Question

If 1 g of water at ${40}^{o}C$ is converted to steam at ${100}^{o}C$, the change in entropy is

• A. $2.303\log \frac{373}{313}ca{l}^o{C}^{-1}$
• B. $\frac{600}{373}ca{l}^o{C}^{-1}$
• C. $\frac{600}{313}ca{l}^o{C}^{-1}$
• D. $\frac{540}{373}+\log \frac{373}{313}ca{l}^o{C}^{-1}$

Answer 1

The correct option is D $\frac{540}{373}+\log \frac{373}{313}ca{l}^o{C}^{-1}$

$\Delta S={\int }_{T_1}^{T_2}\frac{dQ}{T}$ and $Q=msdT$

So, entropy $S_1=0.001\times 1000\ln \frac{373}{313}kj/k$
also $S_1=1\times \ln \frac{373}{313}ca{l}^o{C}^{-1}$

Here the water is changes from liquid to vapour so latent heat of vapourization will tool place which is equals to $539cal/gm$

$S_2=\frac{539}{313}ca{l}^o{C}^{-1}$

so total entropy change will be $S_{total}=\frac{539}{373}+ln\frac{373}{313}ca{l}^o{C}^{-1}$