Water weighing 150g at a rate of 20 degrees was poured into a vessel and heated on a blowtorch to a boil

Water weighing 150g at a rate of 20 degrees was poured into a vessel and heated on a blowtorch to a boil and evaporated completely. how much gasoline was consumed if the efficiency was 70%.

In order to heat water with a mass of m = 150 g = 0.15 kg, poured into a vessel at a temperature of t₀ = 20 ° C, on a blowtorch to boiling, that is, to t = 100 ° C, an amount of heat will be needed:
Q₁ = m₁ ∙ c ∙ (t – t₀), where c = 4200 J / (kg ∙ ° С) is the specific heat capacity of water.
To completely evaporate it later, you will need an amount of heat:
Q₂ = m₁ ∙ L, where L = 2300000 J / kg is the specific heat of vaporization of water.
Suppose that m₂ kg of gasoline was consumed for this, the amount of heat released during its combustion:
Q = m₂ ∙ q, where q = 46,000,000 J / kg is the specific heat of combustion of gasoline.
Knowing that the efficiency of the blowtorch is 70%, we obtain the heat balance equation:
0.7 ∙ Q = Q₁ + Q₂ or 0.7 ∙ m₂ ∙ q = m₁ ∙ c ∙ (t – t₀) + m₁ ∙ L.
Then: m₂ = m₁ ∙ (c ∙ (t – t₀) + L) / (0.7 ∙ q).
Substitute the values ​​of physical quantities in the calculation formula:
m₂ = 0.15 kg ∙ (4200 J / (kg ∙ ° C) ∙ (100 ° C – 20 ° C) + 2300000 J / kg) / (0.7 ∙ 46000000 J / kg);
m₂ ≈ 0.073 kg.
Answer: they spent ≈ 0.073 kg of gasoline.