Mixed 6 liters of cold water, having a temperature of 8 C, with 2 liters of hot water at a temperature

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.

Given:

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.