The water on the gas burner was heated from 20 ° to 60 ° while burning 3 kg of wood. Efficiency 30%. Enter the mass of water.

Let water with a mass of m kg, taken at a temperature of t₀ = 20 ° C, was heated on the burner to t = 60 ° C, this required the amount of heat:

Q₁ = m₁ ∙ c ∙ (t – t₀), where c = 4200 J / (kg ∙ ° C) is the specific heat capacity of water.

Since at the same time m₂ = 3 kg of firewood was consumed, then during their combustion, an amount of heat was released:

Q₂ = m₂ ∙ q, where q = 10000000 J / kg is the specific heat of combustion of firewood.

Knowing that the burner efficiency is 30%, we obtain the heat balance equation:

0.3 ∙ Q₂ = Q₁ or 0.3 ∙ m₂ ∙ q = m₁ ∙ c ∙ (t – t₀).

Then: m₁ = 0.3 ∙ m₂ ∙ q / (c ∙ (t – t₀)).

Substitute the values ​​of physical quantities in the calculation formula:

m₁ = (0.3 ∙ 3 kg ∙ 10000000 J / kg) / (4200 J / (kg ∙ ° С) ∙ (60 ° С – 20 ° С)).

m₁ ≈ 53.57 kg.

Answer: the mass of the heated water was ≈ 53.57 kg.



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