Question

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

• A. 2.303 $log\frac{373}{313}Cal/K$
• B. $\frac{600}{373}Cal/K$
• C. $\frac{600}{313}Cal/K$
• D. $\frac{540}{373}+log\frac{373}{313}Cal/K$

Answer 1

The correct option is D $\frac{540}{373}+log\frac{373}{313}Cal/K$

Change in entropy = $\frac{Q}{T}+C_{p}log\left(\frac{T_2}{T_1}\right)$
Q - Heat supplied at constant temperature = Latent heat of vapourisation of water = 540 Cal/gram
$C_p$ - Specific heat capacity of Water (liquid) = 1 Cal/gram.K
$T_2=373K$
$T_1=313K=T$
So Change in entropy = $\frac{540}{373}+\log \left(\frac{373}{313}\right)\left[cal/K\right]$