A calorimeter containing 0.4 kg of ice at a temperature of –55 ° C was poured with 100 g of water at 15 ° C.

A calorimeter containing 0.4 kg of ice at a temperature of –55 ° C was poured with 100 g of water at 15 ° C. What will be the temperature of the ice formed in the calorimeter?

Data: m1 (ice) = 0.4 kg; t1 (ice temperature) = -55 ºС; m2 (water) = 100 g (0.1 kg); t2 (water temperature) = 15 ºС.

Constants: λ (beats heat of melting of ice) = 3.34 * 10 ^ 5 J / kg; Sv (specific heat capacity of water) = 4200 J / (kg * ºС); Сl (specific heat capacity of ice) = 2100 J / (kg * ºС).

Sl * m1 * (tl – t) = Sv * m2 * (t2 – 0) + λ * m2 + Sl * m2 * (t – 0).

2100 * 0.4 * (55 – t) = 4200 * 0.1 * 15 + 3.34 * 10 ^ 5 * 0.1 + 2100 * 0.1 * t.

46200 – 840t = 6300 + 33400 + 210t.

1050t = 6500.

t = 6500/1050 = 6.19 ºС.