Determine the efficiency of an alcohol lamp if, when heating 150 g of water on it from 20 to 80 ° C

Determine the efficiency of an alcohol lamp if, when heating 150 g of water on it from 20 to 80 ° C, 4 g of alcohol are consumed. The specific heat capacity of water is 4200 J / kg * 0С, the specific heat of combustion of alcohol is 3 * 107 J / kg.

On an alcohol lamp, water was heated with a mass of m₁ = 150 g = 0.15 kg from a temperature t₁ = 20 ° C to a temperature t₂ = 80 ° C, while the water temperature increased by Δt = t₂ – t₁ = 80 ° C – 20 ° C = 60 ° С and the water received the amount of heat Q₁ = m₁ ∙ c ∙ Δt. From the condition of the problem it is known that alcohol with a mass of m₂ = 4 g = 0.004 kg was burned, which released the amount of heat Q₂ = q ∙ m₂. The efficiency is found by the formula: η = Q₁ / Q₂ or η = m₁ ∙ c ∙ Δt / (q ∙ m₂). From the reference books, we determine the specific heat of water c₁ = 4200 J / (kg ∙ ° C), and the specific heat of combustion of alcohol q = 27000000 J / kg. Substituting the values ​​of physical quantities into the calculation formula, we find the efficiency of the spirit lamp:

η = 0.15 kg ∙ 4200 J / (kg ∙ ° С) ∙ 60 ° С / (27,000,000 J / kg ∙ 0.004 kg);

η = 0.35 = 35%.

Answer: The efficiency of the spirit lamp is 35%.