Using a heater with an efficiency of 40%, it is necessary to bring 4 liters of water to the boiling point in an aluminum

| September 3, 2021 | education

Using a heater with an efficiency of 40%, it is necessary to bring 4 liters of water to the boiling point in an aluminum saucepan weighing 2 kg. Determine the consumption of kerosene for heating water and pots, if their initial temperature is 20 degrees.

Efficiency = 40%.

Vw = 4 l = 0.004 m3.

ρw = 1000 kg / m3.

Sv = 4200 J / kg * ° C.

mа = 2 kg.

Ca = 920 J / kg * ° C.

q = 4.6 * 107 J / kg.

Δt = 20 ° C.

mk -?

Efficiency = Qp * 100% / Qz.

Qp = Sv * mw * Δt + Ca * ma * Δt = (Sv * mw + Ca * ma) * Δt = (Sv * Vw * ρw + Ca * ma) * Δt.

Qz = q * mk.

Efficiency = (Sv * Vw * ρw + Ca * ma) * Δt * 100% / q * mk.

mk = (Sv * Vw * ρw + Ca * ma) * Δt * 100% / q * efficiency.

mk = (4200 J / kg * ° C * 0.004 m3 * 1000 kg / m3 + 920 J / kg * ° C * 2 kg) * 20 ° C * 100% / 4.6 * 107 J / kg * 40% = 0.02 kg.

Answer: to heat water in an aluminum pan, it is necessary to burn kerosene with a mass of mk = 0.02 kg.



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