A piece of ice with a mass of m2 = 4 kg, having a temperature of t2 = -20s, was placed in water with a mass of m2 = 10 kg

A piece of ice with a mass of m2 = 4 kg, having a temperature of t2 = -20s, was placed in water with a mass of m2 = 10 kg, having a temperature of t1 = 30c, determine the steady-state temperature and composition of the system after the onset of thermal equilibrium.

m1 = 10 kg.

t1 = 30 ° C.

C1 = 4200 J / kg * ° C.

m2 = 4 kg.

t2 = – 20 ° C.

C2 = 2100 J / kg * ° C.

t3 = 0 ° C.

λ = 3.4 * 10 ^ 5 J / kg.

Let’s find the amount of heat Q1 that will be released when the water is cooled to a temperature of t3 = 0 ° C according to the formula: Q1 = C1 * m1 * (t1 – t3).

Q1 = 4200 J / kg * ° C * 10 kg * (30 ° C – 0 ° C) = 1260000 J.

Let us find the amount of heat Q2, which is necessary to heat ice to a temperature of t3 = 0 ° C, according to the formula: Q2 = C2 * m2 * (t3 – t1).

Q2 = 2100 J / kg * ° C * 4 kg * (0 ° C – (- 20 ° C)) = 168000 J.

Q3 = Q1 – Q2.

Q3 = λ * m3, where m3 is the mass of ice that will melt.

m3 = (Q1 – Q2) / λ.

m3 = (1260000 J – 168000 J) / 3.4 * 10 ^ 5 J / kg = 3.2 kg.

ml = m2 – m3 = 4 kg – 3.2 kg = 0.8 kg.

Answer: the temperature t3 = 0 ° C will be established at which there will be ice ml = 0.8 kg and water mw = 13.2 kg.