# In a mixture consisting of 5 kg of water and 3 kg of ice, 0.2 kg of water vapor was admitted

**In a mixture consisting of 5 kg of water and 3 kg of ice, 0.2 kg of water vapor was admitted at a temperature of 100 ° C. How much water will there be after thermal equilibrium is established?**

Data: mw (mass of water in the mixture) = 5 kg; ml (initial ice mass) = 3 kg; mp (mass of inlet steam) = 0.2 kg; tp (initial steam temperature) = 100 ºС.

Constants: λl (specific heat of melting of ice) = 3.4 * 10 ^ 5 J / kg; Lp (beats heat of condensation of steam) = 2.3 * 10 ^ 6 J / kg.

1) Heat obtained due to condensation and cooling of 0.2 kg of water vapor: Qp = (Lp + Sv * Δt) * mp = (2.3 * 10 ^ 6 + 4200 * 100) * 0.2 = 544 * 10 ^ 3 J.

2) The mass of melted ice: mrl = Qp / λl = 544 * 10 ^ 3 / (3.4 * 10 ^ 5) = 1.6 kg.

3) The resulting mass of water: m = mw + mrl + mp = 5 + 1.6 + 0.2 = 6.8 kg.

Answer: There will be 6.8 kg of water.