Question

100 gm of water is slowly heated from ${27}^{o}C$ to ${87}^{o}C$. Calcualate the change in the entropy of water (special heat capacity of water $=4200J/kg-k$)

• A. 40 J/K
• B. 76.6 J/K
• C. 100 J/K
• D. 28 J/K

Answer 1

The correct option is B 76.6 J/K

Change in entropy $S=m\times C_{p}log\left(\frac{T_2}{T_1}\right)$
where,

m = mass of water (Kg)
$C_p$ = Specific heat capacity of Water (liquid) = 4200 J/Kg.K
$T_2$ = 360 K
$T_1$ = 300 K
So, Change in entropy = $0.1\times 4200\times log\left(\frac{360}{300}\right)=76.575053855J/K=76.6J/K$