In an aluminum calorimeter weighing 0.20 kg, containing 0.34 kg of water at 23.5C, 81.5 g of ice at 0C

In an aluminum calorimeter weighing 0.20 kg, containing 0.34 kg of water at 23.5C, 81.5 g of ice at 0C was dropped. All the ice has melted. Find the temperature established in the colorimeter.

Given: m1 (calorimeter) = 0.2 kg; m2 (water) = 0.34 kg; t1 (water, calorimeter) = 23.5 ºС; m3 (ice) = 81.5 g (0.0815 kg); t0 (ice) = 0 ºС.

Constants: Ca (aluminum) = 920 J / (kg * ºС); Sv (water) = 4200 J / (kg * ºС); λ (ice) = 3.4 * 10 ^ 5 J / kg.

(Ca * m1 + Cw * m2) * (t1 – tp) = λ * m3 + Sv * m3 * (tp – 0).

(920 * 0.2 + 4200 * 0.34) * (23.5 – tr) = 3.4 * 10 ^ 5 * 0.0815 + 4200 * 0.0815 * tr.

1612 * (23.5 – tr) = 27710 + 342.3tr.

37882 – 1612tr = 27710 + 342.3tr.

1954.3tp = 10172.

tр = 10172 / 1954.3 = 5.2 ºС.