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OBJECTIVES:

OBJECTIVES: Classify , by type, the heat changes that occur during melting, freezing, boiling, and condensing. Calculate heat changes that occur during melting, freezing, boiling, and condensing. Heats of Fusion and Solidification.

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OBJECTIVES:

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  1. OBJECTIVES: • Classify, by type, the heat changes that occur during melting, freezing, boiling, and condensing. • Calculate heat changes that occur during melting, freezing, boiling, and condensing.

  2. Heats of Fusion and Solidification • Molar Heat of Fusion (Hfus) - the heat absorbed by one mole of a substance in melting from a solid to a liquid • Molar Heat of Solidification(Hsolid) - heat lost when one mole of liquid solidifies

  3. Heats of Fusion and Solidification • Heat absorbed by a melting solid is equal to heat lost when a liquid solidifies • Thus, Hfus = -Hsolid

  4. Heats of Vaporization and Condensation • When liquids absorb heat at their boiling points, they become vapors. • Molar Heat of Vaporization(Hvap) - the amount of heat necessary to vaporize one mole of a given liquid.

  5. Heats of Vaporization and Condensation • Condensation is the opposite of vaporization. • Molar Heat of Condensation(Hcond) - amount of heat released when one mole of vapor condenses • Hvap = - Hcond

  6. Heats of Vaporization and Condensation • H20(g) H20(l) Hcond = - 40.6kJ/mol • Vaporization (phase change from liquid to gas) of the substance or • Fusion (phase change from solid to liquid) of the substance.

  7. Molar Enthalpy's of Vaporization and Fusion (under standard conditions)

  8. Calculate the total quantity of heat evolved when 10.0g of steam at 200 C is condensed, cooled , and frozen to ice at -50 C. The specific heat capacity of ice and steam are 2.06J/gC and 1.87 J/gC respectively. 1. Cooling steam 1 2 3 4 5

  9. . 1. Cooling steam Q = mcT Q = (10.0g)(1.87J/gC)(100C) Q = 1870 J = 1.870 kJ 1

  10. 2 n(H2O) = mass / molar mass = 10.0g / 18g mol-1 = 0.56 mol H2O Heat of vaporization Hvap (H2O) = 40.6 kJ/mol Hvap (H2O) = 0.56 mol  40.6 kJ/mol = 22.7 kJ 2 condensation

  11. 3 Q = mcT Q = (10.0g)(4.18J/gC)(100C) Q = 4180J = 4.180kJ 3 Cooling of water

  12. 4 n(H2O) = mass / molar mass = 10.0g / 18g mol-1 = 0.56 mol H2O Heat of fusion  Hfus (H2O) = 6.10 kJ/mol  Hfus (H2O) = 0.56 mol  6.01 kJ/mol = 3.37 kJ Water freezing 4

  13. 5 Q = mcT Q = (10.0g)(2.06J/gC)(50C) Q = 1030J = 1.030kJ Cooling ice 5

  14. Calculate the total quantity of heat evolved when 10.0g of steam at 200 C is condensed, cooled , and frozen to ice at -50 C. Total heat = 1 + 2 + 3 + 4 + 5 Total heat = 1.870 kJ + 22.7 kJ + 4.180 kJ + 3.37 kJ + 1.030 kJ = 33.15 kJ

  15. Question: • How much energy is needed to raise the temp of 25 grams of ice from -25°C to 105°C ? • Answer: 90.4 kJ

  16. Solution: • From ice to water Q = mcT Q = (25g)(2.06J/gC)(25C) Q = 12,875 J = 12.9 kJ Solid ice phase n(H2O) = mass / molar mass = 25 g / 18g mol-1 = 1.39 mol H2O Hfus (H2O) = 6.01 kJ/mol Hfus (H2O) = 1.39 mol  6.01 kJ/mol = 8.35 kJ

  17. 0 to 100 Q = mcT Q = (25g)(4.184J/gC)(100C) Q = 10,460 J = 10.4 kJ Liquid to steam n(H2O) = mass / molar mass = 25g / 18g mol-1 = 1.39 mol H2O Hfus (H2O) = 40.6kJ/mol Hfus (H2O) = 1.39 mol  40.6 kJ/mol = 56.4 kJ Above 100 Q = mcT Q = (25g)(1.87J/gC)(5C) Q = 2,337.5 J = 2.34 kJ Ans: 12.9 kJ + 8.35 kJ + 10.4 kJ + 56.4 kJ + 2.34 kJ = 90.4 kJ

  18. Questions: • 1. Find the amount of heat released when 3.0 kg of water goes from 55ºC to -35ºC. • 2. Find the amount of heat absorbed when 400. g ice goes from -20.0ºC to 10.0ºC. • 3. Find the amount of heat absorbed when 75.0 g ice goes from -40.ºC to 110ºC.

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