2. Calorimetery

3. Do Now #23… What is the difference between the reactant and products in terms of energy levels for a exothermic reaction?

4. Objective… I will be able to explain the energy flow from one component in a system to another component of a system when the energy flow is known in 4 out of 5 attempts.

5. Vocabulary… Thermal Energy: Bond Energy: System (in terms of chemistry): Change of phase (chemistry related):

6. Phases of systems Solid: Liquid: Gas: Plasma:

7. 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)

8. 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

9. 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

10. 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

11. Do Now #24… Explain the difference between the energy transfer from a system of boiling water compared to the system of a glass of water with ice.

12. Objective… Be able to explain the flow of thermal energy from one system to the next and explain what will occur to the substance (system) receiving the thermal energy.

13. ` 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

14. 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

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

16. A Bomb Calorimeter

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

18. 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 

19. 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 

20. 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 –water has a very high heat capacity Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

21. 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 www.nisd.net/communicationsarts/pages/chem

22. 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 www.nisd.net/communicationsarts/pages/chem

23. Do Now #25… What is the difference between heat of vaporization verses heat of fusion.

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

25. 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”.

26. Learning Check… A ______ occurs when ice turn into water. ________ of _______ is the amount of heat needed to melt 1 gram of substance. Heating ______ tells us when a substance will change phases. ________ is the amount of energy needed to raise 1 gram of substance 1 degree C.

27. Learning Check… A Phase Change occurs when ice turn into water. Heat of Fusion is the amount of heat needed to melt 1 gram of substance. Heating curve tells us when a substance will change phases. Heat Capacity is the amount of energy needed to raise 1 gram of substance 1 degree C.

28. 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) 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  T x C p, liquid C p = 4.184 J/g o C Heat = m x  T x C p, solid C p (ice) = 2.077 J/g o C Heat = m x  T x C p, gas C p (steam) = 1.87 J/g o C

29. Calculating Energy Changes - Heating Curve for Water Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 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

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

31. 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

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

33. Do Now #26… Write down the equation for Thermal Energy and explain what each part is and means.

34. Objective… I will be able to calculate the amount of energy that a system loses or gains in Joules based upon the change of temperature of that system in 3 out of 4 examples.

35. Read the following In thermodynamics, thermal energy refers to the internal energy present in a system due to its temperature.[1] The average translational kinetic energy possessed by free particles in a system of free particles in thermodynamic equilibrium (as measured in the frame of reference of the center of mass of that system) may also be referred to as the thermal energy per particle.

36. [2] In thermodynamics it is often most convenient and precise to think of heat as the transfer of energy, just as work is also a transfer of energy. Heat and work therefore depend on the path of transfer and are not state functions, whereas internal energy is a state function.

37. 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) ?

38. 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) ? ?

39. 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

40. 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.

41. Specific Heat Water and silver do not transfer heat equally well. Water has a specific heat C p = 4.184 J/g o C Silver has a specific heat C p = 0.235 J/g o C What does that mean? It requires 4.184 Joules of energy to heat 1 gram of water 1 o C and only 0.235 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!

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

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

44. 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 298.15 K Table of Specific Heats

45. 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.

46. 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

47. Do Now #27… What is the difference between molar heat of vaporization and heat of vaporization?What is the difference between molar heat of vaporization and heat of vaporization? What is the difference between molar heat of fusion and heat of fusion?What is the difference between molar heat of fusion and heat of fusion?

48. 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 of fusion

49. 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.

50. “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

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

52. 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 = 4.184 Joules 1 BTU (British Thermal Unit) – amount of heat needed to raise 1 pound of water 1 o F.

53. C p (ice) = 2.077 J/g o C It takes 2.077 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) = 20.77 Joules X Joules to raise 10 gram ice 10 o C. (10 o C)(10 g)(2.077 J/g o C) = 207.7 Joules Heat = (specific heat) (mass) (change in temperature) q = Cp. m.  T Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 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

54. Heat = (specific heat) (mass) (change in temperature) q = Cp. m.  T GivenT i = -30 o C T f = -20 o C q = 207.7 Joules Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 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

55. 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) - [(0.4495 J/g o C) (X g) (42 o C - 500 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) 205.9 X = 22091 X = 107.3 g Fe

56. 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 0.129 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 - 785 o C)] = (4.184 J/g o C) (323 g) (T f - 15 o C)] 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

57. 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 + 26208 = 246.8 T f - 3208 29416 = 610.8 T f T f = 48.2 o C T = 72 o C mass = 87 g

58. 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)] = 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  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

59. 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) (  T)] + (-C v, H 2 O ) (mass) + (C p, H 2 O ) (mass) (  T) = [(C p, H 2 O ) (mass) (  T)] - [ - 748 + -56400 + 104.5T f - 10460] = 997T f - 7980 - [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 - 116 o C)] q A = - 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) (  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 = 997T f - 7980 - [q A + q B + q C ] = q D 748 + 56400 - 104.5T f + 10460 = 997T f - 7980 67598 - 104.5T f = 997T f - 7979 75577 = 1102T f 1102 T f = 68.6 o C Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 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) 238.4 g

60. 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) (  T) + (-C v, H 2 O ) (mass) + (C p, H 2 O ) (mass) (  T)] = (C p, H 2 O ) (mass) (  T) - [ - 748 + -56400 + 104.5T f - 10460] = 997T f - 7980 - [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 - 116 o C) + - [ - 748 J (C v, H 2 O ) (mass) + (-2256 J/g) (25 g) + + - 56400 J (C p, H 2 O ) (mass) (  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 T f = 68.6 o C Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 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) 238.4 g = 997T f - 7980

61. A 322 g sample of lead (specific heat = 0.138 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)] - 2044 + 44.44 T i = 23197 44.44 T i = 25241 T i = 568 o C Pb T f = 46 o C

62. 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)] 458.2 m = - 17339 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 100.3 m 458.2 m q Total = q A + q B + q C 458.2

63. Endothermic Reaction Energy + Reactants  Products +  H Endothermic Reaction progress Energy Reactants Products Activation Energy

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

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

66. 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 http://www.unit5.org/chemistry/Matter.html

67. 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) + 242 kJ  H = -242 kJ Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 211