The mass of water vapor is …. kg, if during the condensation and cooling of the resulting water to 20˚C
The mass of water vapor is …. kg, if during the condensation and cooling of the resulting water to 20˚C, an amount of heat equal to 5MJ is released, the specific heat of water is 4200J / (kg * ˚C), the specific heat of vaporization of water is 2.3 MJ / kg
Given:
Q = 5 MJ = 5 * 10 ^ 6 Joule is the energy released during condensation and cooling of water vapor;
c = 4200 J / (kg * C) – specific heat capacity of water;
q = 2.3 MJ / kg = 2.3 * 10 ^ 6 J / kg is the specific heat of vaporization of water.
T2 = 20 degrees Celsius.
It is required to determine the mass of water vapor m (kilogram).
Steam condensation begins at a temperature of T1 = 100 degrees Celsius, taking this into account we get:
Q = Qcondensing + Qcooling
Q = q * m + c * m * (T1 – T) = m * (q + c * (T1 – T), hence:
m = Q / (q + c * (T1 – T) = 5 * 10 ^ 6 / (2.3 * 10 ^ 6 – 4200 * (100 – 20)) =
= 5 * 10 ^ 6 / (2.3 * 10 ^ 6 – 4200 * 80) = 5 * 10 ^ 6 / (2.3 * 10 ^ 6 – 336000) =
= 5,000,000 / 1964,000 = 2.5 kilograms.
Answer: the mass of water vapor is 2.5 kilograms.
