1 / 15

Thermochemical Calculations

Thermochemical Calculations. CA Standards. Units for Measuring Heat. The Joule is the SI system unit for measuring heat:. The calorie is the heat required to raise the temperature of 1 gram of water by 1 Celsius degree. Specific Heat.

stephanie-cross
Download Presentation

Thermochemical Calculations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Thermochemical Calculations

  2. CA Standards

  3. Units for Measuring Heat The Joule is the SI system unit for measuring heat: The calorie is the heat required to raise the temperature of 1 gram of water by 1 Celsius degree

  4. Specific Heat The amount of heat required to raise the temperature of one gram of substance by one degree Celsius.

  5. Calculations Involving Specific Heat OR cp = Specific Heat Q = Heat lost or gained T = Temperature change m = Mass

  6. The amount of heat required to raise the temperature of one gram of substance by one degree Celsius. Specific Heat

  7. Latent Heat of Phase Change The energy that must be absorbed in order to convert one mole of solid to liquid at its melting point. Molar Heat of Fusion Molar Heat of Solidification The energy that must be removed in order to convert one mole of liquid to solid at its freezing point.

  8. Latent Heat of Phase Change #2 The energy that must be absorbed in order to convert one mole of liquid to gas at its boiling point. Molar Heat of Vaporization Molar Heat of Condensation The energy that must be removed in order to convert one mole of gas to liquid at its condensation point.

  9. 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 0C to liquid water at 0C? Mass of ice Molar Mass of water Heat of fusion

  10. 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

  11. 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

  12. 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

  13. 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

  14. 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 T = -11oC -60 mass = 38 g Heat = mass xDt x Cp, solid ice -80 -100 Time T = 56oC mass = 214 g - LOSE heat = GAIN heat A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC. Find the system's final temperature. D water cools B warm water C A melt ice warm ice C D A B - [(Cp,H2O(l)) (mass) (DT)] = (Cp,H2O(s)) (mass) (DT) + (Cf) (mass) + (Cp,H2O(l)) (mass) (DT) -[(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

  15. Heat of Solution TheHeat of Solutionis the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent.

More Related