A calorimeter in which 100 grams of ice at a temperature of 0 degrees was admitted 100 grams

A calorimeter in which 100 grams of ice at a temperature of 0 degrees was admitted 100 grams of water vapor at 100 degrees. What is the temperature in the calorimeter?

ml = 100 g = 0.1 kg.

mp = 100 g = 0.1 kg.

t1 = 0 ° C.

t2 = 100 ° C.

q = 3.4 * 10 ^ 5 J / kg.

λ = 2.3 * 10 ^ 6 J / kg.

С = 4200 J / kg * ° C.

t -?

When steam condenses and water cools down, the amount of heat Qp is released, which is determined by the formula: Qp = λ * mp.

Qp = 2.3 * 10 ^ 6 J / kg * 0.1 kg = 230,000 J.

To melt ice and heat the resulting water, you need the amount of heat Ql = q * ml + C * ml * (t2 – t1).

Ql = 3.4 * 10 ^ 5 J / kg * 0.1 kg + 4200 J / kg * ° C * 0.1 kg * (100 ° C – 0 ° C) = 76000 J.

We see that only mp “= Ql / λ can condense.

mp “= 76000 J / 2.3 * 10 ^ 6 J / kg = 0.033 kg.

Answer: the temperature in the calorimeter will be set at t2 = 100 ° C, and it will contain mw = 0.133 kg of water and mp “= 0.067 kg of steam.