In a vessel with water 500 g at a temperature of 80 degrees, poured 300 g of water

In a vessel with water 500 g at a temperature of 80 degrees, poured 300 g of water at a temperature of 30 degrees which is the established temperature.

mg = 500 g = 0.5 kg.

tg = 80 ° C.

mx = 300 g = 0.3 kg.

tx = 30 ° C.

t -?

When mixing water of different temperatures, a process of heat exchange occurs, as a result, water with a higher temperature gives off a certain amount of heat to water with a lower temperature.

Qg = C * mg * (tg – t).

Qx = C * mx * (t – tx).

Qg = Qx.

C * mg * (tg – t) = C * mх * (t – tх) – heat balance equation.

mg * tg – mg * t = mх * t – mх * tх.

mg * tg + mx * tx = mg * t + mx * t.

t = (mg * tg + mx * tx) / (mg + mx).

t = (0.5 kg * 80 ° C + 0.3 kg * 30 ° C) / (0.5 kg + 0.3 kg) = 61.25 ° C.

Answer: the temperature in the vessel is t = 61.25 ° C.