What is the mass of kerosene that must be burned in order to heat 2 liters of water from a temperature of 20 degrees Celsius to boiling water, 50% of the combustion cycle is obtained.
Data: V (volume of water) = 2 l = 2 * 10 ^ -3 m ^ 3; t0 (temperature before heating) = 20 ºС; t (final heating temperature) = 100 ºС; η (efficiency) = 50% or 0.5.
Reference data: ρw (water density) = 1000 kg / m ^ 3; C (specific heat capacity of water) = 4200 J / (kg * K); q (specific heat of combustion of kerosene) = 4.6 * 10 ^ 7 J / kg.
Solution: η = C * ρw * V * (t – t0) / q * m2 and m2 = C * ρw * V * (t – t0) / (η * q).
Let’s calculate: m2 = 4200 * 1000 * 2 * 10 ^ -3 * (100 – 20) / (0.5 * 4.6 * 10 ^ 7) = 0.0292 kg or 29.2 g.
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