In 500 g of water at a temperature of 15 ° C, 75 g of water vapor is injected at 100 ° C. Find the final temperature

In 500 g of water at a temperature of 15 ° C, 75 g of water vapor is injected at 100 ° C. Find the final temperature of the water in the vessel. (The specific heat capacity of water is 4200 J / kg “° C; the specific heat of vaporization is 2.3 • 105 J / kg.)

Given:

m1 = 500 grams = 0.5 kilograms is the mass of water;

t0 = 15 ° Celsius – initial water temperature;

m2 = 75 grams = 0.075 kilograms – the mass of water vapor;

t1 = 100 ° Celsius – water vapor temperature;

c = 4200 J / (kg * C) – specific heat capacity of water;

q = 2.3 * 105 J / kg is the specific heat of vaporization of water.

It is required to determine t (degree Celsius) – the final temperature of the mixture.

The energy required to heat the water will be equal to the sum of the energy released when the water vapor is cooled and the resulting water is cooled to the temperature of the mixture, that is:

Q1 = Q2 + Q2;

c * m1 * (t – t0) = q * m2 + c * m2 * (t1 – t);

c * m1 * t – c * m1 * t0 = q * m2 + c * m2 * t1 – c * m2 * t;

c * m1 * t + c * m2 * t = q * m2 + c * m2 * t1 + c * m1 * t0;

t * c * (m1 + m2) = q * m2 + c * m2 * t1 + c * m1 * t0;

t = (q * m2 + c * m2 * t1 + c * m1 * t0) / (c * (m1 + m2));

t = (230,000 * 0.075 + 4200 * 0.075 * 100 + 4200 * 0.5 * 15) / (4200 * (0.5 + 0.075));

t = (17250 + 31500 + 31500) / (4200 * 0.575);

t = 80250/2415 = 33.2 ° Celsius.

Answer: The temperature of the mixture will be 33.2 ° C.