How much kerosene should be consumed so that 2 kg of ice taken at -10 degrees can be turned

How much kerosene should be consumed so that 2 kg of ice taken at -10 degrees can be turned into water at 20 degrees, if the efficiency of kerosene is 40%.

ml = 2 kg.

Cl = 2100 J / kg * ° C.

Cw = 4200 J / kg * ° C.

t1 = -10 ° C.

t2 = 0 ° C.

t3 = 20 ° C.

λ = 3.4 * 105 J / kg.

qk = 4.6 * 107 J / kg.

Efficiency = 40%.

mk -?

To turn ice into water, it is first necessary to heat the ice to the melting temperature t2, at this temperature it is necessary to transfer the ice into water, and heat the resulting water.

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

Qk = qk * mk.

Efficiency = Qw * 100% / Qk.

Efficiency = (Cl * ml * (t2 – t1) + λ * ml + Cw * ml * (t3 – t2)) * 100% / qк * mк.

mk = Cl * ml * (t2 – t1) + λ * ml + Cw * ml * (t3 – t2)) * 100% / qc * efficiency.

mk = 2100 J / kg * ° C * 2 kg * (0 ° C – (-10 ° C)) + 3.4 * 105 J / kg * 2 kg + 4200 J / kg * ° C * 2 kg * ( 20 ° C – 0 ° C)) * 100% / 4.6 * 107 J / kg * 40% = 0.048 kg.

Answer: to obtain water from ice, it is necessary to burn mk = 0.048 kg of kerosene.