To 5 kg of water at 12 degrees, hot water was added at 50 degrees, having received a mixture with a temperature

To 5 kg of water at 12 degrees, hot water was added at 50 degrees, having received a mixture with a temperature of 30 degrees, how much water was added.

Q = С * m * (tк – tн), where С is the specific heat capacity of water, m is the mass of water, tк is the final water temperature, tн is the initial water temperature.
The amount of heat given off by hot water:
Q1 = С * m1 * (t1 – t3), where t1 = 50 ° С, t3 = 30 ° С.
The amount of heat received by cold water:
Q2 = С * m2 * (t3 – t2), where m2 = 5 kg, t2 = 30 ° С.
Q1 = Q2.
С * m1 * (t1 – t3) = С * m2 * (t3 – t2).
m1 * (t1 – t3) = m2 * (t3 – t2).
m1 * (50 – 30) = 5 * (30 – 12).
20m1 = 90
m1 = 90/20 = 4.5 kg.