How much energy (in joules) is required to evaporate 0.0005 kg of liquid ammonia to vapor at the same temperature?

685.6 J

Explanation:

The latent heat of vaporization of ammonia is

L = 1371.2 kJ/kg

mass of ammonia, m = 0.0005 Kg

Heat = mass x latent heat of vaporization

H = 0.0005 x 1371.2

H = 0.686 kJ

H = 685.6 J

Thus, the amount of heat required to vaporize the ammonia is 685.6 J.

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