It is required to heat 80 liters of water taken at a temperature of 6 ° C to 35 ° C by passing steam.

It is required to heat 80 liters of water taken at a temperature of 6 ° C to 35 ° C by passing steam. How much 100-degree steam is required for this?

The steam will give off heat during condensation and when the condensate is cooled to water temperature.
The heat of the cooled steam will heat the water:
L * Mp + c * Mp * (Tp – T) = s * Mw * (T – T₀),
where L is the specific heat of condensation,
Mп – steam mass,
Mw is the mass of water.
с – specific heat capacity of water,
Tp – steam temperature,
T – water temperature after heating,
T₀ is the initial water temperature.
Mp * (L + s (Tp – T) = s * Mw * (T – T₀)
Mp = s * Mw * (T – T₀) / (L + s (Tp – T)) = (4183 J / (kg • deg) * 80 kg (35 ° C – 6 ° C)) / (2260000 J / kg + 4183 J / (kg • deg) * (100 ° С – 35 ° С)) = 3.83 kg.