
A geyser, operating on LPG (liquefied petroleum gas) heats water flowing at the rate of 3.0 litres per minutes, from $27^\circ {\rm{C}}$ to $77^\circ {\rm{C}}$. If the heat of combustion of LPG is 40,000 J/g, how much fuel, in g, is consumed per minute? (Specific heat capacity of water is 4200 J/kg-K)
A.15.25
B.15.50
C.15.75
D.16.00
Answer
561.9k+ views
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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

