A geyser heats water flowing at the rate of 3 kg per minute from `27^(@)C` to `77^(@)C` . If the geyser operates on a gas burner, what is the rate of consumption of fuel if the heat of combustion is `4 xx 10^(4) J//g?` Given specific heat of water is `4.2 xx 10^(3) J//kg//K`.

Explanation

Water is flowing at a rate of 3.0 litre/min.
The geyser heats the water, raising the temperature from 27°C to 77°C.
Initial temperature, `T_(1) = 27°C`
Final temperature, `T_(2) = 77°C`
thereforeRise in temperature. `DeltaT=T_(2)-T_(1)=77-27=50^(@)C`
Heat of combustion = `4 × 10^(4)J//g`
Specific heat of water,`C=4.2" J" " g"^(-1)"" ^(@)C^(-1) "`
Mass of flowing water, m = 3.0 litre/min = 3000 g/min
Total heat used, `DeltaQ = mc DeltaT`
`= 3000 × 4.2 × 50`
`= 6.3 × 10^(5) J/min `
`therefore` Rate of consumption `=(6.3xx10^(5))/(4xx10^(4))=15.75" g/min"`