An iron tank weighing 5 kg contains 20 kg of water and 6 kg of ice at 0 degrees. How much water vapor

An iron tank weighing 5 kg contains 20 kg of water and 6 kg of ice at 0 degrees. How much water vapor at 100 degrees should be introduced into the tank to melt the ice and heat the water to 70 degrees.

To determine the mass of water vapor introduced into the tank, we use the equality: mp * (L + Cw * (tp – t)) = λl * ml + Sv * (ml + mw) * t + Czh * mb * t, from where we express: mp = (λl * ml + Sv * (ml + mv) * t + Czh * mb * t) / (L + Cw * (tp – t)).

Constants and variables: λl – beats. heat of melting of ice (λl = 34 * 10 ^ 4 J / kg); ml is the mass of ice in the tank (ml = 6 kg); Cv – beats heat capacity of water (Cw = 4.2 * 10 ^ 3 J / (kg * K)); mw is the mass of water before steam injection (mw = 20 kg); t – equilibrium temperature (t = 70 ºС); Szh – beats heat capacity of iron (Czh = 460 J / (kg * K)); mb is the mass of the iron tank (mb = 5 kg); L – beats. heat of condensation of water vapor (L = 2.3 * 10 ^ 6 J / kg); tp is the temperature of the steam before it is introduced into the tank (tp = 100 ºС).

Calculation: mp = (34 * 10 ^ 4 * 6 + 4.2 * 10 ^ 3 * (6 + 20) * 70 + 460 * 5 * 70) / (2.3 * 10 ^ 6 + 4.2 * 10 ^ 3 * (100 – 70)) = 4.06 kg.

Answer: You need to enter 4.06 kg of water vapor into the tank.