How much gasoline needs to be burned in order to get boiling water from 3 kg of ice at a temperature

How much gasoline needs to be burned in order to get boiling water from 3 kg of ice at a temperature of -10 degrees? Consider heat loss to be negligible.

Data: m1 (ice mass) = 3 kg; t0 (initial ice temperature) = -10 ºС; heat loss is not taken into account.

Constants: Сl = 2100 J / (kg * K); tmelt = 0 ºС; λ = 3.4 * 10 ^ 5 J / kg; Cw = 4200 J / (kg * K); tboil = 100 ºС; qb = 4.6 * 10 ^ 7 J / kg.

Let us express the mass of gasoline burned from the equality: qb * m2 = m1 * (Cl * (tm – t0) + λ + Sv * (tboil – tm)), whence m2 = m1 * (Sl * (tm – t0) + λ + Sv * (tboil – tmelt)) / qb.

Calculation: m2 = 3 * (2100 * (0 – (-10)) + 3.4 * 10 ^ 5 + 4200 * (100 – 0)) / (4.6 * 10 ^ 7) = 0.051 kg (51 g ).