What amount of alcohol, taken at a temperature of 78 ℃, boiled away if 18 kJ of heat was reported to it?

Initial data: t0 (initial temperature of alcohol) = 78 ºС (corresponds to the boiling point of ethyl alcohol), Q (heat that was reported to alcohol) = 18 kJ or 18 * 10 ^ 3 J.

Reference values: L (specific heat of vaporization) = 0.9 * 10 ^ 6 J / kg.

We express the mass of the boiled off alcohol from the following formula: Q = L * m, whence m = Q / L.

Let’s do the calculation:

m = 18 * 10 ^ 3 / (0.9 * 10 ^ 6) = 0.02 kg or 20 g.

Answer: Boil off 20 grams of ethyl alcohol.