A piece of ice 1 kg with a temperature of 30 grams of Celsius is thrown into water weighing 2 kg

A piece of ice 1 kg with a temperature of 30 grams of Celsius is thrown into water weighing 2 kg with a temperature of 30 degrees Celsius, what is the final temperature?

Data: m1 (water) = 2 kg; t1 (water) = 30 ºС; m2 (ice) = 1 kg; t2 (ice) = -30 ºС.

Constants: Сl = 2100 J / (kg * ºС); λ = 3.4 * 10 ^ 5 J / kg; Sv = 4200 J / (kg * ºС).

1) Cooling water to 0 ºС: Q1 = Cw * m1 * (t1 – 0) = 4200 * 2 * 30 = 252 000 J.

2) Ice heating up to 0 ºС: Q2 = Cl * m2 * (0 – t2) = 2100 * 1 * 30 = 63,000 J (the remainder is ΔQ = Q1 – Q2 = 189,000 J).

3) Ice melting: Q3 = λ * m2 = 3.4 * 105 * 1 = 340,000 J> 189,000 J (final temperature 0 ºС and a piece of ice weighing 0.444 kg will remain).