Question

Heat produced by burning $2kg$ of a fuel, whose calorific value is $22000kJ{kg}^{-1}$, is utilized to boil certain mass of water which is at 34${}^{\circ }C$. Find the mass of water taken.

(Specific heat capacity of water is $4200J{kg}^{-1}{K}^{-1}$ ).

• A. $159kg$
• B. $172kg$
• C. $183kg$
• D. $100kg$

Answer 1

The correct option is A $159kg$

Mass of fuel $\left(m\right)=2kg$
Calorific value of the fuel $\left(S\right)=22000kJ{kg}^{-1}$
$=22\times {10}^3\times {10}^{3}J{kg}^{-1}=22\times {10}^{6}J{kg}^{-1}$
Initial temperature of water $\left(t_1\right)=34C$
Final temperature of water $\left(t_2\right)=100C$
$\therefore$ Rise in temperature $=(t_2-t_1)$
$=(100-34)C=66C$
Specific heat capacity of water $\left(S\right)=4200J{kg}^{-1}{C}^{-1}$
Mass of water $\left(m\right)=$ ?
Amount of heat energy produced on burning the fuel $\left(Q\right)=$ mass of fuel $\times$ calorific value.
$=2kg\times 22\times {10}^{6}J{kg}^{-1}=44\times {10}^{6}J$
Here, Heat lost by fuel $=$ Heat absorbed by water.
$44\times {10}^{6}J=m_w\times s_w\times \Delta t$
$\therefore 44\times {10}^{6}J=m_w\times 4200\times 66$
$m_w=\frac{44\times {10}^6}{66\times 42\times {10}^2}=\frac{1}{63}\times {10}^{4}kg=\frac{100}{63}\times {10}^{2}kg=1.59\times {10}^{2}kg$
Hence mass of water $=159kg$