Question

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 kgm${}^{-3}$ and 0.6 kgm${}^{-3}$ respectively. The latent heat of vaporisation of water is $2.25\times {10}^{6}Jk{g}^{-1}$. What will be increase in energy?

• A. $3\times {10}^{5}J$
• B. $4\times {10}^{6}J$
• C. $2.08\times {10}^{6}J$
• D. None of these

Answer 1

The correct option is C $2.08\times {10}^{6}J$

Volume of $1kg$ of water $=\frac{m}{\rho }$ $=\frac{1kg}{1000kg/m^3}$ $=0.001m^3$

Volume of $1kg$ of steam $=\frac{m}{{\rho }^{\prime }}$ $=\frac{1kg}{0.6kg/m^3}$ $=\frac{10}{6}{m}^3$

Hence, work done by system $=P\Delta V$

$=100kPa\times 1.73m^3$

$=1.7\times {10}^{5}J$

Heat given $=mL=2.25\times {10}^{6}J$

Hence, increase in energy of water $=2.25\times {10}^{6}J-1.7\times {10}^{5}J$

$=2.08\times {10}^{6}J$