How much kerosene must be burned in order to melt 0.5 kg of ice on a stove, the temperature of which is 0 ° С

How much kerosene must be burned in order to melt 0.5 kg of ice on a stove, the temperature of which is 0 ° С, and the resulting water to heat up to 100 ° С? The efficiency of the primus is 30%.

ml = 0.5 kg.

t1 = 0 ° C.

t2 = 100 ° C.

λk = 4.6 * 10 ^ 7 J / kg.

ql = 3.3 * 10 ^ 5 J / kg.

Cw = 4200 J / kg * ° C.

Efficiency = 30%.

mk -?

Efficiency = Qw * 100% / Qk.

The amount of heat Qk, which is released during the combustion of kerosene, is determined by the formula: Qk = λk * mk, where λk is the specific heat of combustion of kerosene, mk is the mass of kerosene that has burned.

The amount of heat Qw, which is necessary for melting ice and heating water, is determined by the formula: Qw = ql * ml + Cw * ml * (t2 – t1), where Cw is the specific heat capacity of water, ml is the mass of ice, t1, t2 are the initial and final temperature, ql – specific heat of ice melting.

Efficiency = (ql * ml + Cw * ml * (t2 – t1)) * 100% / λk * mk.

mk = (ql * ml + Cw * ml * (t2 – t1)) * 100% / λk * efficiency.

mk = (3.3 * 10 ^ 5 J / kg * 0.5 kg + 4200 J / kg * ° C * 0.5 kg * (100 ° C – 0 ° C)) * 100% / 4.6 * 10 ^ 7 J / kg * 30% = 0.027 kg.

Answer: it is necessary to burn mk = 0.027 kg of kerosene in a kerosene stove.