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Calorimetry Heat Transfer Al m = 20 g T = 40 o C SYSTEM Surroundings m = 20 g T = 20 o C 20 g (40 o C) 20 g (20 o C)30 o C Block A Block B Final Temperature.

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Presentation on theme: "Calorimetry Heat Transfer Al m = 20 g T = 40 o C SYSTEM Surroundings m = 20 g T = 20 o C 20 g (40 o C) 20 g (20 o C)30 o C Block A Block B Final Temperature."— Presentation transcript:

1

2 Calorimetry

3 Heat Transfer Al m = 20 g T = 40 o C SYSTEM Surroundings m = 20 g T = 20 o C 20 g (40 o C) 20 g (20 o C)30 o C Block A Block B Final Temperature Assume NO heat energy is lost to the surroundings from the system. What will be the final temperature of the system ? a) 60 o C b) 30 o C c) 20 o C d) ?

4 Heat Transfer Al m = 20 g T = 40 o C SYSTEM Surroundings m = 10 g T = 20 o C 20 g (40 o C) 20 g (20 o C)30.0 o C Block A Block B Final Temperature Assume NO heat energy is lost to the surroundings from the system. 20 g (40 o C) 10 g (20 o C)33.3 o C What will be the final temperature of the system ? a) 60 o C b) 30 o C c) 20 o C d) ? ?

5 Heat Transfer Al m = 20 g T = 20 o C SYSTEM Surroundings m = 10 g T = 40 o C 20 g (40 o C) 20 g (20 o C)30.0 o C Block A Block B Final Temperature Assume NO heat energy is lost to the surroundings from the system. 20 g (40 o C) 10 g (20 o C)33.3 o C 20 g (20 o C) 10 g (40 o C)26.7 o C

6 Heat Transfer m = 75 g T = 25 o C SYSTEM Surroundings m = 30 g T = 100 o C 20 g (40 o C) 20 g (20 o C)30.0 o C Block A Block B Final Temperature 20 g (40 o C) 10 g (20 o C)33.3 o C 20 g (20 o C) 10 g (40 o C)26.7 o C Ag H2OH2O Real Final Temperature = 26.6 o C Why? Weve been assuming ALL materials transfer heat equally well.

7 Specific Heat Water and silver do not transfer heat equally well. Water has a specific heat C p = J/g o C Silver has a specific heat C p = J/g o C What does that mean? It requires Joules of energy to heat 1 gram of water 1 o C and only Joules of energy to heat 1 gram of silver 1 o C. 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!

8 loses heat Calorimetry m = 75 g T = 25 o C SYSTEM Surroundings m = 30 g T = 100 o C Ag H2OH2O T final = 26.6 o C

9 Calorimetry m = 75 g T = 25 o C SYSTEM Surroundings m = 30 g T = 100 o C Ag H2OH2O

10 240 g of water (initially at 20 o C) are mixed with an unknown mass of iron (initially at 500 o C). When thermal equilibrium is reached, the system has a temperature of 42 o C. Find the mass of the iron. Calorimetry Problems 2 question #5 Fe T = 500 o C mass = ? grams T = 20 o C mass = 240 g LOSE heat = GAIN heat - - [(C p, Fe ) (mass) ( T)] = (C p, H 2 O ) (mass) ( T) - [( J/g o C) (X g) (42 o C o C)] = (4.184 J/g o C) (240 g) (42 o C - 20 o C)] Drop Units: - [(0.4495) (X) (-458)] = (4.184) (240 g) (22) X = X = g Fe

11 A 97 g sample of gold at 785 o C is dropped into 323 g of water, which has an initial temperature of 15 o C. If gold has a specific heat of J/g o C, what is the final temperature of the mixture? Assume that the gold experiences no change in state of matter. Calorimetry Problems 2 question #8 Au T = 785 o C mass = 97 g T = 15 o C mass = 323 g LOSE heat = GAIN heat - - [(C p, Au ) (mass) ( T)] = (C p, H 2 O ) (mass) ( T) - [(0.129 J/g o C) (97 g) (T f o C)] = (4.184 J/g o C) (323 g) (T f - 15 o C)] Drop Units: - [(12.5) (T f o C)] = (1.35x 10 3 ) (T f - 15 o C)] T f x 10 3 = 1.35 x 10 3 T f x x 10 4 = 1.36 x 10 3 T f T f = 22.1 o C

12 If 59 g of water at 13 o C are mixed with 87 g of water at 72 o C, find the final temperature of the system. Calorimetry Problems 2 question #9 T = 13 o C mass = 59 g LOSE heat = GAIN heat - - [(C p, H 2 O ) (mass) ( T)] = (C p, H 2 O ) (mass) ( T) - [(4.184 J/g o C) (59 g) (T f - 13 o C)] = (4.184 J/g o C) (87 g) (T f - 72 o C)] Drop Units: - [(246.8) (T f - 13 o C)] = (364.0) (T f - 72 o C)] T f = 364 T f = T f T f = 48.2 o C T = 72 o C mass = 87 g

13 A 322 g sample of lead (specific heat = J/g o C) is placed into 264 g of water at 25 o C. If the system's final temperature is 46 o C, what was the initial temperature of the lead? Calorimetry Problems 2 question #12 Pb T = ? o C mass = 322 g T i = 25 o C mass = 264 g LOSE heat = GAIN heat - - [(C p, Pb ) (mass) ( T)] = (C p, H 2 O ) (mass) ( T) - [(0.138 J/g o C) (322 g) (46 o C - T i )] = (4.184 J/g o C) (264 g) (46 o C- 25 o C)] Drop Units: - [(44.44) (46 o C - T i )] = (1104.6) (21 o C)] T i = T i = T i = 568 o C Pb T f = 46 o C

14 Calorimetry with Change in State

15 A 38 g sample of ice at -11 o C is placed into 214 g of water at 56 o C. Find the system's final temperature. Calorimetry Problems 2 question #10 ice T = -11 o C mass = 38 g T = 56 o C mass = 214 g LOSE heat = GAIN heat - - [(C p, H 2 O ) (mass) ( T)] = (C p, H 2 O ) (mass) ( T) + (C f ) (mass) + (C p, H 2 O ) (mass) ( T) - [(4.184 J/g o C)(214 g)(T f - 56 o C)] = (2.077 J/g o C)(38 g)(11 o C) + (333 J/g)(38 g) + (4.184 J/g o C)(38 g)(T f - 0 o C) - [(895) (T f - 56 o C)] = (159) (T f )] T f = T f T f = T f T f = 34.7 o C = 1054 T f Temperature ( o C) Time H = mol x H fus H = mol x H vap Heat = mass x t x C p, liquid Heat = mass x t x C p, gas Heat = mass x t x C p, solid A B C D warm ice melt ice warm water water cools

16 25 g of 116 o C steam are bubbled into kg of water at 8 o C. Find the final temperature of the system. Calorimetry Problems 2 question #11 - [(C p, H 2 O ) (mass) ( T)] + (C v, H 2 O ) (mass) + (C p, H 2 O ) (mass) ( T) = [(C p, H 2 O ) (mass) ( T)] - [ T f ] = 997T f [q A + q B + q C ] = q D q A = [(C p, H 2 O ) (mass) ( T)] q A = [(2.042 J/g o C) (25 g) (100 o o C)] q A = J q B = (C v, H 2 O ) (mass) q A = (2256 J/g) (25 g) q A = J q C = [(C p, H 2 O ) (mass) ( T)] q C = [(4.184 J/g o C) (25 g) (T f o C)] q A = 104.5T f q D = (4.184 J/g o C) (238.4 g) (T f - 8 o C) q D = - 997T f [q A + q B + q C ] = q D T f = 997T f T f = 997T f = 1102T f 1102 T f = 68.6 o C Temperature ( o C) Time H = mol x H fus H = mol x H vap Heat = mass x t x C p, liquid Heat = mass x t x C p, gas Heat = mass x t x C p, solid A B C D (1000 g = 1 kg) g

17 A sample of ice at –12 o C is placed into 68 g of water at 85 o C. If the final temperature of the system is 24 o C, what was the mass of the ice? Calorimetry Problems 2 question #13 H2OH2O T = -12 o C mass = ? g T i = 85 o C mass = 68 g GAIN heat = - LOSE heat [ q A + q B + q C ] = - [(C p, H 2 O ) (mass) ( T)] m = m = 37.8 g ice T f = 24 o C q A = [(C p, H 2 O ) (mass) ( T)] q C = [(C p, H 2 O ) (mass) ( T)] q B = (C f, H 2 O ) (mass) q A = [(2.077 J/g o C) (mass) (12 o C)] q B = (333 J/g) (mass) q C = [(4.184 J/g o C) (mass) (24 o C)] [ q A + q B + q C ] = - [(4.184 J/g o C) (68 g) (-61 o C)] 24.9 m 333 m m m q Total = q A + q B + q C 458.2

18 A Coffee Cup Calorimeter Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 302 Thermometer Styrofoam cover Styrofoam cups Stirrer Thermometer Glass stirrer Cork stopper Two Styrofoam ® cups nested together containing reactants in solution

19 Bomb Calorimeter thermometer stirrer full of water ignition wire steel bomb sample

20 A Bomb Calorimeter


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