Mixed 0.8 kg of water having a temperature of 25 degrees, with 0.2 kg of boiling water. What is the temperature of the mixture?

Mixed 0.8 kg of water having a temperature of 25 degrees, with 0.2 kg of boiling water. What is the temperature of the mixture? How much heat did the boiling water give off? How much heat did the water receive?

mх = 0.8 kg.

tx = 25 ° C.

mk = 0.2 kg.

tк = 100 ° C.

C = 4200 J / kg * ° C.

t -?

Qx -?

Qk -?

When mixing water of different temperatures tх and tк, due to heat exchange, the temperature t will be established. Boiling water will give Qk, and cold water will receive the amount of heat Qx.

Qk = Qx.

Boiling water, upon cooling, will give up the amount of heat Qk, which we express by the formula: Qk = C * mk * (tk – t).

Cold water will take the amount of heat Qx, which we express by the formula: Qx = C * mx * (t – tx).

C * mk * (tk – t) = C * mx * (t – tx).

mk * tk – mk * t = mх * t – mх * tх.

(mk + mx) * t = mk * tk + mx * tx.

t = (mk * tk + mx * tx) / (mk + mx).

t = (0.2 kg * 100 ° C + 0.8 kg * 25 ° C) / (0.2 kg + 0.8 kg) = 40 ° C.

Qк = 4200 J / kg * ° C * 0.2 kg * (100 ° C – 40 ° C) = 50400 J.

Qx = 4200 J / kg * ° C * 0.8 kg * (40 ° C – 25 ° C) = 50400 J.

Answer: after mixing, the temperature will be set to t = 40 ° C, Qx = Qc = 50400 J.