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Chapter 13

Chapter 13. Solids and Liquids. Molar Heat of Vaporization. The amount of heat energy required to vaporize one mole of liquid at its boiling point. Joules are the standard unit to measure heat energy. Molar heat of vaporization for water is 40.7 kJ/mole .

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Chapter 13

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  1. Chapter 13 Solids and Liquids

  2. Molar Heat of Vaporization • The amount of heat energy required to vaporize one mole of liquid at its boiling point. • Joules are the standard unit to measure heat energy. • Molar heat of vaporization for water is 40.7 kJ/mole. • Ex1: How much heat energy would be required to vaporize 5.00 moles of H2O?

  3. 45.0g H2O x 1 mol H2O = 2.5 mol H2O 18.0g H2O Molar Heat of Vaporization • Ex 2. How much heat energy would be required to vaporize 45.0g of H2O? • Step 1: convert to moles • Step 2: calculate heat 2.5 mol H2O x 40.7 KJ/mol = 102 KJ

  4. Molar Heat of Vaporization • When a liquid evaporates, it absorbs energy. Energy is used to overcome attractive forces.The energy doesn’t increase the average energy of the particles, so the temperature doesn’t change. • When a liquid evaporates, it takes energy from its surroundings that’s why alcohol feels cool to the skin. • It’s also why we get cold when getting out of the shower

  5. Molar Heat of Vaporization Heat of vaporization - Hvap Definition – amount of energy needed to vaporize a unit of substance (mass or moles) Formula - q = (Hvap) x (mass) mass = gram or mole • Ex3 - How much heat does it take to vaporize 50.0 g of water at 100.0 C0? q = (Hvap) x (mass) q = 40.7 KJ X 50 g H2O X 1 mol 1 mol 18.0 g H2O q = 113 KJ

  6. Molar Heat of Fusion • The amount of heat energy required to melt one mole of a solid at its melting point. • The molar heat of fusion of water is 6.02 kJ/mole. • Ex1: How much energy would be required to melt 12.75 moles of ice? 12.75 moles ice x 6.02 KJ/mol = 76.8 KJ

  7. Molar Heat of Fusion • Ex2: How much energy would be required to melt 6.48 x 1020 kg of ice? • Step 1: Convert Kg to g 6.48 x 1020 Kg x 1000 g = 6.48 x 1023 g • Step 2: Calculate energy (convert g to mol) 6.48 x 1023g H2O x 1 mol x 6.02 KJ = 2.16 x 1023 KJ 1 Kg 18 g mol

  8. 2.9 x 10 4 J x 1 KJ = 29 KJ 1000 J X 18 g = 86.8 g ice 1 mol Molar Heat of Fusion Heat of Fusion - Hfus Definition - heat required to change a unit of substance from solid to liquid Formula - q = (Hfus) x (mass) mass = g or mole • Ex3: - How much ice can be melted by 2.9 x 104 J? • Step 1: Convert J to KJ • Step 2: Calculate amount q = (Hfus) x (mass) Mass = q / Hfus Mass = 29 KJ / 6.02 KJ/mol = 4.82 mol

  9. Phase Change Diagram Flat sections at boiling/melting why? All energy input is directed at changing phase, so there is no increase in temp.

  10. Temperature and Phase Changes • 3 formulas to use: • q = mCpt for sections A, C, E • q = mHfus for section B • q = mHvap for section D

  11. Temperature and Phase Changes • If 75 g H2O is at -5oC and is heated to 105oC. How much total heat in joules is required? (Convert to calories after)

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