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## Calorimetry

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**D(PE)**(Products) Burning of a Match System Surroundings (Reactants) Potential energy Energy released to the surrounding as heat Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 293**Endothermic Reaction**Reactant + Energy Product Surroundings System Conservation of Energy in a Chemical 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. Surroundings Energy System Before reaction After reaction Myers, Oldham, Tocci, Chemistry, 2004, page 41**Exothermic Reaction**Reactant Product + Energy Surroundings System Conservation of Energy in a Chemical 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. Surroundings System Energy Before reaction After reaction Myers, Oldham, Tocci, Chemistry, 2004, page 41**H2O(s) + heat H2O(l)**H2O(l) H2O(s) + heat melting freezing System Direction of Heat Flow Surroundings EXOthermic qsys < 0 ENDOthermic qsys > 0 System Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 207**Caloric Values**Food joules/grams calories/gram Calories/gram Protein 17 000 4000 4 Fat 38 000 9000 9 Carbohydrates 17 000 4000 4 1calories = 4.184 joules 1000 calories = 1 Calorie "science" "food" Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 51**Experimental Determination of Specific Heat of a Metal**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.**Thermometer**Styrofoam cover Styrofoam cups Stirrer A Coffee Cup Calorimeter Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 302**“loses” heat**Calorimetry Surroundings SYSTEM Tfinal = 26.7oC H2O Ag m = 75 g T = 25oC m = 30 g T = 100oC**Calorimetry**Surroundings SYSTEM H2O Ag m = 75 g T = 25oC m = 30 g T = 100oC**140**DH = mol xDHvap DH = mol xDHfus 120 100 80 60 Heat = mass xDt x Cp, gas 40 Temperature (oC) 20 Heat = mass xDt x Cp, liquid 0 -20 -40 -60 Heat = mass xDt x Cp, solid -80 -100 Time Cp(ice) = 2.077 J/g oC It takes 2.077 Joules to raise 1 gram ice 1oC. X Joules to raise 10 gram ice 1oC. (10 g)(2.077 J/g oC) = 20.77 Joules X Joules to raise 10 gram ice 10oC. (10oC)(10 g)(2.077 J/g oC) = 207.7 Joules q = Cp . m .DT Heat = (specific heat) (mass) (change in temperature)**140**DH = mol xDHvap DH = mol xDHfus 120 100 80 60 Heat = mass xDt x Cp, gas 40 Temperature (oC) 20 Heat = mass xDt x Cp, liquid 0 -20 -40 -60 Heat = mass xDt x Cp, solid -80 -100 Time Given Ti= -30oC Tf = -20oC q = Cp . m .DT Heat = (specific heat) (mass) (change in temperature) q = 207.7 Joules**T = 500oC**Fe T = 20oC mass = 240 g 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. mass = ? grams - LOSE heat = GAIN heat - [(Cp,Fe) (mass) (DT)] = (Cp,H2O) (mass) (DT) - [(0.4495 J/goC) (X g) (42oC - 500oC)] = (4.184 J/goC) (240 g) (42oC - 20oC)] - [(0.4495) (X) (-458)] = (4.184) (240 g) (22) Drop Units: 205.9 X = 22091 X = 107.3 g Fe Calorimetry Problems 2 question #5**T = 785oC**mass = 97 g Au T = 15oC mass = 323 g - LOSE heat = GAIN heat 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. - [(Cp,Au) (mass) (DT)] = (Cp,H2O) (mass) (DT) - [(0.129 J/goC) (97 g) (Tf - 785oC)] = (4.184 J/goC) (323 g) (Tf - 15oC)] Drop Units: - [(12.5) (Tf - 785oC)] = (1.35 x 103) (Tf - 15oC)] -12.5 Tf + 9.82 x 103 = 1.35 x 103 Tf - 2.02 x 104 3 x 104 = 1.36 x 103 Tf Tf = 22.1oC Calorimetry Problems 2 question #8**T = 72oC**T = 13oC mass = 87 g mass = 59 g - LOSE heat = GAIN heat If 59 g of water at 13oC are mixed with 87 g of water at 72oC, find the final temperature of the system. - [(Cp,H2O) (mass) (DT)] = (Cp,H2O) (mass) (DT) - [(4.184 J/goC) (87 g) (Tf - 72oC)] = (4.184 J/goC) (59 g) (Tf - 13oC) Drop Units: - [(364.0) (Tf - 72oC)] = (246.8) (Tf - 13oC) -364 Tf + 26208 = 246.8 Tf - 3208 29416 = 610.8 Tf Tf = 48.2oC Calorimetry Problems 2 question #9**Ti = 25oC**mass = 264 g Pb 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? T = ? oC mass = 322 g Pb Tf = 46oC - LOSE heat = GAIN heat - [(Cp,Pb) (mass) (DT)] = (Cp,H2O) (mass) (DT) - [(0.138 J/goC) (322 g) (46oC - Ti)] = (4.184 J/goC) (264 g) (46oC- 25oC)] Drop Units: - [(44.44) (46oC - Ti)] = (1104.6) (21oC)] - 2044 + 44.44 Ti = 23197 44.44 Ti = 25241 Ti = 568oC Calorimetry Problems 2 question #12