The efficiency of the kerosene stove is 30%. How much kerosene needs to be burned to heat 3 liters

The efficiency of the kerosene stove is 30%. How much kerosene needs to be burned to heat 3 liters of water from 15 degrees of boiling?

V = 3 l = 3 * 10-3 m3.

ρ = 1000 kg / m3.

t1 = 15 ° C.

t2 = 100 ° C.

C = 4200 J / kg * ° C.

q = 4.6 * 106 J / kg.

mk -?

Let us write down the definition for efficiency: efficiency = Qp * 100% / Qz, where Qp is the useful amount of heat that goes to heating water, Qz is the consumed amount of heat that is released during the combustion of kerosene.

Qп = С * m * (t2 – t1) = С * ρ * V * (t2 – t1).

Qz = q * mk.

Efficiency = С * ρ * V * (t2 – t1) * 100% / q * mk.

mk = C * ρ * V * (t2 – t1) * 100% / q * efficiency.

mk = 4200 J / kg * ° C * 1000 kg / m3 * 3 * 10-3 m3 * (100 ° C – 15 ° C) * 100% / 4.6 * 106 J / kg * 30% = 0.0078 kg.

Answer: to heat water to the boiling point, it is necessary to burn kerosene mk = 0.0078 kg.