Calculate how much water vapor, t which is 100 ° C, must be admitted to an aluminum

Calculate how much water vapor, t which is 100 ° C, must be admitted to an aluminum calorimeter weighing 100 g, in which there is 150 g of snow at t -20 ° C, in order for all the snow to melt?

t1 = 100 ° C.

mk = 100 g = 0.1 kg.

Ca = 920 J / kg * ° C.

msn = 150 g = 0.15 kg.

Csn = 2100 J / kg * ° C.

t2 = – 20 ° C.

t3 = 0 ° C.

Cw = 4200 J / kg * ° C.

r = 2.3 * 106 J / kg.

q = 3.4 * 105 J / kg.

mp -?

Qp = r * mp + Cw * mp * (t1 – t3).

Qsn = mk * Ca * (t3 – t2) + Csn * msn * (t3 – t2) + q * msn.

Let’s write down the heat balance equation: Qп = Qсн.

r * mp + Cv * mp * (t1 – t3) = mk * Ca * (t3 – t2) + Csn * msn * (t3 – t2) + q * msn.

mp = (mk * Ca * (t3 – t2) + Csn * msn * (t3 – t2) + q * msn) / (r + Cw * (t1 – t3)).

mp = (0.1 kg * 920 J / kg * ° C * (0 ° C – (- 20 ° C)) + 2100 J / kg * ° C * 0.15 kg * (0 ° C – (- 20 ° C)) + 3.4 * 105 J / kg * 0.15 kg) / (2.3 * 106 J / kg + 4200 J / kg * ° C * (100 ° C – 0 ° C)) = 0.022 kg.

Answer: it is necessary to admit mp = 0.022 kg of water vapor into the calorimeter.