Answer
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Hint: Here, we have to find the rate of consumption of fuel in heating water, so, we have to use the formula of $Q = mC\Delta T$, where Q is the total used heat, m is the mass, C is specific heat and $\Delta T$ is temperature difference.
Complete step by step solution:
Here, the flow of water to the geyser is 3.0 litres per minutes and the geyser heats water from $27^\circ {\rm{C}}$ to $77^\circ {\rm{C}}$. The specific heat capacity of the water is given as 4200 J/kg-K . So, we are provided all the values that are required to calculate heat using the formula $Q = mC\Delta T$.
The mass of water=3 l/min
Now, we have converted litre to g. We know that 1 litre=1000 g. So, the mass of water (m) is 3000 g/min
Now, we have converted initial and final temperatures to Kelvin.
${T_1} = 27^\circ {\rm{C}}$
${T_2} = 77^\circ {\rm{C}}$
The difference of temperatures $\left( {\Delta T} \right)$
$ = 77^\circ {\rm{C}} - {\rm{27}}^\circ {\rm{C}} = {\rm{50}}^\circ {\rm{C}}$
The specific heat (C) of water is 4200 J/kg-K or $4.2\,{\rm{J/g}}^\circ {\rm{C}}$
Now, we have to put all the values in the formula of Q.
$Q = mC\Delta T$
$ \Rightarrow Q = 3000 \times 4.2 \times 50$
$ \Rightarrow Q = 6.3 \times {10^5}\,{\rm{J}}/\min $
Now, we have to calculate the consumption of fuel in one minute. The heat of combustion of LPG is 40,000 J/g or.
So, fuel consumption in one minute $ = \dfrac{Q}{{{\rm{Heat}}\,{\rm{of}}\,{\rm{combustion}}}}$
$ \Rightarrow {\rm{Fuel}}\,\,{\rm{consumption}} = \dfrac{{6.3 \times {{10}^5}\,{\rm{J}}/\min }}{{4 \times {{10}^4}{\rm{J}}/{\rm{g}}}} = 15.75\,g/\min $
So, the correct answer is Option C.
Note: Always remember that Calorimetry is the science of measurement of changes in state variables of a body to derive the transfer of heat associated with changes of its state due, for example, to physical changes, chemical reactions or phase transitions under specified constraints. Calorimeter is the device that is used for measurement of the heat changes in chemical and physical processes.
Complete step by step solution:
Here, the flow of water to the geyser is 3.0 litres per minutes and the geyser heats water from $27^\circ {\rm{C}}$ to $77^\circ {\rm{C}}$. The specific heat capacity of the water is given as 4200 J/kg-K . So, we are provided all the values that are required to calculate heat using the formula $Q = mC\Delta T$.
The mass of water=3 l/min
Now, we have converted litre to g. We know that 1 litre=1000 g. So, the mass of water (m) is 3000 g/min
Now, we have converted initial and final temperatures to Kelvin.
${T_1} = 27^\circ {\rm{C}}$
${T_2} = 77^\circ {\rm{C}}$
The difference of temperatures $\left( {\Delta T} \right)$
$ = 77^\circ {\rm{C}} - {\rm{27}}^\circ {\rm{C}} = {\rm{50}}^\circ {\rm{C}}$
The specific heat (C) of water is 4200 J/kg-K or $4.2\,{\rm{J/g}}^\circ {\rm{C}}$
Now, we have to put all the values in the formula of Q.
$Q = mC\Delta T$
$ \Rightarrow Q = 3000 \times 4.2 \times 50$
$ \Rightarrow Q = 6.3 \times {10^5}\,{\rm{J}}/\min $
Now, we have to calculate the consumption of fuel in one minute. The heat of combustion of LPG is 40,000 J/g or.
So, fuel consumption in one minute $ = \dfrac{Q}{{{\rm{Heat}}\,{\rm{of}}\,{\rm{combustion}}}}$
$ \Rightarrow {\rm{Fuel}}\,\,{\rm{consumption}} = \dfrac{{6.3 \times {{10}^5}\,{\rm{J}}/\min }}{{4 \times {{10}^4}{\rm{J}}/{\rm{g}}}} = 15.75\,g/\min $
So, the correct answer is Option C.
Note: Always remember that Calorimetry is the science of measurement of changes in state variables of a body to derive the transfer of heat associated with changes of its state due, for example, to physical changes, chemical reactions or phase transitions under specified constraints. Calorimeter is the device that is used for measurement of the heat changes in chemical and physical processes.
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