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# A geyser water flowing at the rate of 4 litre per minute from $30^\circ {\text{C}}$ to $85^\circ {\text{C}}$. If the geyser operates on gas burner then the amount of heat used per minute is:A. $9.24 \times {10^5}\,{\text{J}}$B. $6.24 \times {10^7}\,{\text{J}}$C. $9.24 \times {10^7}\,{\text{J}}$D. $6.24 \times {10^5}\,{\text{J}}$

Last updated date: 18th Jun 2024
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Hint:Use the formula for heat exchanged by the substance. This formula gives the relation between the amount of heat exchanged, mass of the substance, specific heat of the substance and change in temperature of the substance. First determine the mass of the water flowing per minute from the volume flow rate and then use the formula for the amount of heat exchanged.

Formulae used:
The heat exchanged $\Delta Q$ by the material is given by
$\Delta Q = mc\Delta T$ …… (1)
Here, $m$ is the mass of the material, $c$ is the specific heat of the material and $\Delta T$ is the change in temperature of the material.
The density $\rho$ of an object is given by
$\rho = \dfrac{m}{V}$ …… (2)
Here, $m$ is the mass of the object and $V$ is the volume of the object.

Complete step by step answer:
We have given that the volume flow rate of the water from the geyser is 4 litre per minute.
$V = 4\,{\text{lit/min}}$
The temperature of the water changes from $30^\circ {\text{C}}$ to $85^\circ {\text{C}}$.
${T_i} = 30^\circ {\text{C}}$
$\Rightarrow {T_f} = 85^\circ {\text{C}}$
Let us first calculate the mass of water from the geyser flowing per minute.
We know that the density of the water is $1000\,{\text{kg/}}{{\text{m}}^3}$.
$\rho = 1000\,{\text{kg/}}{{\text{m}}^3}$
Let us convert the unit of the volume flow rate from litre per minute to meter cube per minute.
$V = \left( {4\,\dfrac{{{\text{lit}}}}{{{\text{min}}}}} \right)\left( {\dfrac{{{{10}^{ - 3}}\,{{\text{m}}^{ - 3}}}}{{1\,{\text{lit}}}}} \right)$
$\Rightarrow V = 4 \times {10^{ - 3}}\,{{\text{m}}^{\text{3}}}{\text{/min}}$
Hence, the volume flow rate of the water is $4 \times {10^{ - 3}}\,{{\text{m}}^{\text{3}}}{\text{/min}}$.
Rearrange equation (2) for mass $m$.
$m = \rho V$
Substitute $1000\,{\text{kg/}}{{\text{m}}^3}$ for $\rho$ and $4 \times {10^{ - 3}}\,{{\text{m}}^{\text{3}}}{\text{/min}}$ for $V$ in the above equation.
$m = \left( {1000\,{\text{kg/}}{{\text{m}}^3}} \right)\left( {4 \times {{10}^{ - 3}}\,{{\text{m}}^{\text{3}}}{\text{/min}}} \right)$
$\Rightarrow m = 4\,{\text{kg/min}}$
Hence, the mass of the flowing water from the geyser is $4\,{\text{kg/min}}$.

The change in temperature of the flowing water is
$\Delta T = {T_f} - {T_i}$
Substitute $85^\circ {\text{C}}$ for ${T_f}$ and $30^\circ {\text{C}}$ for ${T_i}$ in the above equation.
$\Delta T = 85^\circ {\text{C}} - 30^\circ {\text{C}}$
$\Rightarrow \Delta T = 55^\circ {\text{C}}$
Hence, the change in temperature of the water is $55^\circ {\text{C}}$.
The specific heat of the water is $4200\,{\text{J/kg/}}^\circ {\text{C}}$.
$c = 4200\,{\text{J/kg/}}^\circ {\text{C}}$
Let now calculate the amount of heat used per minute using equation (1).
Substitute $4\,{\text{kg/min}}$ for $m$, $4200\,{\text{J/kg/}}^\circ {\text{C}}$ for $c$ and $55^\circ {\text{C}}$ for $\Delta T$ in equation (1).
$\Delta Q = \left( {4\,{\text{kg/min}}} \right)\left( {4200\,{\text{J/kg/}}^\circ {\text{C}}} \right)\left( {55^\circ {\text{C}}} \right)$
$\Rightarrow \Delta Q = 924000\,{\text{J/min}}$
$\therefore \Delta Q = 9.24 \times {10^5}\,{\text{J/min}}$
Therefore, the amount of heat used per minute is $9.24 \times {10^5}\,{\text{J}}$.

Hence, the correct option is A.

Note:The students should keep in mind that we have asked to determine the amount of heat used per minute. We obtain the amount of heat used for the heating of water with the unit joule per minute. But since we are asked to calculate the amount of heat used per minute, the final unit of the heat used will be only joule and not joule per minute.
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