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Heat in changes of state

Heat in changes of state. Section 11.3. After reading Section 11.3, you should know:. The difference between fusion, solidification, vaporization and condensation How to calculate the heat changes for each type of reaction. Definitions.

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Heat in changes of state

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  1. Heat in changes of state Section 11.3

  2. After reading Section 11.3, you should know: • The difference between fusion, solidification, vaporization and condensation • How to calculate the heat changes for each type of reaction

  3. Definitions • Molar heat of fusion (∆Hfus) – the heat absorbed by one mole of a substance in melting from a solid to a liquid at constant temperature. • Table 11.5 (pg 308) • Molar heat of solidification (∆Hsolid) – the heat released when one mole of a liquid solidifies at constant temperature. • Determined from Table 11.5 (pg 308)

  4. Definitions • Molar heat of vaporization (∆Hvap) – the amount of heat absorbed when one mole of a given liquid is vaporized. • Table 11.5 (pg 308) • Molar heat of condensation (∆Hcond) – the amount of heat released when one mole of a vapor condenses. • Determined from Table 11.5 (pg 308)

  5. Definitions • Molar heat of solution (∆Hsoln) – the heat change caused by the dissolution of one mole of a substance; can be absorbed or released. • Molar heat of combustion(ΔHcomb) – heat of reaction for the complete burningof one mole of a substance. • Table 11.4 (pg 305)

  6. How they all relate (values on pg 308) Hsolid and Hfus are opposites Hsolid = - Hvap H2O(s) --> H2O(l) Hfus = 6.01 kJ / mol H2O(l) --> H2O(s)Hsolid = - 6.01 kJ / mol Hvap and Hcond are opposites, too! Hvap = - Hcond H2O(l) --> H2O(g) Hvap = 40.7 kJ / mol H2O(g) --> H2O(l)Hcond = - 40.7 kJ / mol

  7. Heating Curve for Water ∆Hvap ∆Hfus

  8. Constants for H20 • ∆Hfus = 6.01 kJ/mol • ∆Hvap = 40.7 kJ/mol • ∆Hsolid = − 6.01 kJ/mol • ∆Hcond= −40.7 kJ/mol • Cp (ice) = 2.06 J/goC • Cp (water) = 4.18 J/goC • Cp (steam) = 2.02 J/goC

  9. Sample Problem • How many grams of ice at 0oC and 101.3 kPa could be melted by the addition of 2.25 kJ of heat? Known Values:Unknowns: Melting = ∆Hfusion grams of ice = ? ∆Hfusion (H2O) = 6.01 kJ/mol 2.25 kJ of heat are absorbed ? Grams = 2.25 kJ 1 mol 18.0 grams = 6.74 grams 6.01 kJ 1 mol

  10. After reading Section 11.3, you should know: • The difference between fusion, solidification, vaporization and condensation • How to calculate the heat changes for each type of reaction

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