Vw = 80 l = 0.08 m ^ 3.
t1 = 6 ° C.
t2 = 35 ° C.
ρ = 1000 kg / m ^ 3.
С = 4200 J / kg * ° С.
λ = 2.3 * 10 ^ 6 J / kg.
To heat water, you need the amount of thermal energy Q, which is determined by the formula: Q = C * mw * (t2 – t1), where C is the specific heat capacity, mw is the mass of water, t2, t1 are the final and initial water temperatures.
We find the mass of water mw by the formula: mw = Vw * ρ.
Q = С * Vв * ρ * (t2 – t1).
The same amount of heat Q should be released by steam during condensation.
The amount of thermal energy Q during condensation is determined by the formula: Q = λ * mp, where λ is the specific heat of vaporization, mp is the mass of the vapor that has condensed.
С * Vв * ρ * (t2 – t1) = λ * mp.
mп = С * Vв * ρ * (t2 – t1) / λ.
mp = 4200 J / kg * ° C * 0.08 m ^ 3 * 1000 kg / m ^ 3 * (35 ° C – 6 ° C) / 2.3 * 10 ^ 6 J / kg = 4.24 kg.
Answer: you need mp = 4.24 kg of steam.
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