How much energy is required to convert 3 kg alcohol, taken at a temperature of 10 degrees, into steam?

Data: m (mass of alcohol) = 3 kg; t (temperature at which the alcohol was located) = 10 ºС.

Reference values: C (specific heat of alcohol) = 2500 J / (kg * K); tpap (temperature of the beginning of vaporization of ethyl alcohol) = 78.3 ºС; L (specific heat of vaporization of alcohol) = 0.9 * 10 ^ 6 J / kg.

The required amount of energy is determined by the formula: Q = Q1 + Q2 = C * m * (tpar – t) + L * m.

Calculation: Q = 2500 * 3 * (78.3 – 10) + 0.9 * 10 ^ 6 * 3 = 3 212 250 J ≈ 3.2 MJ.

Answer: Requires 3.2 MJ of heat.