3 liters of boiling water were poured into a saucepan with 2 liters of water at a temperature of 25 degrees Celsius. What is the water temperature?
V1 = 2 liters = 0.002 m3 – the volume of water in the pan;
T1 = 25 degrees Celsius – the temperature of the water in the pan;
V2 = 3 liters = 0.003 m3 – boiling water volume;
T2 = 100 degrees Celsius – boiling water temperature.
It is required to determine T (degree Celsius) – the temperature of the mixture.
According to the law of thermodynamics, the amount of heat required to heat cold water will be equal to the amount of heat released when boiling water is cooled. Then:
Q1 = Q2;
c * m1 * (T – T1) = c * m2 * (T2 – T), where c is the specific heat of water, m1 and m2 are the masses of cold water and boiling water;
V1 * ro * (T – T1) = V2 * ro * (T2 – T), where ro is the density of water;
V1 * (T – T1) = V2 * (T2 – T);
V1 * T – V1 * T1 = V2 * T2 – V2 * T;
V1 * T + V2 * T = V2 * T2 + V1 * T1;
T * (V1 + V2) = V2 * T2 + V1 * T1;
T = (V2 * T2 + V1 * T1) / (V1 + V2) = (0.003 * 100 + 0.002 * 25) / (0.003 + 0.002) =
= (0.3 + 0.05) / 0.005 = 0.35 / 0.005 = 70 degrees Celsius.
Answer: The temperature of the mixture will be 70 degrees Celsius.