An iron calorimeter weighing 200 grams has 100 grams of ice at a temperature of -15 degrees

An iron calorimeter weighing 200 grams has 100 grams of ice at a temperature of -15 degrees, water vapor is released in the calorimeter, weighing 50 grams, which has a temperature of 100 degrees. What temperature is set in the calolimeter?

mk = 200 g = 0.2 kg.

ml = 100 g = 0.1 kg.

t1 = – 15 ° C.

t2 = 0 ° C.

Cl = 2100 J / kg * ° C.

Cg = 460 J / kg * ° C.

Cw = 4200 J / kg * ° C.

λl = 3.4 * 105 J / kg.

t3 = 100 ° C.

mp = 50 g = 0.05 kg.

q = 2.3 * 106 J / kg.

t -?

During the condensation of steam, the amount of heat Qp will be released, which is determined by the formula: Qp = q * mp.

Qp = 2.3 * 10 ^ 6 J / kg * 0.05 kg = 115000 J.

For heating ice in the calorimeter and melting it, the required amount of heat Ql is expressed by the formula: Ql = Czh * mk * (t2 – t1) + Cl * ml * (t2 – t1) + λl * ml.

Ql = 460 J / kg * ° C * 0.2 kg * (0 ° C – (- 15 ° C)) + 2100 J / kg * ° C * 0.1 kg * (0 ° C – (- 15 ° C)) + 3.4 * 105 J / kg * 0.1 kg = 38530 J.

The amount of heat Q, which goes to heating the water, which is obtained from ice, is expressed by the formula: Q = Qp – Ql = 115000 J – 38530 J = 76470 J.

Let’s find the required amount of heat for heating water according to the formula: Q = Czh * mk * (t3 – t2) + Cw * ml * (t3 – t2) = (Czh * mk + Cw * ml) * (t3 – t2).

Q = (460 J / kg * ° C * 0.2 kg + 4200 J / kg * ° C * 0.1 kg) * (100 ° C – 0 ° C) = 51200 J.

We see that for heating water, the required amount of heat is less than that released during steam condensation.

Answer: in the calorimeter, the temperature becomes t3 = 100 ° C.