How much kerosene must be burned in order to heat water weighing 500 g

How much kerosene must be burned in order to heat water weighing 500 g to a boil in a steel pan weighing 200 g, if their initial temperature was 20 ° C.

Problem data: m1 (water mass) = 500 g = 0.5 kg; m2 (weight of a steel pan) = 200 g = 0.2 kg; t0 (initial temperature of water, pot) = 20 ºС.

Constants: t (boiling point of water, n.o.) = 100 ºС; C1 (specific heat capacity of water) = 4200 J / (kg * K); C2 (specific heat capacity of steel) = 500 J / (kg * K); q (beats heat of combustion of kerosene) = 4.6 * 10 ^ 7 J / kg.

Solution: (C1 * m1 + C2 * m2) * (t – t0) = q * m3.

m3 = (C1 * m1 + C2 * m2) / q = (4200 * 0.5 + 500 * 0.2) * (100 – 20) / (4.6 * 10 ^ 7) = 0.00382 kg (3 , 82 g).