How much cold water with a temperature of 8 degrees should be added to 80 liters

How much cold water with a temperature of 8 degrees should be added to 80 liters of hot water with a temperature of 90 degrees to get a bath with a temperature of 35 degrees.

Given:

V1 = 80 liters – hot water volume;

T1 = 90 degrees Celsius – hot water temperature;

T2 = 8 degrees Celsius – cold water temperature;

T3 = 35 degrees Celsius – required mixture temperature.

It is required to determine the volume of cold water V2 to obtain the mixture temperature T3.

According to the laws of thermodynamics, the energy released when cooling hot water will be used to heat cold water, that is:

Q1 = Q2;

m1 * c * (T1 – T3) = m2 * c * (T3 – T2);

V1 * ro * c * (T1 – T3) = V2 * ro * (T3 – T2);

V1 * (T1 – T3) = V2 * (T3 – T2);

V2 = V1 * (T1 – T3) / (T3 – T2) = 80 * (90 – 35) / (35 – 8) = 80 * 55/27 = 4400/27 = 163 liters.

Answer: you need cold water in a volume of 163 liters.