Q104 / 180 · JIPMER 2015 · Physics
1 kg of water is converted into steam at the same temperature and at 1 atm ( 100 kPa ). The density of water and steam are $1000 \mathrm{kgm}^{-3}$ and $0.6 \mathrm{kgm}^{-3}$ respectively. The latent heat of vaporisation of water is $2.25 \times 10^6 \mathrm{Jkg}^{-1}$. What will be increase in energy?
- A$3 \times 10^5 \mathrm{~J}$
- B$4 \times 10^6 \mathrm{~J}$
- Correct answerC$2.08 \times 10^6 \mathrm{~J}$
- DNone of these
Solution
The volume of 1 kg water $=\frac{1}{100} \mathrm{~m}^3$
and volume of 1 kg steam
$=\frac{1}{0.6} \mathrm{~m}^3$
The increase in volume
$\begin{aligned} & =\frac{1}{0.6} \mathrm{~m}^3-\frac{1}{1000} \mathrm{~m}^3 \\ & =(1.7-0.001) \mathrm{m}^3 \\ & \cong 1.7 \mathrm{~m}^3\end{aligned}$
The work done by steam is
$\begin{aligned} W & =(100 \mathrm{kPa})\left(1.7 \mathrm{~m}^3\right) \\ & =1.7 \times 10^5 \mathrm{~J}\end{aligned}$
The change in intemal energy
$\begin{aligned} \Delta U & =\Delta Q-\Delta W \\ & =2.25 \times 10^6 \mathrm{~J}-1.7 \times 10^5 \mathrm{~J} \\ & =208 \times 10^6 \mathrm{~J}\end{aligned}$
Hence, the correct answer is option: 'C'.