0.2 kg of water vapor at 100 degrees Celsius was introduced into water taken at 15 degrees Celsius.

0.2 kg of water vapor at 100 degrees Celsius was introduced into water taken at 15 degrees Celsius. Find the initial mass of water if, after condensation of steam, a temperature of 318 K.

To solve the problem, we use the formula for the heat balance equation:

Qwater = Qpair;

Qwater = s * m * (t2 – t1);

Qpair = Q1 + Q2, Q1 = r * m – steam condensation, Q2 = c * m * (t2 – t1) – steam cooling;

Q1 = 2,300,000 * 0.2 = 460,000 J;

Final temperature resulting from mixing water and steam:

T = 318 K = 318 – 273 = 45 C;

Q2 = 4200 * 0.2 * (100 – 45) = 46200 J;

Qpair = 460000 + 46200 = 506200 J;

mwater = Qpair / (c * (t2-t1));

mwater = 506200 / (4200 * (45-15)) = 4 kg.