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

Answer 1

Well, it's at the same temperature, so you just need to multiply the amount of ammonia by the "heat of vaporization" of ammonia -- and be sure to get the units to cancel out. For example, if it were 62 mg of WATER, the calculation would be 62mg x (0.001g/mg) x (540 cal/g) x (4.184 J/cal).

Answer 2

This gives the heat of evaporation in Joules per gram:

https://en.wikipedia.org/wiki/Enthalpy_of_vaporization

Convert kilograms to grams and then multiply by the value for Ammonia