Question

The entropy change when 1 kg of water is heated from ${27}^{o}C$ to ${200}^{o}C$ forming super heated steam under constant pressure is:

Given; specific heat of water $=4180J/kgK$

Specific heat of steam $=1670+0.49TJ/kgK$

Latent heat of vapourization = $23\times {10}^{5}J/kg$

• A. $7512.50J$
• B. $7532.50J$
• C. $7322.10J$
• D. $7522.50J$

Answer 1

The correct option is D $7522.50J$

$\Delta S=2.303n\times C_p\times \log \frac{T_2}{T_1}[where,minkgand{C}_{p}inJ/kg]$

Entropy change for heating water from ${27}^{o}C$ to ${100}^{o}C$.

$\Delta S_1=2.303\times \frac{1000}{18}\times \frac{4180\times 18}{1000}\times \log \frac{373}{300}$

$=910.55J$

Entropy change for heating $1$ kg ${H_{2}O}_{\left(l\right)}$ to $1$ kg steam at ${100}^{o}C$.

$\Delta S_2={\int }_{373}^{473}\frac{{nC}_p.dT}{T}=m{\int }_{373}^{473}\frac{(1670+0.49T)dt}{T}$

(n is replaced by m since value of $C_p$ in ${3kg}^{-1}{K}^{-1}$

$m{\int }_{373}^{473}\frac{1670dT}{T}+m{\int }_{373}^{473}0.49dT$

$m\times 1670\times 2.303\times \left[\log T\right]\begin{array}{c}473\\ 373\end{array}+m\times 0.49\left[T\right]\begin{array}{c}473\\ 373\end{array}$

$=1\times 1670\times 2.303\times \log \frac{473}{373}+1\times 0.49\times 100$

$=396.73+49=445.73J$

Entropy change for heating $1$ kg steam at ${100}^{o}C$ to $1$ kg steam at ${200}^{o}C$.

$\Delta S_3=\frac{q}{T}=\frac{23\times {10}^5}{373}=6166.2J{K}^{-1}$

$\therefore$ Total entropy change$=\Delta S_1+\Delta S_2+\Delta S_3=910.55+6166.22+445.73$

$=7522.50J$

Hence, the correct option is $D$