An aluminum pan weighing 500 grams contains 300 grams of ice at a temperature of -10 degrees.

An aluminum pan weighing 500 grams contains 300 grams of ice at a temperature of -10 degrees. How much heat is required to turn this ice into water with a temperature of 50 degrees.

Given: m1 (saucepan) = 500 g (0.5 kg); m2 (ice) = 300 g (0.3 kg); t0 (initial temperature) = -10 ºС; t (end temperature) = 50 ºС.

Constants: Сl (specific heat capacity of ice) = 2100 J / (kg * ºС); Ca (specific heat capacity of aluminum) = 920 J / (kg * ºС); λ (specific heat of melting of ice) = 3.4 * 10 ^ 5 J / kg; Sv (specific heat capacity of water) = 4200 J / (kg * ºС).

Q = (Сl * m2 + Ca * m1) * (0 – t0) + λ * m2 + (Sv * m2 + Ca * m1) * t = (2100 * 0.3 + 920 * 0.5) * (0 + 10) + 3.4 * 105 * 0.3 + (4200 * 0.3 + 920 * 0.5) * 50 = 193230 J.