What volume of water can be heated from 20 * C to the boiling point due to the heat obtained by burning 0.5

What volume of water can be heated from 20 * C to the boiling point due to the heat obtained by burning 0.5 liters of kerosene? Consider that all the amount of water released during the combustion of kerosene goes to heating the water. boiling t * = 100 * С

t1 = 20 ° C

t2 = 100 ° C.

Vk = 0.5 l = 0.5 * 10 ^ -3 m ^ 3.

C = 4200 J / kg * ° C.

q = 4.6 * 10 ^ 7 J / kg.

ρк = 800 kg / m ^ 3.

ρw = 1000 kg / m ^ 3.

Vв -?

The amount of heat Q, which is released during the combustion of kerosene, is expressed by the formula: Q = q * mk.

We find the mass of kerosene by the formula: mк = ρк * Vк.

Q = q * ρк * Vк.

The amount of heat Q, which is necessary for heating water, is expressed by the formula: Q = C * mw * (t2 – t1).

We express the mass of heated water by the formula: mw = ρw * Vw.

Q = C * ρw * Vw * (t2 – t1).

The heat balance equation will have the form: q * ρк * Vк = C * ρв * Vв * (t2 – t1).

Vw = q * ρk * Vk / C * ρw * (t2 – t1).

Vw = 4.6 * 10 ^ 7 J / kg * 800 kg / m ^ 3 * 0.5 * 10 ^ -3 m ^ 3/4200 J / kg * ° C * 1000 kg / m ^ 3 * (100 ° C – 20 ° C) = 55 * 10 ^ -3 m ^ 3 = 55 l.