What amount of heat must be imparted to an alcohol with a mass of 10 g

What amount of heat must be imparted to an alcohol with a mass of 10 g taken at a temperature of 18 s in order to heat it to the boiling point and evaporate.

Data: m (mass of alcohol) = 10 g; t (initial temperature) = 18 ºС.

Reference data: tboil (boiling point) = 78 ºС; C (specific heat) = 2500 J / (kg * K); L (specific heat of vaporization) = 0.9 MJ / kg = 0.9 * 10 ^ 6 J / kg;.

SI system: m = 10 g = 10 * 10 ^ -3 kg.

The heat that was reported to alcohol: Q = Q1 (heating) + Q2 (vaporization) = C * m * (tboil – t) + L * m = (C * (tboil – t) + L) * m.

Let’s calculate: Q = (2500 * (78 – 18) + 0.9 * 10 ^ 6) * 10 * 10 ^ -3 = 10 500 J = 10.5 kJ.