They boiled water in a kettle on a gas stove for 15 minutes. How much gas burns in 1 s if the kettle contains 3

They boiled water in a kettle on a gas stove for 15 minutes. How much gas burns in 1 s if the kettle contains 3 liters of water at a temperature of 20. The heat capacity of the kettle and other losses should be neglected.

T1 = 15 min = 900 s.

Vw = 3 l = 0.003 m3.

ρw = 1000 kg / m3.

t1 = 20 ° C.

t2 = 100 ° C.

Cw = 4200 J / kg * ° С.

Vg = 0.1 m3.

ρg = 0.8 kg / m3.

r = 4.4 * 10 ^ 7 J / kg.

mg1 -?

Since the loss of thermal energy, according to the condition of the problem, can be neglected, then the entire amount of heat that is released during the combustion of gas Qg goes to heating water Qw: Qg = Qw.

When gas is burned, the following amount of heat Qg is released: Qg = r * mg.

mg = Qg / r.

mg1 = mg / T = Qg / r * T.

The amount of heat Qw, which goes to heating water, is expressed by the formula: Qw = Cw * mw * (t2 – t1) = Cw * Vw * ρw * (t2 – t1).

Qw = 4200 J / kg * ° C * 0.003 m3 * 1000 kg / m3 * (100 ° C – 20 ° C) = 1008000 J.

mg1 = 1008000 J / 4.4 * 10 ^ 7 J / kg * 900 s = 0.0000255 kg.

Answer: when heating water in kettles, natural gas with a mass of mg1 = 0.0000255 kg burns every second.