Water weighing 400 grams and a temperature of 10 degrees was admitted to 10 grams of water vapor

Water weighing 400 grams and a temperature of 10 degrees was admitted to 10 grams of water vapor with a temperature of 100 degrees. Determine what temperature will be established in the vessel.

mw = 400 g = 0.4 kg.

tv = 10 ° C.

mp = 10 g = 0.01 kg.

tp = 100 ° C.

C = 4200 J / kg * ° C.

λ = 2.3 * 10 ^ 6 J / kg.

t -?

When mixed, the steam will first condense into water and the resulting water will cool. The amount of heat will be released from the steam: Q = λ * mp + C * mp * (tp – t), where λ is the specific heat of vaporization of water, mp is the mass of steam, C is the specific heat capacity of water, tp is the steam temperature, t is the temperature of the resulting water.

The water will receive an amount of heat from the steam and heat up.

The amount of heat during heating is determined by the formula: Q = C * mw * (t – tv).

λ * mp + C * mp * (tp – t) = C * mw * (t – tv).

λ * mp + C * mp * tp – C * mp * t = C * mw * t – C * mw * tv.

λ * mp + C * mp * tp + C * mw * tv = C * mp * t + C * mw * t.

λ * mp + C * mp * tp + C * mw * tv = (mp + mw) * C * t.

t = (λ * mp + C * mp * tp + C * mw * tv) / (mp + mw) * C.

t = (2.3 * 10 ^ 6 J / kg * 0.01 kg + 4200 J / kg * ° C * 0.01 kg * 100 ° C + 4200 J / kg * ° C * 0.4 kg * 10 ° C) / (0.01 kg + 0.4 kg) * 4200 J / kg * ° C = 25.5 ° C.

Answer: the temperature will be set at t = 25.5 ° C.