What mass of alcohol can be evaporated when 27 kJ of heat is applied to it?

What mass of alcohol can be evaporated when 27 kJ of heat is applied to it? Alcohol temperature 78 degrees (specific heat of vaporization of alcohol 0.9 MJ / kg)

Initial data: Q (the amount of heat that was brought to alcohol) = 27 kJ (27 * 10 ^ 3 J); t (temperature of alcohol) = 78 ºС (this value coincides with the temperature of the beginning of vaporization of ethyl alcohol).

Reference data: according to the condition Lc (specific heat of vaporization of alcohol) = 0.9 MJ / kg (0.9 * 10 ^ 6 J / kg).

We express the mass of the evaporated alcohol from the formula: Q = Lc * m, whence m = Q / Lc.

Calculation: m = 27 * 10 ^ 3 / (0.9 * 10 ^ 6) = 0.03 kg (30 g).

Answer: It will turn out to evaporate 30 grams of ethyl alcohol.