Calorimetry

Energy released to the surrounding as heat Burning of a Match System Surroundings D(PE) (Reactants) Potential energy Energy released to the surrounding as heat (Products) Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 293

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. Endothermic Reaction Reactant + Energy Product Surroundings Surroundings System Energy System Before reaction After reaction Myers, Oldham, Tocci, Chemistry, 2004, page 41

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. Exothermic Reaction Reactant Product + Energy Surroundings Surroundings System System Energy Before reaction After reaction Myers, Oldham, Tocci, Chemistry, 2004, page 41

Direction of Heat Flow Surroundings ENDOthermic EXOthermic qsys > 0 System ENDOthermic qsys > 0 EXOthermic qsys < 0 System H2O(s) + heat  H2O(l) melting H2O(l)  H2O(s) + heat freezing 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.

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

A Bomb Calorimeter

Calorimetry “loses” heat 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

q = Cp . m . DT Cp(ice) = 2.077 J/g oC Temperature (oC) 40 20 -20 -40 -60 -80 -100 120 100 80 60 140 Time DH = mol x DHfus DH = mol x DHvap Heat = mass x Dt x Cp, liquid Heat = mass x Dt x Cp, gas Heat = mass x Dt x Cp, solid 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)

Temperature (oC) 40 20 -20 -40 -60 -80 -100 120 100 80 60 140 Time DH = mol x DHfus DH = mol x DHvap Heat = mass x Dt x Cp, liquid Heat = mass x Dt x Cp, gas Heat = mass x Dt x Cp, solid q = Cp . m . DT Heat = (specific heat) (mass) (change in temperature) Given Ti = -30oC Tf = -20oC q = 207.7 Joules

Find the mass of the iron. 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) (DT)] = (Cp,H2O) (mass) (DT) - [(0.4495 J/goC) (X g) (42oC - 500oC)] = (4.184 J/goC) (240 g) (42oC - 20oC)] Drop Units: - [(0.4495) (X) (-458)] = (4.184) (240 g) (22) 205.9 X = 22091 X = 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) (DT)] = (Cp,H2O) (mass) (DT) Drop Units: - [(0.129 J/goC) (97 g) (Tf - 785oC)] = (4.184 J/goC) (323 g) (Tf - 15oC)] - [(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

If 59 g of water at 13oC are mixed with 87 g of water at 72oC, find the final temperature of the system. T = 13oC mass = 59 g T = 72oC mass = 87 g LOSE heat = GAIN heat - - [(Cp,H2O) (mass) (DT)] = (Cp,H2O) (mass) (DT) Drop Units: - [(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 = 610.8 Tf Tf = 48.2oC Calorimetry Problems 2 question #9

A 322 g sample of lead (specific heat = 0 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 Pb mass = 322 g Ti = 25oC mass = 264 g Pb Tf = 46oC - LOSE heat = GAIN heat - [(Cp,Pb) (mass) (DT)] = (Cp,H2O) (mass) (DT) Drop Units: - [(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 = 25241 Ti = 568oC Calorimetry Problems 2 question #12