A piece of ice weighing 100 grams and a temperature of -10 degrees Celsius was poured into 1.5 kg

A piece of ice weighing 100 grams and a temperature of -10 degrees Celsius was poured into 1.5 kg of molten lead at the melting temperature. How much water will turn into steam if the lead has cooled to a temperature of 27 degrees Celsius?

ml = 100 gr = 0.1 kg.

mw = 1.5 kg.

Cw = 140 J / kg * “C.

Cl = 2100 J / kg * “C.

Cw = 4200 J / kg * “C.

λw = 0.25 * 10 ^ 5 J / kg.

λl = 3.4 * 10 ^ 5 J / kg.

kw = 2.3 * 10 ^ 6 J / kg.

t1 = -10 “C.

t2 = 0 “C.

t3 = 327 “C.

t4 = 27 “C.

t5 = 100 “C.

mv -?

Q = λsv * msv + Csv * msv * (t3 – t4).

Q = Cl * ml * (t2 – t1) + λl * ml + Cw * ml * (t5 – t2) + kw * mw.

λw * mw + Cw * mw * (t3 – t4) = Cl * ml * (t2 – t1) + λl * ml + Cw * ml * (t5 – t2) + kw * mw.

mw = (λw * mw + Cw * mw * (t3 – t4) – Cl * ml * (t2 – t1) – λl * ml – Cw * ml * (t5 – t2)) / kw.

mw = (0.25 * 10 ^ 5 J / kg * 1.5 kg + 140 J / kg * “C * 1.5 kg * (327” C – 27 “C) – 2100 J / kg *” C * 0.1 kg * (0 “C + 10” C) – 3.4 * 10 ^ 5 J / kg * 0.1 kg – 4200 J / kg * “C * 0.1 kg * (100” C – 0 “C)) / 2.3 * 10 ^ 6 J / kg = 0.0097 kg.

Answer: mw = 0.0097 kg.