Question

Calculate the increase in internal energy of $1$ $kg$ of water at ${100}^{\circ }C$ when it is converted into steam at the same temperature and at $1$ $atm$ ($100$ $kPa$). The density of water and steam are $1000kg{m}^{-3}$ and $0.6kg{m}^{-3}$ respectively. The latent heat of vapourisation of water $=2.25\times {10}^{6}Jk{g}^{-1}$

• A. $4.16\times {10}^{6}J$
• B. $2.08\times {10}^{6}J$
• C. $1.04\times {10}^{6}J$
• D. $3.04\times {10}^{6}J$

Answer 1

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

The volume of $1$$kg$ of water
$=\frac{1}{1000}{m}^3$
and $1$ $kg$ of steam $=\frac{1}{0.6}{m}^3$
The increase in volume
$=\frac{1}{0.6}{m}^3-\frac{1}{1000}{m}^3=(1.7-0.001)m^3=1.7m^3$
The work done by the system is $p\Delta V$
$=\left(100kPa\right)(1.7m^3)=1.7\times {10}^{5}J$
The heat given to convert $1$ $kg$ of water into steam
$=2.25\times {10}^{6}J$
The change in internal energy is
$\Delta U=\Delta Q-\Delta W=2.25\times {10}^{6}J-1.7\times {10}^{6}J=2.08\times {10}^{6}J$