Question

In the process:
$H_{2}O\left(s,-10^{\circ }C,1atm\right)\to H_{2}O\left(l,10^{\circ }C,1atm\right)$
$C_P$ for ice $=9calde{g}^{-1}mo{l}^{-1},C_P$ for $H_{2}O=18calde{g}^{-1}mo{l}^{-1}$. Latent heat of fusion of ice $=$ $1440calmol{}^{-1}$ at $0^{\circ }C$. The entropy change for the above process is $6.258calde{g}^{-1}mo{l}^{-1}$ Give the total number of steps in which the third law of thermodynamics is used

Answer 1

Answer: 2

Step 1. (using the third law of thermodynamics):
(For changing $H_{2}O(s)\left(-10^{\circ }C,1atm\right)\to$ $H_{2}O\left(s,0^{\circ }C1atm\right)$
$\Delta S_1=\underset{-10}{\overset{0}{\int }}n\frac{C_P}{T}dT=1\times 9\times 2.3\times \log \frac{273}{263}$
$=0.336calde{g}^{-1}mo{l}^{-1}$
Step 2 (using the second law of thermodynamics):
$H_{2}O(s)\left(0^{\circ }C,1atm\right)\to H_{2}O(l)\left(0^{\circ }C,1atm\right)$
$\Delta S_2=\frac{q_{rev}}{T}=\frac{1440}{273}=5.258calde{g}^{-1}mo{l}^{-1}$
Step 3(using the third law of thermodynamics):
$H_{2}O(l)\left(0^{\circ }C,1atm\right)\to H_{2}O(l)\left(10^{\circ }C,1atm\right)$
$\Delta S_3=\underset{0}{\overset{10}{\int }}n\frac{C_P}{T}dT=1\times 18\times 2.3\times \log \frac{283}{273}$
$=0.647calde{g}^{-1}mo{l}^{-1}$
$\Delta S=\Delta S_1+\Delta S_2+\Delta S_3$
$=0.336+5.258+0.647$
$=6.258calde{g}^{-1}mo{l}^{-1}$