How much do you need to mix hot water with a temperature of 80 ° C and cold water with a temperature of 20 ° C

How much do you need to mix hot water with a temperature of 80 ° C and cold water with a temperature of 20 ° C to get 60 liters of water with a temperature of 40 ° C?

t1 = 80 ° C.

t2 = 20 ° C.

V = 60 l = 0.06 m3.

t3 = 40 ° C.

C = 4200 J / kg * ° С.

ρ = 1000 kg / m3.

mg -?

mx -?

The amount of heat Qg, which hot water will give when mixing, is expressed by the formula: Qg = C * mg * (t1 – t3).

The amount of heat Qx that cold water will receive when mixing is expressed by the formula: Qx = C * mx * (t3 – t2).

The heat balance equation will look like: C * mg * (t1 – t3) = C * mх * (t3 – t2).

mg * (t1 – t3) = mх * (t3 – t2).

mg + mх = ρ * V.

mg = ρ * V – mх.

(ρ * V – mх) * (t1 – t3) = mх * (t3 – t2).

ρ * V * (t1 – t3) – mх * (t1 – t3) = mх * (t3 – t2).

ρ * V * (t1 – t3) = mх * (t1 – t3) + mх * (t3 – t2).

mх = ρ * V * (t1 – t3) / ((t1 – t3) + (t3 – t2)).

mх = 1000 kg / m3 * 0.06 m3 * (80 ° С – 40 ° С) / ((80 ° С – 40 ° С) + (40 ° С – 20 ° С)) = 40 kg.

mg = 1000 kg / m3 * 0.06 m3 – 40 kg = 20 kg.

Answer: mх = 40 kg, mg = 20 kg.