In a stove with an efficiency of 40%, 3 g of kerosene burns out every minute.

In a stove with an efficiency of 40%, 3 g of kerosene burns out every minute. How long does it take to heat 1.5 liters of water on it from 10 to 100 degrees?

To calculate the duration of water heating on the used stove, we apply the formula: t = Sv * m * (tк – t0) / (N * η) = Sv * mw * (tk – t0) / (Qk / t1 * η) = Sv * mw * (tk – t0) / (qk * mk / t1 * η).

Data: St. – beats. heat capacity of water (Sv = 4200 J / (kg * K)); mw is the mass of heated water (for 1.5 liters of water mw = 1.5 kg); tк – final heating temperature (tк = 100 ºС); t0 – initial temperature (t0 = 10 ºC); qk – beats. heat of combustion of kerosene (qk = 46 * 10 ^ 6 J / kg); mk is the mass of kerosene (mk = 3 g = 0.003 kg); t1 is the duration of kerosene combustion (t1 = 1 min = 60 s); η – efficiency of the used primus (η = 40% = 0.4).

Calculation: t = Sv * mw * (tk – t0) / (qk * mk / t1 * η) = 4200 * 1.5 * (100 – 10) / (46 * 10 ^ 6 * 0.003 / 60 * 0.4 ) = 616.3 s ≈ 10.27 min.

Answer: The water on the used stove will heat up in 616.3 s.