Ice weighing 2.5 kg at a temperature of -20 degrees Celsius was melted, then this water was heated to 100 degrees

Ice weighing 2.5 kg at a temperature of -20 degrees Celsius was melted, then this water was heated to 100 degrees Celsius and then completely evaporated. How much energy was required for this, how much kerosene was burned?

m = 2.5 kg.

t1 = -20 ° C.

t2 = 0 ° C.

t3 = 100 ° C.

Cl = 2100 J / kg * ° C.

q = 3.4 * 10 ^ 5 J / kg.

Cw = 4200 J / kg * ° C.

λ = 2.3 * 10 ^ 6 J / kg.

k = 4.6 * 10 ^ 7 J / kg.

mk -?

Q = Q1 + Q2 + Q3 + Q4.

Q1 = Cl * m * (t2 – t1).

Q1 = 2100 J / kg * ° C * 2.5 kg * (0 ° C – (- 20 ° C)) = 105000 J.

Q2 = q * m.

Q2 = 3.4 * 10 ^ 5 J / kg * 2.5 kg = 850,000 J.

Q3 = Cв * m * (t3 – t2).

Q3 = 4200 J / kg * ° C * 2.5 kg * (100 ° C – 0 ° C) = 1050000 J.

Q4 = λ * m.

Q4 = 2.3 * 10 ^ 6 J / kg * 2.5 kg = 5750000 J.

Q = 105000 J + 8500000 J + 1050000 J + 5750000 J = 7755000 J.

Q = k * mk.

mk = Q / k.

mk = 7755000 J / 4.6 * 10 ^ 7 J / kg = 0.17 kg.

Answer: Q = 7755000 J, mk = 0.17 kg.