The boiler of the experimental steam engine has a volume of 10 liters and is initially completely filled with water.

The boiler of the experimental steam engine has a volume of 10 liters and is initially completely filled with water. The boiler is heated by burning coal, while a power of 10 kW is supplied to it. The generated steam does work in the steam engine, and 95% of its mass is returned to the boiler in the form of water at a temperature of 20 ° C. Determine how long (in seconds) the boiler will be half empty. The heat of vaporization of water is 2.26 MJ / kg, the specific heat of water is 4200 J / (kg • ºС). The answer is a whole number, if necessary round to whole numbers.

Let water with mass m evaporate in a second. Then 0.95m of water returns per second. Power P is spent on evaporating and heating these masses of water:
P = 0.95mc * (100 ° C – 20 ° C) + λm,
where c is the specific heat of water, λ is the specific heat of vaporization.
10000 J / s = 0.95 * 4200 J / kg * deg * 80 ° C * m + 2.26 * 106 J / kg * m
m = 10000 J / s / (0.95 * 4200 J / kg * deg * 80 ° C + 2.26 * 106 J / kg) = 0.003877 kg / s.
5% of m is finally evaporated:
m1 = 0.05 * 0.003877 kg / s = 0.000194 kg / s
Evaporation time for 5 liters of water (5 kg):
t = 5 kg / 0.000194 kg / s = 25793.14 s ≈ 7 h 10 m.
Answer: 7 h 10 m.