How much heat is needed to bring a mixture of 6 kg of water and 9 kg of ice taken at 0 degrees to a boil?

Data: m1 (mass of water in the mixture) = 6 kg; m2 (mass of ice in the mixture) = 9 kg; t (mixture temperature) = 0 ºС.

Constants: λ (specific heat of melting of ice) = 34 * 10 ^ 4 J / kg; C (specific heat of water) = 4200 J / (kg * ºС); tboil (temperature of the beginning of boiling) = 100 ºС.

1) The transition of ice into water: Q1 = λ * m2 = 34 * 10 ^ 4 * 9 = 306 * 10 ^ 4 J.

2) Heating the total mass of water to boiling: Q2 = C * (m1 + m2) * (tboil – t) = 4200 * (6 + 9) * (100 – 0) = 6300000 J.

3) Heat: Q = Q1 + Q2 = 306 * 104 + 6300000 = 9.36 MJ.