Question

The increase in internal energy of $1$ kg of water at ${100}^{0}C$ when it is converted into steam at the same temperature and at $1atm$ $\left(100kPa\right)$ will be [The density of water and steam are $1000kg/m^3$ and $0.6kg/m^3$ respectively. The latent heat of vapourisation of water is $2.25\times {10}^{6}J/kg$] :

• A. $2.08\times {10}^{6}J$
• B. $4\times {10}^{7}J$
• C. $3.27\times {10}^{8}J$
• D. $5\times {10}^{9}J$

Answer 1

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

$L_V=2.25\times {10}^{-6}$ $J/kg$

$m=1kg$

$p=1\times {10}^{2}pa$

volume of water ($V$)$={10}^{-3}{m}^3$

volume of steam ($V$)$=\frac{1}{0.6}{m}^3$

Increase in volume$=(\frac{1}{0.6}-\frac{1}{1000})=1.7m^3$

Work done by system$=P\left(AV\right)=(1\times {10}^5)\left(1.7\right)=1.7\times {10}^{5}j$

$△Q=m{l}_v=2.25\times {10}^{6}j$

change in internal energy $△U=△Q-△W$

$△U=2.25\times {10}^6-1.7\times {10}^5$

$△U=2.25\times {10}^6-0.17\times {10}^6$

$△U=2.08\times {10}^{6}j$