A geyser heats water flowing at a rate of 2 . 0   kg per minute from 30 ° C to 70 ° C . If…

Given here, mass flow rate of water, $\frac{dm}{dt}=2 \mathrm{kg} {\min }^{-1}$

The geyser heats the water, raising the temperature from ${\theta }_1=30°\mathrm{C}$ to ${\theta }_2=70°\mathrm{C}$.

Total heat utilised in raising the temperature, $\frac{\Delta Q}{\Delta t}=\frac{dm}{dt}\times s\Delta \theta$

$\frac{\Delta Q}{\Delta t}=\frac{2}{60}\times 4.2\times 40=\frac{m}{t}\times L$

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

Now, rate of consumption of fuel is $\frac{m}{t}=\frac{2\times 4.2\times 40}{60\times 8\times {10}^3}$

$=\frac{4.2}{6}\times {10}^{-3}=0.7\times {10}^{-3} \mathrm{kg} {\sec }^{-1}$

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