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Thermochemical Calculations CA Standards Students know energy is released when a material condenses or freezes and is absorbed when a material evaporates or melts. Students know how to solve problems involving heat flow and temperature changes, using known values of specific heat and latent heat of phase change. Units for Measuring Heat The Joule is the SI system unit for measuring heat: 1 kg m 1 Joule1 newton m eter 2 s 2 The calorie is the heat required to raise the temperature of 1 gram of water by 1 Celsius degree 1calorie 4.18 Joules Specific Heat The amount of heat required to raise the temperature of one gram of substance by one degree Celsius. 5 4 6 5 7 4 3 8 3 2 9 2 1 11 6 7 8 9 1 10 Calculations Involving Specific Heat Q m T c p OR Q cp m T cp = Specific Heat Q = Heat lost or gained T = Temperature change m = Mass Specific Heat The amount of heat required to raise the temperature of one gram of substance by one degree Celsius. Substance Specific Heat (J/g·K) Water (liquid) 4.18 Ethanol (liquid) 2.44 Water (solid) 2.06 Water (vapor) 1.87 Aluminum (solid) 0.897 Carbon (graphite,solid) 0.709 Iron (solid) 0.449 Copper (solid) 0.385 Mercury (liquid) 0.140 Lead (solid) 0.129 Gold (solid) 0.129 Latent Heat of Phase Change Molar Heat of Fusion The energy that must be absorbed in order to convert one mole of solid to liquid at its melting point. Molar Heat of Solidification The energy that must be removed in order to convert one mole of liquid to solid at its freezing point. Latent Heat of Phase Change #2 Molar Heat of Vaporization The energy that must be absorbed in order to convert one mole of liquid to gas at its boiling point. Molar Heat of Condensation The energy that must be removed in order to convert one mole of gas to liquid at its condensation point. 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 0C to liquid water at 0C? 60 g H 2O 1 m olH 2O 6.009kJ 20 kiloJoules 18.02 g H 2O 1 m olH 2O Mass of ice Molar Mass of water Heat of fusion A 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? Pb T = ? oC mass = 322 g Ti = 25oC mass = 264 g Tf = 46oC Pb - LOSE heat = GAIN heat - [(Cp,Pb) (mass) (T)] Drop Units: = (Cp,H O) (mass) (T) 2 - [(0.138 J/goC) (322 g) (46oC - Ti)] = (4.184 J/goC) (264 g) (46oC- 25oC)] - [(44.44) (46oC - Ti)] = (1104.6) (21oC)] - 2044 + 44.44 Ti = 23197 44.44 Ti = Ti = 25241 568oC Calorimetry Problems 2 question #12 240 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. Fe T = 500oC mass = ? grams T = 20oC mass = 240 g - LOSE heat = GAIN heat - [(Cp,Fe) (mass) (T)] = (Cp,H O) (mass) (T) 2 - [(0.4495 J/goC) (X g) (42oC - 500oC)] Drop Units: - [(0.4495) (X) (-458)] 205.9 X X = = = (4.184 J/goC) (240 g) (42oC - 20oC)] (4.184) (240 g) (22) 22091 = 107.3 g Fe Calorimetry Problems 2 question #5 A 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. Au T = 785oC mass = 97 g T = 15oC mass = 323 g - LOSE heat = GAIN heat - [(Cp,Au) (mass) (T)] Drop Units: = (Cp,H O) (mass) (T) 2 - [(0.129 J/goC) (97 g) (Tf - 785oC)] = - [(12.5) (Tf - 785oC)] = -12.5 Tf + 9.82 x 103 3 x 104 = Tf = = (4.184 J/goC) (323 g) (Tf - 15oC)] (1.35 x 103) (Tf - 15oC)] 1.35 x 103 Tf - 2.02 x 104 1.36 x 103 Tf 22.1oC Calorimetry Problems 2 question #8 If 59 g of water at 13oC are mixed with 87 g of water at 72oC, find the final temperature of the system. T = 72oC mass = 87 g T = 13oC mass = 59 g - LOSE heat = GAIN heat - [(Cp,H O) (mass) (T)] = 2 Drop Units: (Cp,H O) (mass) (T) 2 - [(4.184 J/goC) (87 g) (Tf - 72oC)] = (4.184 J/goC) (59 g) (Tf - 13oC) - [(364.0) (Tf - 72oC)] = (246.8) (Tf - 13oC) -364 Tf + 26208 = 246.8 Tf - 3208 29416 = Tf = 610.8 Tf 48.2oC Calorimetry Problems 2 question #9 ice T = -11oC mass = 38 g A T = 56oC mass = 214 g D D water cools B warm water C melt ice warm Temperature (oC) A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC. Find the system's final temperature. 140 120 100 80 60 40 20 0 -20 -40 -60 -80 -100 H = mol x Hvap H = mol x Hfus Heat = mass x t x Cp, gas Heat = mass x t x Cp, liquid Heat = mass x t x Cp, solid Time - LOSE heat = GAIN heat A B C - [(Cp,H2O(l)) (mass) (T)] = (Cp,H2O(s)) (mass) (T) + (Cf) (mass) + (Cp,H2O(l)) (mass) (T) -[(4.184 J/goC)(214g)(Tf-56oC)] = (2.077J/goC)(38g)(11oC) + (333J/g)(38g) + (4.184J/goC)(38g)(Tf-0oC) - [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)] - 895 Tf + 50141 = 868 + 12654 + 159 Tf - 895 Tf + 50141 = 13522 + 159 Tf 36619 = 1054 Tf Tf = 34.7oC Calorimetry Problems 2 question #10 Heat of Solution The Heat of Solution is the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent. Substance Heat of Solution (kJ/mol) NaOH -44.51 NH4NO3 +25.69 KNO3 +34.89 HCl -74.84