How much alcohol is needed to turn 200 g of lead taken at a temperature of 27 ° C into a liquid at the melting

How much alcohol is needed to turn 200 g of lead taken at a temperature of 27 ° C into a liquid at the melting temperature in a steel container weighing 100 g?

Data: m1 (lead) = 0.2 kg; t0 (initial temp. lead, capacity) = 27 ºС; m2 (capacity) = 0.1 kg; tmelt (lead) = 327 ºС; C1 (heat capacity of lead) = 140 J / (kg * K); C2 (heat capacity of steel) = 500 J / (kg * K); λ (specific heat of melting lead) = 0.25 * 10 ^ 5 J / kg; q (beats heat of combustion of alcohol) = 27 * 10 ^ 6 J / kg.

Q = Q1 + Q2 + Q3 = C1 * m1 * (tm – t0) + λ * m1 + C2 * m2 * (tm – t0) = 140 * 0.2 * (327 – 27) + 0.25 * 10 ^ 5 * 0.2 + 500 * 0.1 * (327 – 27) = 28400 J.

m3 = Q / q = 28400 / (27 * 10 ^ 6) = 0.001 kg = 1 g.