Mixed 24 liters of water at 12 ° C and 40 liters of water at 80 ° C. Determine the final temperature

Mixed 24 liters of water at 12 ° C and 40 liters of water at 80 ° C. Determine the final temperature of the mixture if during mixing the heat loss was 420 kJ.

V1 = 24 l = 24 * 10 ^ -3 m ^ 3.

t1 = 12 ° C.

V2 = 40 l = 40 * 10 ^ -3 m ^ 3.

t2 = 80 ° C.

Qп = 420 kJ = 420,000 J.

С = 4200 J / kg * ° С.

ρ = 1000 kg / m ^ 3.

t -?

Let us write down the heat balance equation: Q1 + Qp = Q2, where Q1 is the amount of heat that goes to heat water with a volume of V1 from temperature t1 to t, Q2 is the amount of heat that the volume of water V2 gives off during cooling from temperature t2 to t.

Q1 = C * m1 * (t – t1).

Q2 = C * m2 * (t2 – t).

We express the mass of water m by the formula: m = V * ρ.

The heat balance equation will take the form: С * V1 * ρ * (t – t1) + Qп = С * V2 * ρ * (t2 – t).

С * V1 * ρ * t – С * V1 * ρ * t1 + Qп = С * V2 * ρ * t2 – С * V2 * ρ * t.

С * V1 * ρ * t + С * V2 * ρ * t = С * V1 * ρ * t1 – Qп + С * V2 * ρ * t2.

С * ρ * t * (V1 + V2) = С * V1 * ρ * t1 – Qп + С * V2 * ρ * t2.

t = (С * V1 * ρ * t1 – Qп + С * V2 * ρ * t2) / С * ρ * (V1 + V2).

t = (4200 J / kg * ° C * 24 * 10 ^ -3 m ^ 3 * 1000 kg / m ^ 3 * 12 ° C – 420,000 J + 4200 J / kg * ° C * 40 * 10 ^ -3 m ^ 3 * 1000 kg / m ^ 3 * 80 ° C) / 4200 J / kg * ° C * 1000 kg / m ^ 3 * (24 * 10 ^ -3 m ^ 3 + 40 * 10 ^ -3 m ^ 3 ) = 53 ° C.

Answer: the water temperature will be set at t = 53 ° С.