Question

The increase in internal energy of 1 kg of water at ${100}^{0}C$ when it is converted into steam at the same temperture and at 1 atm (100 kPa) 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 ${10}^6$ 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 $2.08\times {10}^{6}J$

Latent heat of vaporisation of water , $△H$$=2.25\times {10}^{6}J/kg$
$△H=2.25\times {10}^{6}J/kg$
and ,
Work done $=-P_{ext}(V_2-V_1)$


(a) Now, volume of water $V=\left(\frac{mass}{density}\right)$ $=\left(\frac{m}{d}\right)=\frac{1}{1000}{m}^3,since1m^3=1000L$
$=1L$
(b) Volume of steam =$\left(\frac{mass}{density}\right)$ $\frac{1}{0.6}{m}^3=\frac{1}{0.6}\times 1000L=1666.67L$
So, using the relationship,
$△H=△U+P△V$
$2.25\times {10}^6=△U+1[1666.67-1]1.01325\times {10}^5$
$△U=22.5\times {10}^5-1.68\times {10}^5=20.8\times {10}^5=2.08\times {10}^{6}J$