# Ice taken at a temperature of -20 degrees was turned into steam at a temperature of 100 degrees.

Ice taken at a temperature of -20 degrees was turned into steam at a temperature of 100 degrees. At the same time, they burned 620 g of firewood. Find a mass of ice

t1 = -20 ° C.

t2 = 0 ° C.

t3 = 100 ° C.

Cl = 2100 J / kg * ° C.

Cw = 4200 J / kg * ° C.

q = 3.4 * 105 J / kg.

r = 2.3 * 106 J / kg.

λ = 1 * 107 J / kg.

md = 620 g = 0.62 kg.

ml -?

Ql = Cl * ml * (t2 – t1) + q * ml + Cw * ml * (t3 – t2) + r * ml = (Cl * (t2 – t1) + q * + Cw * (t3 – t2) + r) * ml.

Qd = λ * md.

Since the entire amount of heat during the combustion of firewood Qd goes to heating the ice Ql, the heat balance equation will take the form: Qd = Ql.

(Cl * (t2 – t1) + q * + Cw * (t3 – t2) + r) * ml = λ * md.

ml = λ * md / Cl * (t2 – t1) + q * + Cw * (t3 – t2) + r).

ml = 1 * 107 J / kg * 0.62 kg / (2100 J / kg * ° C * (0 ° C – (-20 ° C)) + 3.4 * 105 J / kg * + 4200 J / kg * ° C * (100 ° C – 0 ° C) + 2.3 * 106 J / kg) = 2 kg.

Answer: the ice mass was ml = 2 kg.

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