Mixed 6 liters of cold water, having a temperature of 8 C, with 2 liters of hot water at a temperature of 80 C. Determine the temperature of the sweep.
T1 = 8 degrees Celsius – cold water temperature;
V1 = 6 liters = 0.006 cubic meters – the volume of cold water;
T2 = 80 degrees Celsius – hot water temperature;
V2 = 2 liters = 0.002 cubic meters – the volume of hot water.
It is required to determine T (degree Celsius) – the temperature of the mixture.
According to the laws of thermodynamics:
Q1 = Q2;
c * m1 * (T – T1) = c * m2 * (T2 – T), where c is the specific heat capacity of water;
m1 * (T – T1) = m2 * (T2 – T);
ro * V1 * (T – T1) = ro * V2 * (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) = (0.002 * 80 + 0.006 * 8) / (0.002 + 0.006) =
= (0.16 + 0.048) / 0.008 = 0.208 / 0.008 = 26 degrees Celsius.
Answer: The temperature of the mixture will be 26 degrees Celsius.