How much natural gas q = 44 MJ / kg needs to be burned to convert 22 kg of ice taken at -10 C into steam?

How much natural gas q = 44 MJ / kg needs to be burned to convert 22 kg of ice taken at -10 C into steam? Specific heat of melting of ice = 340 J / kg; specific heat of ice = c = 2100 J / kg C, water c = 4200 J / kg C; specific heat of vaporization L = 2.2 MJ / kg

q = 44 MJ / kg = 44 * 10 ^ 6 J / kg.

ml = 22 kg.

t1 = -10 ° C.

t2 = 0 ° C.

t3 = 100 ° C.

λ = 340 kJ / kg = 3.4 * 10 ^ 5 J / kg.

Cl = 2100 J / kg * ° C.

Sv = 4200 J / kg * ° C.

L = 2.2 MJ / kg = 2.2 * 10 ^ 6 J / kg.

mg -?

Q = q * mg.

Q = Cl * ml * (t2 – t1) + λ * ml + Cw * ml * (t3 – t2) + L * ml.

q * mg = Cl * ml * (t2 – t1) + λ * ml + Cw * ml * (t3 – t2) + L * ml.

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

mg = (2100 J / kg * ° C * 22 kg * (0 ° C – (-10 ° C) + 3.4 * 10 ^ 5 J / kg * 22 kg + 4200 J / kg * ° C * 22 kg * (100 ° C – 0 ° C) + 2.2 * 10 ^ 6 J / kg * 22 kg) / 44 * 10 ^ 6 J / kg = 1.5 kg.

Answer: to turn ice into steam, it is necessary to burn mg = 1.5 kg of natural gas.