# How many kilograms of centigrade steam is required to heat 80 liters of water from 6 to 35 ° C?

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.

mp -?

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.