3 years ago

Determine the amount of heat (in kj) required to vaporize 1.55 kg of water at its boiling point. for water, δhvap

1 answer:

5 0

Heat required to vaporize 1 mol of water from water at 100C to steam at 100C = 40.7 kJ
1 mol of water weighs = 18.015g
1.55 kg = 1550/18.015 mol = 86.03 mol

Heat required to vaporize :
= 86.03 mol x 40.7 kJ

= 3501.421 kJ

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