A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 10^{4} J/g?

#### Solution 1

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

∴Rise in temperature, Δ*T* = *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, Δ*Q* = *mc* Δ*T*

= 3000 × 4.2 × 50

= 6.3 × 10^{5 }J/min

∴ Rate of consumption = `(6.3 xx 0^5)/(4xx10^4) = 15.75 "g/min"`

#### Solution 2

Volume of water heated = 3.0 litre per minute Mass of water heated, m = 3000 g per minute Increase in temperature,

`triangle T = 77^@C - 27^@C =50^@C`

Specific heat of water, `c = 4.2 Jg^(-1) ""^@ C^(-1)`

Amount of heat used = `Q = mctriangleT`

or `Q = 3000 "gmin"^(-1) xx 4.2 Jg^(-1) ""^@C^(-1)xx50^@C`

`= 63 xx 10^4 J min^(-1)`

Rate of combustion of fuel = `(63xx10^4 J min^(-1))/(4.0xx10^4 Jg^(-1))` = 15.75 `"g min"^(-1)`