A piece of ice weighing 4 kg at a temperature of -20 degrees Celsius was immersed in water having a temperature

A piece of ice weighing 4 kg at a temperature of -20 degrees Celsius was immersed in water having a temperature of 80 degrees Celsius. The mass of water is 10 kg. What temperature will the water have when all the ice has melted? The specific heat capacity of water is 4200 J / (kg * degree Celsius), the specific heat capacity of ice is 2100 J / (kg * degree Celsius), and the specific heat of melting of ice is 340 kJ / kg.

ml = 4 kg.

t1 = – 20 ° C.

t2 = 80 ° C.

mw = 10 kg.

Cw = 4200 J / kg * ° C.

Cl = 2100 J / kg * ° C.

λ = 340 kJ / kg = 340,000 J / kg.

t -?

Ql = Cl * ml * (t – t1) + λ * ml.

Qw = Cw * mw * (t2 – t).

Ql = Qv.

Cl * ml * (t – t1) + λ * ml = Cw * mw * (t2 – t).

Cl * ml * t – Cl * ml * t1 + λ * ml = Cw * mw * t2 – Cw * mw * t.

Cl * ml * t + Cw * mw * t = Cl * ml * t1 – λ * ml + Cw * mw * t2.

(Cl * ml + Cw * mw) * t = Cl * ml * t1 – λ * ml + Cw * mw * t2.

t = (Cl * ml * t1 – λ * ml + Cw * mw * t2) / (Cl * ml + Cw * mw).

t = (2100 J / kg * ° C * 4 kg * (- 20 ° C) – 340,000 J / kg * 4 kg + 4200 J / kg * ° C * 10 kg * 80 ° C) / (2100 J / kg * ° C * 4 kg + 4200 J / kg * ° C * 10 kg) = 36.4 ° C.

Answer: the water will have a temperature of t = 36.4 ° C.