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A geyser heats water flowing at the rate of 3.0 liter/min from $27^{\circ} \mathrm{C}$ to $77^{\circ} \mathrm{C}$. If the geyser operates on a gas burner, what is the rate of consumption of the fuel, if its heat of combustion is $4.0 \times 10^4 \mathrm{~J} / \mathrm{g}$ ?
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Given, volume of water heated $=3$ liter $/ \mathrm{min}$.
Mass of water heated, $m=3000 \mathrm{~g} / \mathrm{min}$
Rise in temperature, $\Delta \mathrm{T}=77-27=50^{\circ} \mathrm{C}$, Specific heat of water, $\mathrm{c}=4.2 \mathrm{Jg}^{-1}{ }^{\circ} \mathrm{C}^{-1}$
Amount of heat used, $\Delta \mathrm{Q}=\mathrm{mc} \Delta \mathrm{T}$
$=3000 \times 4.2 \times 50=63 \times 10^4 \mathrm{~J} / \mathrm{min}$
Heat of combustion $=4 \times 10^4 \mathrm{~J} / \mathrm{g}$;
Rate of combustion of fuel
$=\frac{63 \times 10^4}{4 \times 10^4}=15.75 \mathrm{~g} / \mathrm{m}$
Mass of water heated, $m=3000 \mathrm{~g} / \mathrm{min}$
Rise in temperature, $\Delta \mathrm{T}=77-27=50^{\circ} \mathrm{C}$, Specific heat of water, $\mathrm{c}=4.2 \mathrm{Jg}^{-1}{ }^{\circ} \mathrm{C}^{-1}$
Amount of heat used, $\Delta \mathrm{Q}=\mathrm{mc} \Delta \mathrm{T}$
$=3000 \times 4.2 \times 50=63 \times 10^4 \mathrm{~J} / \mathrm{min}$
Heat of combustion $=4 \times 10^4 \mathrm{~J} / \mathrm{g}$;
Rate of combustion of fuel
$=\frac{63 \times 10^4}{4 \times 10^4}=15.75 \mathrm{~g} / \mathrm{m}$
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