What amount of heat is required to bring 0.5 kg of alcohol taken at t = 28 g centigrade to boil and evaporate?

Initial data: m (mass of alcohol) = 0.5 kg; t0 (temperature at which the alcohol was located) = 28 ºС; for calculations, we assume that ethyl alcohol is heated to boiling and evaporation.

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

Required amount of heat: Q = Q1 + Q2 = C * m * (tboil – t0) + L * m.

Calculation: Q = 2500 * 0.5 * (78 – 28) + 0.9 * 10 ^ 6 * 0.5 = 512 500 J = 512.5 kJ.