A geyser heats water flowing at a rate of 2 . 0kg per minute from...

Given here, mass flow rate of water, $\dfrac{d m}{d t} = 2kg{min}^{- 1}$

The geyser heats the water, raising the temperature from ${\theta}_{1} = 30 ° C$ to ${\theta}_{2} = 70 ° C$.

Total heat utilised in raising the temperature, $\dfrac{\DeltaQ}{\Deltat} = \dfrac{d m}{d t} \times s\Delta\theta$

$\dfrac{\DeltaQ}{\Deltat} = \dfrac{2}{60} \times 4 . 2 \times 40 = \dfrac{m}{t} \times L$

$L =$ heat of combination$= 8 \times {10}^{3}J{g}^{- 1}$

Now, rate of consumption of fuel is $\dfrac{m}{t} = \dfrac{2 \times 4 . 2 \times 40}{60 \times 8 \times {10}^{3}}$

$= \dfrac{4 . 2}{6} \times {10}^{- 3} = 0 . 7 \times {10}^{- 3}kg{sec}^{- 1}$

$= 42g{min}^{- 1}$