Calorimetry
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Calorimetry Heat Transfer Surroundings Block “A” SYSTEM Al Block “B” 20 g (40oC) 20 g (20oC) Al Final Temperature 30oC 20 g40o C 20 g20o C 30o C (20 g 20 g) m = 20 g T = 40oC m = 20 g T = 20oC What will be the final temperature of the system ? a) 60oC b) 30oC c) 20oC d) ? Assume NO heat energy is “lost” to the surroundings from the system. Heat Transfer ? Surroundings Block “A” SYSTEM Al Al m = 20 g T = 40oC m = 10 g T = 20oC Block “B” Final Temperature 20 g (40oC) 20 g (20oC) 30.0oC 20 g (40oC) 10 g (20oC) 33.3oC 20 g40o C 10 g20o C 33. 3 o C (20 g 10 g) What will be the final temperature of the system ? a) 60oC b) 30oC c) 20oC d) ? Assume NO heat energy is “lost” to the surroundings from the system. Heat Transfer Surroundings Block “A” SYSTEM Al Al m = 20 g T = 20oC Block “B” Final Temperature 20 g (40oC) 20 g (20oC) 30.0oC 20 g (40oC) 10 g (20oC) 33.3oC 20 g (20oC) 10 g (40oC) 26.7oC 20 g20o C 10 g40o C 26. 7 o C (20 g 10 g) m = 10 g T = 40oC Assume NO heat energy is “lost” to the surroundings from the system. Heat Transfer Surroundings Block “A” SYSTEM H2O m = 75 g T = 25oC Ag m = 30 g T = 100oC Block “B” Final Temperature 20 g (40oC) 20 g (20oC) 30.0oC 20 g (40oC) 10 g (20oC) 33.3oC 20 g (20oC) 10 g (40oC) 26.7oC 75 g25o C 30 g100 o C 46o C (75 g 30 g) Real Final Temperature = 26.6oC Why? We’ve been assuming ALL materials transfer heat equally well. Specific Heat • Water and silver do not transfer heat equally well. Water has a specific heat Cp = 4.184 J/goC Silver has a specific heat Cp = 0.235 J/goC • What does that mean? It requires 4.184 Joules of energy to heat 1 gram of water 1oC and only 0.235 Joules of energy to heat 1 gram of silver 1oC. • 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! “loses” heat qAg qH2O Calorimetry Cp m T Cp m T Cp m Tfinal Tinitial Cp m Tf Ti Substitute values into equation. 0.235 J go C30 gx - 100o C 4.184 J go C75 gx - 25o C Surroundings Drop units and solve the algebra. 705 7.05 x 313.8x 7845 8550 320.8x SYSTEM Tfinal = 26.6oC x 26.6o C H2O Ag m = 75 g T = 25oC m = 30 g T = 100oC qAg Calorimetry qH2O Cp m T Cp m T Cp m Tfinal Tinitial Cp m Tf Ti Substitute values into equation. 0.235 J go C30 gx - 100 o C 4.184 J go C75 gx - 25 o C Drop units and solve the algebra. 705 7.05 x 313.8x 7845 320.8x 8550 Surroundings SYSTEM x 26.6 o C H2O Ag m = 75 g T = 25oC m = 30 g T = 100oC 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)] = (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) (T)] = (Cp,H O) (mass) (T) 2 Drop Units: - [(0.129 J/goC) (97 g) (Tf - 785oC)] = (4.184 J/goC) (323 g) (Tf - 15oC)] - [(12.5) (Tf - 785oC)] = (1.35x 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 = 72oC mass = 87 g T = 13oC mass = 59 g - LOSE heat = GAIN heat - [(Cp,H O) (mass) (T)] = (Cp,H O) (mass) (T) 2 Drop Units: 2 - [(4.184 J/goC) (59 g) (Tf - 13oC)] = (4.184 J/goC) (87 g) (Tf - 72oC)] - [(246.8) (Tf - 13oC)] = (364.0) (Tf - 72oC)] -246.8 Tf + 3208 = 364 Tf - 26208 29416 = 610.8 Tf Tf = 48.2oC Calorimetry Problems 2 question #9 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)] = (Cp,H O) (mass) (T) 2 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 (1000 g = 1 kg) 238.4kg g of water at 8oC. Find the final temperature of the system. 25 g of 116oC steam are bubbled into 0.2384 - [qA + qB + qC] = qD - [(Cp,H O) (mass) (T)] + (Cv,H O) (mass) + (Cp,H O) (mass) (T) = [(Cp,H O) (mass) (T)] 2 2 2 2 qD = (4.184 J/goC) (238.4 g) (Tf - 8oC) qD = - 997Tf - 7972 qA = [(Cp,H O) (mass) (T)] qB = (Cv,H O) (mass) qC = [(Cp,H O) (mass) (T)] qA = [(2.042 J/goC) (25 g) (100o - 116oC)] qA = - 816.8 J qA = (2256 J/g) (25 g) qA = - 56400 J qC = [(4.184 J/goC) (25 g) (Tf - 100oC)] qA = 104.5Tf - 10450 2 2 2 - [qA + qB + qC] = qD 816.8 + 56400 - 104.5Tf + 10450 = 997Tf - 7972 67667 - 104.5Tf = 997Tf - 7979 75646 = 1102Tf 1102 1102 A C B Tf = 68.6oC Temperature (oC) - [ - 816.8 - 56400 + 104.5Tf - 10450] = 997Tf - 7972 140 120 100 80 60 40 20 0 -20 -40 -60 -80 -100 H = mol x Hfus H = mol x Hvap Heat = mass x t x Cp, gas Heat = mass x t x Cp, liquid Heat = mass x t x Cp, solid Time D Calorimetry Problems 2 question #11 Thermometer Thermometer A Coffee Cup Calorimeter Glass stirrer Cork stopper Styrofoam cover Styrofoam cups Two Styrofoam ® cups Stirrer nested together containing reactants in solution Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 302 Bomb Calorimeter thermometer stirrer full of water ignition wire steel “bomb” sample A Bomb Calorimeter
