A  B warm ice B  C melt ice (solid  liquid) C  D warm water D  E boil water (liquid  gas) E  D condense steam (gas  liquid) E  F superheat steam Heating Curve for Water (Phase Diagram) 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  T x C p, liquid C p = J/g o C Heat = m x  T x C p, solid C p (ice) = J/g o C Heat = m x  T x C p, gas C p (steam) = 1.87 J/g o C

23 Calculating Energy Changes - Heating Curve for Water Temperature ( o C) Time  H = mol x  H fus  H = mol x  H vap Heat = mass x  t x C p, liquid Heat = mass x  t x C p, gas Heat = mass x  t x C p, solid

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

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

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

27 Heat Transfer Al m = 20 g T = 40 o C SYSTEM Surroundings m = 20 g T = 20 o C 20 g (40 o C) 20 g (20 o C)30 o C 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) ?

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

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

30 Heat Transfer m = 75 g T = 25 o C SYSTEM Surroundings m = 30 g T = 100 o C 20 g (40 o C) 20 g (20 o C)30.0 o C Block “A” Block “B” Final Temperature 20 g (40 o C) 10 g (20 o C)33.3 o C 20 g (20 o C) 10 g (40 o C)26.7 o C Ag H2OH2O Real Final Temperature = 26.7 o C Why? We’ve been assuming ALL materials transfer heat equally well.

31 Specific Heat Water and silver do not transfer heat equally well. Water has a specific heat C p = J/g o C Silver has a specific heat C p = J/g o C What does that mean? It requires Joules of energy to heat 1 gram of water 1 o C and only Joules of energy to heat 1 gram of silver 1 o C. 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!

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

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

34 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)0.129 Specific Heats of Some Common Substances at K Table of Specific Heats

35 The energy that must be absorbed in order to convert one mole of solid to liquid at its melting point. Latent Heat of Phase Change Molar Heat of Fusion The energy that must be removed in order to convert one mole of liquid to solid at its freezing point.

36 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

37 Latent Heat – Sample Problem Problem: The molar heat of fusion of water is 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 of fusion

38 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.

39 “loses” heat Calorimetry m = 75 g T = 25 o C SYSTEM Surroundings m = 30 g T = 100 o C Ag H2OH2O T final = 26.7 o C

40 Calorimetry m = 75 g T = 25 o C SYSTEM Surroundings m = 30 g T = 100 o C Ag H2OH2O

41 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 o C 1 calorie = Joules 1 BTU (British Thermal Unit) – amount of heat needed to raise 1 pound of water 1 o F.

42 C p (ice) = J/g o C It takes Joules to raise 1 gram ice 1 o C. X Joules to raise 10 gram ice 1 o C. (10 g)(2.077 J/g o C) = Joules X Joules to raise 10 gram ice 10 o C. (10 o C)(10 g)(2.077 J/g o C) = Joules Heat = (specific heat) (mass) (change in temperature) q = Cp. m.  T Temperature ( o C) Time  H = mol x  H fus  H = mol x  H vap Heat = mass x  t x C p, liquid Heat = mass x  t x C p, gas Heat = mass x  t x C p, solid

43 Heat = (specific heat) (mass) (change in temperature) q = Cp. m.  T GivenT i = -30 o C T f = -20 o C q = Joules Temperature ( o C) Time  H = mol x  H fus  H = mol x  H vap Heat = mass x  t x C p, liquid Heat = mass x  t x C p, gas Heat = mass x  t x C p, solid

44 240 g of water (initially at 20 o C) are mixed with an unknown mass of iron (initially at 500 o C). When thermal equilibrium is reached, the system has a temperature of 42 o C. Find the mass of the iron. Calorimetry Problems 2 question #5 Fe T = 500 o C mass = ? grams T = 20 o C mass = 240 g LOSE heat = GAIN heat - - [(C p, Fe ) (mass) (  T)] = (C p, H 2 O ) (mass) (  T) - [( J/g o C) (X g) (42 o C o C)] = (4.184 J/g o C) (240 g) (42 o C - 20 o C)] Drop Units: - [(0.4495) (X) (-458)] = (4.184) (240 g) (22) X = X = g Fe

45 A 97 g sample of gold at 785 o C is dropped into 323 g of water, which has an initial temperature of 15 o C. If gold has a specific heat of J/g o C, 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 = 785 o C mass = 97 g T = 15 o C mass = 323 g LOSE heat = GAIN heat - - [(C p, Au ) (mass) (  T)] = (C p, H 2 O ) (mass) (  T) - [(0.129 J/g o C) (97 g) (T f o C)] = (4.184 J/g o C) (323 g) (T f - 15 o C)] Drop Units: - [(12.5) (T f o C)] = (1.35 x 10 3 ) (T f - 15 o C)] T f x 10 3 = 1.35 x 10 3 T f x x 10 4 = 1.36 x 10 3 T f T f = 22.1 o C

46 If 59 g of water at 13 o C are mixed with 87 g of water at 72 o C, 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, H 2 O ) (mass) (  T)] = (C p, H 2 O ) (mass) (  T) - [(4.184 J/g o C) (87 g) (T f - 72 o C)] = (4.184 J/g o C) (59 g) (T f - 13 o C) Drop Units: - [(364.0) (T f - 72 o C)] = (246.8) (T f - 13 o C) -364 T f = T f = T f T f = 48.2 o C T = 72 o C mass = 87 g

47 A 38 g sample of ice at -11 o C is placed into 214 g of water at 56 o C. Find the system's final temperature. Calorimetry Problems 2 question #10 ice T = -11 o C mass = 38 g T = 56 o C mass = 214 g LOSE heat = GAIN heat - - [(C p, H 2 O(l) ) (mass) (  T)] = (C p, H 2 O(s) ) (mass) (  T) + (C f ) (mass) + (C p, H 2 O(l) ) (mass) (  T) - [(4.184 J/g o C)(214 g)(T f - 56 o C)] = (2.077 J/g o C)(38 g)(11 o C) + (333 J/g)(38 g) + (4.184 J/g o C)(38 g)(T f - 0 o C) - [(895) (T f - 56 o C)] = (159) (T f )] T f = T f T f = T f T f = 34.7 o C = 1054 T f Temperature ( o C) Time  H = mol x  H fus  H = mol x  H vap Heat = mass x  t x C p, liquid Heat = mass x  t x C p, gas Heat = mass x  t x C p, solid A B C D warm ice melt ice warm water water cools D B AC

48 25 g of 116 o C steam are bubbled into 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) (  T)] + (-C v, H 2 O ) (mass) + (C p, H 2 O ) (mass) (  T) = [(C p, H 2 O ) (mass) (  T)] - [ T f ] = 997T f [q A + q B + q C ] = q D q A = [(C p, H 2 O ) (mass) (  T)] q A = [(1.87 J/g o C) (25 g) (100 o o C)] q A = J q B = (C v, H 2 O ) (mass) q B = (-2256 J/g) (25 g) q B = J q C = [(C p, H 2 O ) (mass) (  T)] q C = [(4.184 J/g o C) (25 g) (T f o C)] q C = 104.5T f q D = (4.184 J/g o C) (238.4 g) (T f - 8 o C) q D = 997T f [q A + q B + q C ] = q D T f = 997T f T f = 997T f = 1102T f 1102 T f = 68.6 o C Temperature ( o C) Time  H = mol x  H fus  H = mol x  H vap Heat = mass x  t x C p, liquid Heat = mass x  t x C p, gas Heat = mass x  t x C p, solid A B C D (1000 g = 1 kg) g

49 25 g of 116 o C steam are bubbled into 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) (  T) + (-C v, H 2 O ) (mass) + (C p, H 2 O ) (mass) (  T)] = (C p, H 2 O ) (mass) (  T) - [ T f ] = 997T f [q A + q B + q C ] = q D - [(C p, H 2 O ) (mass) (  T) + - [(1.87 J/g o C) (25 g) (100 o o C) + - [ J (C v, H 2 O ) (mass) + (-2256 J/g) (25 g) J (C p, H 2 O ) (mass) (  T)] (4.184 J/g o C) (25 g) (T f o C)] T f ] = (4.184 J/g o C) (238.4 g) (T f - 8 o C) = 997T f T f = 997T f T f = 997T f = 1102T f 1102 T f = 68.6 o C Temperature ( o C) Time  H = mol x  H fus  H = mol x  H vap Heat = mass x  t x C p, liquid Heat = mass x  t x C p, gas Heat = mass x  t x C p, solid A B C D (1000 g = 1 kg) g = 997T f

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

51 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 o C, what was the mass of the ice? Calorimetry Problems 2 question #13 H2OH2O T = -12 o C mass = ? g T i = 85 o C mass = 68 g GAIN heat = - LOSE heat [ q A + q B + q C ] = - [(C p, H 2 O ) (mass) (  T)] m = m = 37.8 g ice T f = 24 o C q A = [(C p, H 2 O ) (mass) (  T)] q C = [(C p, H 2 O ) (mass) (  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) (24 o C)] [ q A + q B + q C ] = - [(4.184 J/g o C) (68 g) (-61 o C)] 24.9 m 333 m m m q Total = q A + q B + q C 458.2

52 Endothermic Reaction Energy + Reactants  Products +  H Endothermic Reaction progress Energy Reactants Products Activation Energy

53 Calorimetry Problems 1 Keys Calorimetry 1

54 Calorimetry Problems 2 Keys Calorimetry 2 Specific Heat Values Calorimetry 2 Specific Heat Values

55 Heat Energy Problems Keys a b c Keys a b c Heat Problems (key) Heat Problemskey Heat Energy of Water Problems (Calorimetry) Specific Heat Problems Heat Energy Problems Heat Problems (key) Heat Problems key Heat Energy of Water Problems (Calorimetry) Specific Heat Problems

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

57 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)  O 2 (g)  H = kJ C(s)  C(g)  H = +720 kJ CO 2 (g)  C(s) + O 2 (g)  H = +390 kJ Smith, Smoot, Himes, pg 141 2O(g)  O 2 (g)  H = kJ C(g) + 2O(g)  CO 2 (g)  H = kJ C(g)  C(s)  H = kJ C(s) + O 2 (g)  CO 2 (g)  H = kJ

58 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 net 5 yard net gain


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