How much kerosene should be consumed to turn 3 kg of ice taken at 0 ° C into steam, if the efficiency

How much kerosene should be consumed to turn 3 kg of ice taken at 0 ° C into steam, if the efficiency of the kerosene on which the ice was heated is 40%?

Given: m1 (ice) = 3 kg; t (temperature of taken ice) = 0 ºС; η (kerosene efficiency) = 40% (0.4).
Reference values: q (specific heat of combustion of kerosene) = 46 * 106 J / kg; λ (specific heat of melting of ice) = 3.4 * 10 ^ 5 J / kg; C (specific heat of water) = 4200 J / (kg * K); L (specific heat of vaporization) = 2.3 * 10 ^ 6 J / kg.
Formula: η = (λ + C * (100 – t) + L) * m1 / (q * m2).
m2 = (λ + C * (100 – t) + L) * m1 / (η * q) = (3.4 * 10 ^ 5 + 4200 * 100 + 2.3 * 10 ^ 6) * 3 / (0 , 4 * 46 * 10 ^ 6) ≈ 0.5 kg.