The jar contains 1 liter of hot water at a temperature of 80ºC. What will the temperature of the water

The jar contains 1 liter of hot water at a temperature of 80ºC. What will the temperature of the water be if you throw 100g of ice at 0ºC into it? The density of water is 1000 kg / m³, the specific heat of water is 4200 J / (kg • ºC), the specific heat of ice is 2100 J / (kg • ºC), the specific heat of melting of ice ..

Vv = 1 l = 1 * 10 ^ -3 m ^ 3.

ρ = 1000 kg / m ^ 3.

t1 = 80 ºC.

ml = 100 g = 0.1 kg.

t2 = 0 ºC.

Sv = 4200 J / kg * ºC.

Сl = 2100 J / kg * ºC.

q = 3.4 * 10 ^ 5 J / kg.

t -?

The amount of thermal energy Qw, which the water gives off during cooling, is determined by the formula: Qw = Sv * Vw * ρ * (t1 – t).

The amount of thermal energy Ql, which is necessary for melting ice and heating water, can be found by the formula:

Ql = Sl * ml * (t – t2) + q * ml.

The heat balance equation will have the form: Sv * Vw * ρ * (t1 – t) = Сl * ml * (t – t2) + q * ml.

Sv * Vv * ρ * t1 – Sv * Vv * ρ * t = Sl * ml * t – Sl * ml * t2 + q * ml.

Sv * Vv * ρ * t1 + Sl * ml * t2 – q * ml = Sl * ml * t + Sv * Vw * ρ * t.

Sv * Vw * ρ * t1 + Sl * ml * t2 – q * ml = (Sl * ml + Sv * Vw * ρ) * t.

t = (Sv * Vв * ρ * t1 + Сl * ml * t2 – q * ml) / (Сl * ml + Сl * Vв * ρ).

t = (4200 J / kg * ºC * 1 * 10 ^ -3 m ^ 3 * 1000 kg / m ^ 3 * 80 ºC + 2100 J / kg * ºC * 0.1 kg * 0 ºC – 3.4 * 10 ^ 5 J / kg * 0.1 kg) / (2100 J / kg * ºC * 0.1 kg + 4200 J / kg * ºC * 1 * 10 ^ -3 m ^ 3 * 1000 kg / m ^ 3) = 68.5 ºC.

Answer: the water temperature will become t = 68.5 ºC.