Heat of Fusion and Solidification
- If ice melts, the ice is the system and the air is the surroundings.
- The ice absorbs heat from the surroundings and begins to melt, the temp of the ice and the water produced stays the same (0oC) until all the ice has melted.
Heat of Fusion/Solidification
- all solids absorb heat as they melt to become liquids.
- The heat absorbed by one mole of a substance in melting at a constant temperature is the molar heat of fusion (ΔHfus)
- The heat lost when one mole of a liquid solidifies at a constant temp is the molar heat of solidification (ΔHsolid).
- The quantity of heat absorbed by a melting solid is exactly the same as the quantity of heat lost when the liquid solidifies.
- ΔHfus = -ΔHsolid
More on Heat of Fusion/Solidification
- The melting of 1 mol of ice at 0oC to 1 mol of water at 0oC requires the absorption of 6.01 kJ of heat. This quantity of heat is the molar heat of fusion. Also the conversion of 1 mol of water at 0oC to 1 mol of ice at 0oC releases 6.01 kJ of heat. This quantity of heat is the molar heat of solidification
- Ie. H2O(s) à H2O(l) ΔHfus = 6.01 kJ/mol
- H2O(l) à H2O(s) ΔHsolid = -6.01 kJ/mol
- How many grams of ice at 0oC and 101.3kPa could be melted by the addition of 2.25 kJ of heat?
- Start with knowns:
- Initial conditions are 0oC and 101.3kPa
- ΔHfus = 6.01 kJ/mol
- ΔH = 2.25 kJ
- Mice = ? g
- The conditions 0oC and 101.3kPa indicate that the standard conditions for the fusion of ice have been met. Use the chemical equation H2O(s) + 6.01 kJ à H2O(l) to find the number of moles of ice that can be melted by the addition of 2.25 kJ of heat. Then convert moles of ice to grams of ice.
- So take known heat and use ratios to solve
- 2.25 kJ x 1 mol ice x 18.0 g ice
- 6.01 kJ 1 mol ice
- = 6.74 g ice
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