120 liters of water were mixed in a bath at a temperature of 10 degrees, with 160 liters

120 liters of water were mixed in a bath at a temperature of 10 degrees, with 160 liters of water at a temperature of 70 degrees. What is the temperature of the resulting mixture?

Vх = 120 l = 0.12 m3.

C = 4200 J / kg * ° C.

tx = 10 ° C.

Vg = 160 l = 0.16 m3.

tg = 70 ° C.

ρ = 1000 kg / m3.

t -?

If we neglect the losses of thermal energy that goes to heating the bath, then hot water, when cooling, will give up as much thermal energy Qg as cold water will take when heating Qx: Qg = Qx.

Qg = C * mg * (tg – t) = C * ρ * Vg * (tg – t).

Qx = C * mx * (tg – t) = C * ρ * Vx * (t – tx).

C * ρ * Vg * (tg – t) = C * ρ * Vx * (t – tx).

Vg * tg – Vg * t = Vx * t – Vx * tx.

Vx * t + Vg * t = Vg * tg + Vx * tx.

The water temperature t, which is established in the bathroom, will be determined by the formula: t = (Vg * tg + Vx * tx) / (Vx + Vg).

t = (0.16 m3 * 70 ° C + 0.12 m3 * 10 ° C) / (0.12 m3 + 0.16 m3) = 44.3 ° C.

Answer: after mixing, the temperature in the bathroom will be t = 44.3 ° C.