2. Unit 3: Thermochemistry Chemistry 3202 1 May 11

3. Unit Outline Temperature and Kinetic Energy Heat/Enthalpy Calculation  Temperature changes (q = mc∆T)  Phase changes (q = n∆H)  Heating and Cooling Curves  Calorimetry (q = C∆T & above formulas) 2 May 11

4. Unit Outline Chemical Reactions  PE Diagrams  Thermochemical Equations  Hess’s Law  Bond Energy STSE: What Fuels You? 3 May 11

5. Temperature and Kinetic Energy Thermochemistry is the study of energy changes in chemical and physical changes eg. dissolving burning phase changes 4 May 11

6. Temperature, T, measures the average kinetic energy of particles in a substance - a change in temperature means particles are moving at different speeds - measured in either Celsius degrees or degrees Kelvin Kelvin = Celsius + 273.15 5 May 11

7. The Celsius scale is based on the freezing and boiling point of water The Kelvin scale is based on absolute zero - the temperature at which particles in a substance have zero kinetic energy. 6 May 11

8. p. 628 7 May 11

9. K 50.15450.15 °C 48-200 8 May 11

10. # of particles 500 K 300 K Kinetic Energy 9 May 11

11. Heat/Enthalpy Calculations system - the part of the universe being studied and observed surroundings - everything else in the universe open system - a system that can exchange matter and energy with the surroundings eg. an open beaker of water a candle burning closed system - allows energy transfer but is closed to the flow of matter. 10 May 11

12. isolated system – a system completely closed to the flow of matter and energy heat - refers to the transfer of kinetic energy from a system of higher temperature to a system of lower temperature. - the symbol for heat is q WorkSheet: Thermochemistry #1 11 May 11

13. Part A: Thought Lab (p. 631) 12 May 12

14. Part B: Thought Lab (p. 631) 13 May 12

15. specific heat capacity – the energy, in Joules (J), needed to change the temperature of one gram (g) of a substance by one degree Celsius (°C). The symbol for specific heat capacity is a lowercase c Heat/Enthalpy Calculations 14 May 12

16. A substance with a large value of c can absorb or release more energy than a substance with a small value of c. ie. For two substances, the substance with the larger c will undergo a smaller temperature change with the same loss or gain of heat. 15 May 12

17. FORMULA q = mc∆T q = heat (J) m = mass (g) c = specific heat capacity ∆T = temperature change = T 2 – T 1 = T f – T i 16 May 12

18. eg. How much heat is needed to raise the temperature of 500.0 g of water from 20.0 °C to 45.0 °C? Solve q = m c ∆T for c, m, ∆T, T 2 & T 1 p. 634 #’s 1 – 4 p. 636 #’s 5 – 8 WorkSheet: Thermochemistry #2 17 May 12

19. heat capacity - the quantity of energy, in Joules (J), needed to change the temperature of a substance by one degree Celsius (°C) The symbol for heat capacity is uppercase C The unit is J/ °C or kJ/ °C 18 May 13

20. FORMULA C = mc q = C ∆T C = heat capacity c = specific heat capacity m = mass ∆T = T 2 – T 1 Your Turn p.637 #’s 11-14 WorkSheet: Thermochemistry #3 19 May 13

21. Enthalpy Changes enthalpy change - the difference between the potential energy of the reactants and the products during a physical or chemical change AKA: Heat of Reaction or ∆H 20 May 18

22. Reaction Progress PE Reactants Products ∆H Endothermic Reaction 21 May 18

23. Reaction Progress PE Reactants Products ∆H Enthalpy Endothermic Reaction 22 May 18

24. ∆H is + Enthalpy Reactants Products Endothermic 23 May 18

25. Enthalpy products ∆H is - Exothermic reactants 24 May 18

26. Enthalpy Changes in Reactions All chemical reactions require bond breaking in reactants followed by bond making to form products Bond breaking requires energy (endothermic) while bond formation releases energy (exothermic) see p. 639 25 May 18

27. 26 May 18

28. Enthalpy Changes in Reactions endothermic reaction - the energy required to break bonds is greater than the energy released when bonds form. ie. energy is absorbed exothermic reaction - the energy required to break bonds is less than the energy released when bonds form. ie. energy is produced 27 May 18

29. Enthalpy Changes in Reactions ∆H can represent the enthalpy change for a number of processes 1. Chemical reactions ∆H rxn – enthalpy of reaction ∆H comb – enthalpy of combustion (see p. 643) 28 May 18

30. 2. Formation of compounds from elements ∆H o f – standard enthalpy of formation The standard molar enthalpy of formation is the energy released or absorbed when one mole of a compound is formed directly from the elements in their standard states. (see p. 642) eg. C(s) + ½ O 2 (g) → CO(g) ΔH f o = -110.5 kJ/mol 29 May 18

31. Use the equation below to determine the ΔH f o for CH 3 OH(l) 2 C(s) + 4 H 2 (g) + O 2 (g) → 2 CH 3 OH(l) + 477.2 kJ 1 C(s) + 2 H 2 (g) + ½ O 2 (g) → 1 CH 3 OH(l) + 238.6 kJ ∆H = -238.6 kJ/mol 30 May 18

32. Use the equation below to determine the ΔH f o for CaCO 3 (s) 2 CaCO 3 (s) + 2413.8kJ → 2 Ca(s) + 2 C(s) + 3 O 2 (g) 2 Ca(s) + 2 C(s) + 3 O 2 (g) → 2 CaCO 3 (s) + 2413.8kJ 1 Ca(s) + 1 C(s) + 1.5 O 2 (g) → 1 CaCO 3 (s) + 1206.9 kJ ∆H = -1206.9 kJ/mol 31 May 18

33. Use the equation below to determine the ΔH f o for PH 3(g) 4 PH 3 (g) → P 4 (s) + 6 H 2 (g) + 21.6 kJ a) +21.6 kJ/mol b) -21.6 kJ/mol c) +5.4 kJ/mol d) -5.4 kJ/mol 32 May 18

34. 3. Phase Changes (p.647) ∆H vap – enthalpy of vaporization (l → g) ∆H fus – enthalpy of melting (fusion: s → l) ∆H cond – enthalpy of condensation (g → l) ∆H fre – enthalpy of freezing (l → s) eg. H 2 O(l)  H 2 O(g) ΔH vap = Hg(l)  Hg(s) ΔH fre = 33 +40.7 kJ/mol -23.4 kJ/mol May 18

35. 34 4. Solution Formation (p.647, 648) ∆H soln – enthalpy of solution eg. ΔH soln, of ammonium nitrate is +25.7 kJ/mol. NH 4 NO 3 (s) + 25.7 kJ → NH 4 NO 3 (aq) ΔH soln, of calcium chloride is −82.8 kJ/mol. CaCl 2 (s) → CaCl 2 (aq) + 82.8 kJ May 18

36. Three ways to represent an enthalpy change: 1. thermochemical equation - the energy term written into the equation. 2.enthalpy term is written as a separate expression beside the equation. 3.enthalpy diagram. 35 May 18

37. eg. the formation of water from the elements produces 285.8 kJ of energy. 1. H 2(g) + ½ O 2(g) → H 2 O (l) + 285.8 kJ 2. H 2(g) + ½ O 2(g) → H 2 O (l) ∆H f = -285.8 kJ/mol thermochemical equation 36 May 18

38. 3. H 2 O (l) H 2(g) + ½ O 2(g) ∆H f = -285.8 kJ/mol Enthalpy (H) enthalpy diagram examples:pp. 641-643 questionsp. 643 #’s 15-18 WorkSheet: Thermochemistry #4 37 May 18

39. Calculating Enthalpy Changes FORMULA: q = n∆H q = heat (kJ) n = # of moles ∆H = molar enthalpy (kJ/mol) 38 May 24

40. eg. How much heat is released when 50.0 g of CH 4 forms from C and H ? (p. 642) q = nΔH = (3.115 mol)(-74.6 kJ/mol) = -232 kJ 39 May 24

41. eg. How much heat is released when 50.00 g of CH 4 undergoes complete combustion? ( p. 643) q = nΔH = (3.115 mol)(-965.1 kJ/mol) = -3006 kJ 40 May 24

42. eg. How much energy is needed to change 20.0 g of H 2 O (l) at 100 °C to steam at 100 °C ? M water = 18.02 g/molΔH vap = +40.7 kJ/mol q = nΔH = (1.110 mol)(+40.7 kJ/mol) = +45.2 kJ 41 May 24

43. ∆H fre and ∆H cond have the opposite sign of the above values. 42 May 24

44. eg. The molar enthalpy of solution for ammonium nitrate is +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves? q = nΔH = (0.4996 mol)(+25.7 kJ/mol) = +12.8 kJ 43 May 24

45. What mass of ethane, C 2 H 6, must be burned to produce 405 kJ of heat? ΔH = -1250.9 kJ q = - 405 kJ m = ? q = nΔH n = 0.3238 mol m = n x M = (0.3238 mol)(30.08 g/mol) = 9.74 g 44 May 24

46. Complete:p. 645; #’s 19 – 23 pp. 648 – 649; #’s 24 – 29 p. 638 #’ 4 – 8 pp. 649, 650 #’s 3 – 8 p. 657, 658 #’s 9 - 18 45 WorkSheet: Thermochemistry #5 May 24

47. 19. (a) -8.468 kJ (b) -7.165 kJ20. -1.37 x10 3 kJ 21. (a) -2.896 x 10 3 kJ (b) -6.81 x10 4 kJ 21. (c) -1.186 x 10 6 kJ22. -0.230 kJ 23. 3.14 x10 3 g24. 2.74 kJ 25.(a) 33.4 kJ (b) 33.4 kJ 26.(a) absorbed (b) 0.096 kJ 27.(a) NaCl(s) + 3.9 kJ/mol → NaCl(aq) (b) 1.69 kJ (c) cool; heat absorbed from water 28. 819.2 g 29. 3.10 x 10 4 kJ 46

48. Heating and Cooling Curves Demo: Cooling of p-dichlorobenzene Time (s)Temperature (°C)Time (s)Temperature (°C) 47 May 25

49. Cooling curve for p-dichlorobenzene KEPEKE solid freezing liquid Temp. (°C ) 50 80 Time 20 48 May 25

50. Heating curve for p-dichlorobenzene Temp. (°C ) 50 20 80 KE PE Time 49 May 25

51. What did we learn from this demo?? During a phase change temperature remains constant and PE changes Changes in temperature during heating or cooling means the KE of particles is changing 50 May 25

52. p. 651 51 May 25

53. p. 652 52 q = n∆H q = mc∆T May 25

54. p. 656 53 q = n∆H q = mc∆T May 25

55. 54 May 25

56. Heating Curve for H 2 0 (s) to H 2 O (g) A 40.0 g sample of ice at -40 °C is heated until it changes to steam and is heated to 140 °C. 1. Sketch the heating curve for this change. 2. Calculate the total energy required for this transition. 55 May 25

57. Time Temp. (°C ) -40 0 100 140 q = mc∆T q = n∆H q = mc∆T q = n∆H 56 May 25

58. Data: c ice = 2.01 J/g.°C c water = 4.184 J/g.°C c steam = 2.01 J/g.°C ΔH fus = +6.02 kJ/mol ΔH vap = +40.7 kJ/mol 57 May 25

59. warming ice: (from -40 ºC to 0 ºC) q = mc∆T = (40.0)(2.01)(0 - - 40) = 3216 J warming water: (from 0 ºC to 100 ºC) q = mc∆T = (40.0)(4.184)(100 – 0) = 16736 J 58 May 26

60. warming steam: (from 100 ºC to 140 ºC) q = mc∆T = (40.0)(2.01)(140 -100) = 3216 J 59 n = 40.0 g 18.02 g/mol = 2.22 mol moles of water: May 26

61. melting ice: (fusion) q = n∆H = (2.22 mol)(6.02 kJ/mol) = 13.364 kJ boiling water: (vaporization) q = n∆H = (2.22 mol)(40.7 kJ/mol) = 90.354 kJ 60 May 26

62. Total Energy 90.354 kJ 13.364 kJ 3216 J 16736 J 127 kJ 61 May 27

63. Practice p. 655: #’s 30 – 34 pp. 656: #’s 1 - 9 p. 657 #’s 2, 9 p. 658 #’s 10, 16 – 20 30.(b) 3.73 x10 3 kJ 31.(b) 279 kJ 32.(b) -1.84 x10 -3 kJ 33.(b) -19.7 kJ -48.77 kJ 34. -606 kJ WorkSheet: Thermochemistry #6 62 May 27

64. Law of Conservation of Energy (p. 627) The total energy of the universe is constant ∆E universe = 0 Universe = system + surroundings ∆E universe = ∆E system + ∆E surroundings ∆E universe = ∆E system + ∆E surroundings = 0 OR ∆E system = -∆E surroundings OR q system = -q surroundings First Law of Thermodynamics 63 May 30

65. Calorimetry (p. 661) calorimetry - the measurement of heat changes during chemical or physical processes calorimeter - a device used to measure changes in energy 2 types of calorimeters 1. constant pressure or simple calorimeter (coffee-cup calorimeter) 2. constant volume or bomb calorimeter. 64 May 30

66. Simple Calorimeter p.661 65 May 30

67. a simple calorimeter consists of an insulated container, a thermometer, and a known amount of water simple calorimeters are used to measure heat changes associated with heating, cooling, phase changes, solution formation, and chemical reactions that occur in aqueous solution 66 May 30

68. to calculate heat lost or gained by a chemical or physical change we apply the first law of thermodynamics: q system = -q calorimeter Assumptions: - the system is isolated - c (specific heat capacity) for water is not affected by solutes - heat exchange with calorimeter can be ignored 67 May 30

69. eg. A simple calorimeter contains 150.0 g of water. A 5.20 g piece of aluminum alloy at 525 °C is dropped into the calorimeter causing the temperature of the calorimeter water to increase from 19.30°C to 22.68°C. Calculate the specific heat capacity of the alloy. 68 May 30

70. aluminum alloywater m = 5.20 gm = 150.0 g T 1 = 525 ºCT 1 = 19.30 ºC T 2 = ºCT 2 = 22.68 ºC FIND c for Alc = 4.184 J/g.ºC q sys = - q cal mcΔT = - mc ΔT (5.20)(c)(22.68 - 525 ) = -(150.0)(4.184)(22.68 – 19.30) -2612 c = -2121 c = 0.812 J/g.°C 69 May 31 22.68

71. eg. The temperature in a simple calorimeter with a heat capacity of 1.05 kJ/°C changes from 25.0 °C to 23.94 °C when a very cold 12.8 g piece of copper was added to it. Calculate the initial temperature of the copper. (c for Cu = 0.385 J/g.°C) 70 May 31

72. copper m = 12.8 g T 2 = ºC c = 0.385 J/g.°C FIND T 1 for Cu q sys = - q cal mc  T = - C  T (12.8)(0.385)(23.94 – T 1 ) = -(1050)(23.94 – 25.0) 4.928 (23.94 – T 1 ) = 1113 23.94 – T 1 = 1113/4.928 23.94 – T 1 = 225.9 T 1 = -202 ºC calorimeter C = 1.05 kJ/°C T 1 = 25.00 ºC T 2 = 23.94 ºC 71 May 31 23.94

73. Homework p. 664, 665 #’s 1b), 2b), 3 & 4 p. 667, #’s 5 - 7 72 May 31

74. p. 665 # 4.b) (60.4)(0.444)(T 2 – 98.0) = -(125.2)(4.184)(T 2 – 22.3) 26.818(T 2 – 98.0) = -523.84(T 2 – 22.3) 26.818T 2 - 2628.2 = -523.84T 2 + 11681 550.66T 2 = 14309.2 T 2 = 26.0 °C 73

75. 6. System (Mg) m = 0.50 g = 0.02057 mol Find ΔH 74 Calorimeter v = 100 ml so m = 100 g c = 4.184 T 2 = 40.7 T 1 = 20.4 7. System ΔH = -53.4 kJ/mol n = CV = (0.0550L)(1.30 mol/L) = 0.0715 mol Calorimeter v = 110 ml so m = 110 g c = 4.184 T 1 = 21.4 Find T 2 q Mg = -q cal nΔH = -mcΔT

76. Bomb Calorimeter 75 June 1

77. Bomb Calorimeter used to accurately measure enthalpy changes in combustion reactions the inner metal chamber or bomb contains the sample and pure oxygen an electric coil ignites the sample temperature changes in the water surrounding the inner “bomb” are used to calculate ΔH 76

78. to accurately measure ΔH you need to know the heat capacity (kJ/°C) of the calorimeter. must account for all parts of the calorimeter that absorb heat C total = C water + C thermom. + C stirrer + C container NOTE: C is provided for all bomb calorimetry calculations 77

79. eg.A technician burned 11.0 g of octane in a steel bomb calorimeter. The heat capacity of the calorimeter was calibrated at 28.0 kJ/°C. During the experiment, the temperature of the calorimeter rose from 20.0 °C to 39.6 °C. What is the enthalpy of combustion for octane? 78

80. system (octane)calorimeter m = 11.0 g T 2 = 39.6 ºC T 1 = 20.0 ºC q sys = - q cal n ΔH = -CΔT (0.09627) ΔH = - (28.0)(39.6 – 20.0) ΔH = -5700 kJ/mol 79 n = 11.0 g 114.26 g/mol = 0.09627 mol C = 28.0 kJ/ºC Find ΔH comb May 31

81. eg. 1.26 g of benzoic acid, C 6 H 5 COOH (s), is burned in a bomb calorimeter. The temperature of the calorimeter and contents increases from 23.62 °C to 27.14 °C. Calculate the heat capacity of the calorimeter. (∆H comb = -3225 kJ/mol) 80 benzoic acidcalorimeter T 1 = 23.62 ºC T 2 = 27.14 ºC Find C m = 1.26 g ΔH comb = -3225 kJ/mol May 31

82. Homework p. 675 #’s 8 – 10 WorkSheet: Thermochemistry #7 81 q sys = - q cal n ΔH = -CΔT (0.01032) ΔH = - (C)(27.14 – 23.62) C = 9.45 kJ/ ºC n = 1.26 g 122.13 g/mol = 0.01032 mol May 31

83. Hess’s Law of Heat Summation the enthalpy change (∆H) of a physical or chemical process depends only on the beginning conditions (reactants) and the end conditions (products) ∆H is independent of the pathway and/or the number of steps in the process ∆H is the sum of the enthalpy changes of all the steps in the process 82 June 2

84. eg. production of carbon dioxide Pathway #1: 2-step mechanism C (s) + ½ O 2(g) → CO (g) ∆H = -110.5 kJ CO (g) + ½ O 2(g) → CO 2(g) ∆H = -283.0 kJ C (s) + O 2(g) → CO 2(g) ∆H = -393.5 kJ 83 June 2

85. eg. production of carbon dioxide Pathway #2: formation from the elements C (s) + O 2(g) → CO 2(g) ∆H = -393.5 kJ 84 June 2

86. Using Hess’s Law We can manipulate equations with known ΔH to determine an unknown enthalpy change. NOTE: Reversing an equation changes the sign of ΔH. If we multiply the coefficients we must also multiply the ΔH value. 85 June 2

87. eg. Determine the ΔH value for: H 2 O (g) + C (s) → CO (g) + H 2(g) using the equations below. C (s) + ½ O 2(g) → CO (g) ΔH = -110.5 kJ H 2(g) + ½ O 2(g) → H 2 O (g) ΔH = -241.8 kJ 86 reverse ? multiply ? June 2

88. eg. Determine the ΔH value for: 4 C (s) + 5 H 2(g) → C 4 H 10(g) using the equations below. ΔH (kJ) C 4 H 10(g) + 6½ O 2(g) → 4 CO 2(g) + 5 H 2 O (g) -110.5 H 2(g) + ½ O 2(g) → H 2 O (g) -241.8 C (s) + O 2(g) → CO 2(g) -393.5 Switch Multiply by 5 Multiply by 4 87 June 2

89. 4 CO 2(g) + 5 H 2 O (g) → C 4 H 10(g) + 6½ O 2(g) +110.5 5( H 2(g) + ½ O 2(g) → H 2 O (g) -241.8 ) 4( C (s) + O 2(g) → CO 2(g) -393.5 ) Ans: -2672.5 kJ 4 CO 2(g) + 5 H 2 O (g) → C 4 H 10(g) + 6½ O 2(g) +110.5 5 H 2(g) + 2½ O 2(g) → 5 H 2 O (g) -1209.0 4C (s) + 4 O 2(g) → 4 CO 2(g) -1574.0 88 June 2

90. Practice pg. 681 #’s 11-14 WorkSheet: Thermochemistry #8 89 June 2

91. Review ∆H o f (p. 642, 684, & 848) The standard molar enthalpy of formation is the energy released or absorbed when one mole of a substance is formed directly from the elements in their standard states. ∆H o f = 0 kJ/mol for elements in the standard state The more negative the ∆H o f, the more stable the compound 90 June 3

92. Use the formation equations below to determine the ΔH value for: C 4 H 10(g) + 6½ O 2(g) → 4 CO 2(g) + 5 H 2 O (g) ΔH f (kJ/mol) 4 C (s) + 5 H 2(g) → C 4 H 10(g) -2672.5 H 2(g) + ½ O 2(g) → H 2 O (g) -241.8 C (s) + O 2(g) → CO 2(g) -393.5 Using Hess’s Law and ΔH f 91 June 3

93. Using Hess’s Law and ΔH f ΔH rxn = ∑ΔH f (products) - ∑ΔH f (reactants) eg. Use ΔH f, to calculate the enthalpy of reaction for the combustion of glucose. C 6 H 12 O 6(s) + 6 O 2(g) → 6 CO 2(g) + 6 H 2 O (g) 92 June 3

94. ΔH rxn = ∑ΔH f (products) - ∑ΔH f (reactants) ΔH f CO 2 (g)-393.5 kJ/mol H 2 O(g)-241.8 kJ/mol C 6 H 12 O 6 (s)-1274.5 kJ/mol C 6 H 12 O 6(s) + 6 O 2(g) → 6 CO 2(g) + 6 H 2 O (g) ΔH rxn = [6(-393.5) + 6(-241.8)] – [1(-1274.5)+ 6(0)] = [-2361 + -1450.8] - [-1274.5 + 0] = - 2537.3 kJ 93 June 3

95. Use the molar enthalpy of formation to calculate ΔH for this reaction: Fe 2 O 3(s) + 3 CO (g) → 3 CO 2(g) + 2 Fe (s) ΔH rxn = [3(-393.5) + 2(0) ] – [3(-110.5)+ 1(-824.2)] = [-1180.5 + 0] - [-331.5 + -824.20] = - 24.8 kJ p. 688 #’s 21 & 22 94 −824.2 kJ/mol −110.5 kJ/mol −393.5 kJ/mol June 3

96. Eg. The combustion of phenol is represented by the equation below: C 6 H 5 OH (s) + 7 O 2(g) → 6 CO 2(g) + 3 H 2 O (g) If ΔH comb = -3059 kJ/mol, calculate the heat of formation for phenol. 95 June 7 −393.5 kJ/mol −241.8 kJ/mol ΔH comb = -27.4 kJ/mol

97. Bond Energy Calculations (p. 688) The energy required to break a bond is known as the bond energy. Each type of bond has a specific bond energy (BE). (table p. 847) Bond Energies may be used to estimate the enthalpy of a reaction. 96

98. Bond Energy Calculations (p. 688) ΔH rxn = ∑BE (reactants) - ∑BE (products) eg. Estimate the enthalpy of reaction for the combustion of ethane using BE. 2 C 2 H 6(g) + 7 O 2(g) → 4 CO 2(g) + 6 H 2 O (g) Hint: Drawing the structural formulas for all reactants and products will be useful here. 97

99. C-C = 347 C-H = 338 O=O = 498 C=O = 745 H-O = 460 p. 690 #’s 23,24,& 26 p. 691 #’s 3, 4, 5, & 7 = -3244 kJ 98 [2(347) + 2(6)(338) + 7(498)] - [4(2)(745) + 6(2)(460)] + 7 O = O CC → 4 O=C=O + 6 H-O-H 2 8236 - 11480

100. Energy Comparisons Phase changes involve the least amount of energy with vaporization usually requiring more energy than melting. Chemical changes involve more energy than phase changes but much less than nuclear changes. Nuclear reactions produce the largest ΔH  eg. nuclear power, reactions in the sun 99

101. STSE What fuels you? (Handout) 100

102. aluminum alloywater m = 5.20 gm = 150.0 g T 1 = 525 ºCT 1 = 19.30 ºC T 2 = ºCT 2 = 22.68 ºC FIND c for Alc = 4.184 J/g.ºC q sys = - q cal mc  T = - mc  T (5.20)(c)(22.68 - 525 ) = -(150.0)(4.184)(22.68 – 19.30) -2612 c = -2121 c = 0.812 J/g.°C 101

103. copper m = 12.8 g T 2 = ºC c = 0.385 J/g.°C FIND T 1 for Cu q sys = - q cal mc  T = - C  T (12.8)(0.385)(23.94 – T 1 ) = -(1050)(23.94 – 25.0) 4.928 (23.94 – T 1 ) = 1113 23.94 – T 1 = 1113/4.928 23.94 – T 1 = 225.9 T 1 = -202 ºC calorimeter C = 1.05 kJ/°C T 1 = 25.00 ºC T 2 = 23.94 ºC 102

104. qheatJ or kJ cSpecific heat capacity J/g.ºC CHeat capacitykJ/ ºC or J/ ºC ΔHΔHMolar heat or molar enthalpy kJ/mol 103