How much water taken at a temperature of 0 can be converted into steam due to the energy

How much water taken at a temperature of 0 can be converted into steam due to the energy released during the combustion of 50 g of alcohol?

Given: m1 (mass of consumed alcohol) = 50 g (in SI m1 = 0.05 kg); t0 (initial water temperature) = 0 ºС.

Constants: qc (heat of combustion of alcohol) = 2.7 * 10^7 J / kg; Cw (heat capacity of water) = 4200 J / (kg * K); tpap (vaporization temperature) = 100 ºС; L (heat of vaporization) = 2.3 * 10^6 J / kg.

Let us express the mass of the evaporated water: qс * m1 = m2 * (Sv * (tp – t0) + L) and m2 = qc * m1 / (Sv * (tp – t0) + L).

Calculation: m2 = 2.7 * 10^7 * 0.05 / (4200 * (100 – 0) + 2.3 * 10^6) = 0.496 kg (496 g).