Question

The entropy change associated with the conversion of $1kg$ of ice at $273K$ to water vapours at $383K$ is:
[Given that : Specific heat of water liquid and water vapour are $4.2kJ{K}^{-1}k{g}^{-1}$ and $2.0kJ{K}^{-1}k{g}^{-1}$; latent heat of fusion and vapourisation of water are $344kjk{g}^{-1}$ and $2491kJk{g}^{-1}$, respectively].
[Take : $log273=2.436,log373=2.572,log383=2.583]$

• A. $7.90kJk{g}^{-1}{K}^{-1}$
• B. $2.64kJk{g}^{-1}{K}^{-1}$
• C. $8.49kJk{g}^{-1}{K}^{-1}$
• D. $9.26kJk{g}^{-1}{K}^{-1}$

Answer 1

Answer: (D)

$9.26kJk{g}^{-1}{K}^{-1}$

$\Delta S=\frac{\Delta H_{Transition}}{T}$

$\Delta S=nCln\frac{T_f}{T_i}=msln\frac{T_f}{T_i}$

n= no. of moles, m= mass, C=molar heat capacity, s= specific heat capacity

Phase change path

$H_{2}O\left(s\right)\stackrel{\Delta S_1}{\to }{H}_{2}O\left(l\right)\stackrel{\Delta S_2}{\to }{H}_{2}O\left(l\right)\stackrel{\Delta S_3}{\to }{H}_{2}O\left(g\right)\stackrel{\Delta S_4}{\to }{H}_{2}O\left(g\right)$

$\Delta s_1=\frac{\Delta H_{fusion}}{273}=\frac{334}{273}=1.22$

$\Delta s_2=4.2ln\frac{363}{273}=1.31$

$\Delta s_3=\frac{\Delta H_{vap}}{2491}373=6.67$

$\Delta s_4=2.0ln\frac{383}{373}=0.05$

$\Delta s_{total}=\Delta s_1+\Delta s_2+\Delta s_3+\Delta s_4=9.25kJk{g}^{-1}{K}^{-1}$