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

Chemistry

1 answer:

laila [671]2 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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