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Anni [7]
3 years ago
6

How much energy is required to vaporize 14.27 grams of water at its boiling point? *

Chemistry
1 answer:
laila [671]3 years ago
6 0

Energy required to vaporize : 32.3 kJ

<h3>Further explanation </h3>

The heat to change the phase can be formulated :

Q = mLf (melting/freezing)

Q = mLv (vaporization/condensation)

Lf=latent heat of fusion

Lv=latent heat of vaporization

Mass of water = 14.27 g

Latent heat of vaporization ( boiling point of 100 ºC) :  2260 J/g

Energy required :

\tt Q=m.Lv\\\\Q=14.27\times 2260\\\\Q=32250.2~J\approx 32.3~kJ

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