What mass of water can be heated from 20 C to the boiling point on a lamp in which 200 g of kerosene burns, and its efficiency is 12%?
t1 = 20 ° C.
t2 = 100 ° C.
mk = 200 g = 0.2 kg.
C = 4200 J / kg * ° C.
λ = 4.6 * 10 ^ 7 J / kg.
Efficiency = 12%.
When kerosene is burned, the amount of thermal energy Qk is released in the lamp: Qk = λ * mk, where λ is the specific heat of combustion of kerosene, mk is the mass of kerosene that has burned.
The required thermal energy for heating water Qw is expressed by the formula: Qw = C * mw * (t2 – t1), where C is the specific heat capacity of water, mw is the mass of heated water, t2, t1 are the final and initial temperatures.
Since water is heated to boiling point, t2 = 100 ° C.
But since only 12% of the heat released during combustion goes to heating the water, then Qc * efficiency / 100% = Qw.
λ * mk * efficiency / 100% = C * mw * (t2 – t1).
mv = λ * mk * efficiency / 100% * C * (t2 – t1).
mw = 4.6 * 10 ^ 7 J / kg * 0.2 kg * 12% / 100% * 4200 J / kg * ° * (100 ° C – 20 ° C) = 3.3 kg.
Answer: on a kerosene lamp, you can heat up to the boiling point mw = 3.3 kg of water.
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