On a stove with an efficiency of 50%, 2L water was heated from 10 degrees to boiling and 100g

On a stove with an efficiency of 50%, 2L water was heated from 10 degrees to boiling and 100g was evaporated. How much coal was burned?

Given: η (efficiency) = 50% (0.5); V (water volume) = 2 l (0.002 m3); t0 (initial temp.) = 10 ºС; m2 (mass of evaporated water) = 100 g (0.1 kg).

Constants: ρ (density) = 1000 kg / m3; С (heat capacity) = 4200 J / (kg * ºС); tк (boiling) = 100 ºС; L (heat of steam) = 2260 * 10 ^ 3 J / kg; q (heat of combustion) = 2.7 * 10 ^ 7 J / kg.

η = (C * ρ * V * (tк – t0) + L * m2) / (q * m3) and m3 = (C * ρ * V * (tк – t0) + L * m2) / (η * q ) = (4200 * 1000 * 0.002 * (100 – 10) + 2260 * 10 ^ 3 * 0.1) / (0.5 * 2.7 * 10 ^ 7) ≈ 0.073 kg (73 g).