How many kJ of heat are needed to completely vaporize 43.2 g of H2O? The heat of
ID: 831345 • Letter: H
Question
How many kJ of heat are needed to completely vaporize 43.2 g of H2O? The heat of vaporization for water at the boiling point is 40.6 kJ/mole.
How many kJ of heat are needed to completely melt 3.30 moles of H2O, given that the water is at its melting point? The heat of fusion for water is 6.02 kJ/mole.
How many kJ of heat are needed to completely vaporize 1.30 moles of H2O? The heat of vaporization for water at the boiling point is 40.6 kJ/mole.
How many joules of heat are needed to completely vaporize 43.2 grams of water at its boiling point?
Given ?Hvap = 40.6 kJ/mol
Explanation / Answer
(1) Moles of H2O = mass/molar mass of H2O
= 43.2/18.02 = 2.397 mol
Heat required = moles of H2O x heat of vaporization
= 2.397 x 40.6
= 97.3 kJ
(2) Heat required = moles of H2O x heat of fusion
= 3.30 x 6.02
= 19.9 kJ
(3) Heat required = moles of H2O x heat of vaporization
= 1.30 x 40.6
= 52.8 kJ
(4) Moles of H2O = mass/molar mass of H2O
= 43.2/18.02 = 2.397 mol
Heat required = moles of H2O x heat of vaporization
= 2.397 x 40.6
= 97.3 kJ
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