Question

In the process:
$H _2 O \left( s ,-10^{\circ} C , 1 atm \right) \rightarrow H _2 O \left(l, 10^{\circ} C , 1\, atm \right)$
$C_P$ for ice $=9\, cal\, deg ^{-1} \, mol ^{-1}, C_P$ for $H _2 O =18 \, cal\, deg ^{-1}\, mol ^{-1}$. Latent heat of fusion of ice $=$ $1440\, cal \, mol { }^{-1}$ at $0^{\circ} C$. The entropy change for the above process is $6.258 \, cal \, deg ^{-1} \, mol ^{-1}$ Give the total number of steps in which the third law of thermodynamics is used

Answer 1

Step 1. (using the third law of thermodynamics):
(For changing $H _2 O ( s )\left(-10^{\circ} C , 1\, atm \right) \rightarrow$ $H _2 O \left( s , 0^{\circ} C 1 \,atm \right)$
$\Delta S_1=\int\limits_{-10}^0 n \frac{C_P}{T} d T=1 \times 9 \times 2.3 \times \log \frac{273}{263}$
$=0.336 \,cal \,deg ^{-1} \,mol ^{-1}$
Step 2 (using the second law of thermodynamics):
$H _2 O ( s )\left(0^{\circ} C , 1 atm \right) \rightarrow H _2 O (l)\left(0^{\circ} C , 1 \,atm \right)$
$\Delta S_2=\frac{q_{ rev }}{T}=\frac{1440}{273}=5.258\, cal\, deg ^{-1}\, mol ^{-1}$
Step 3(using the third law of thermodynamics):
$H _2 O (l)\left(0^{\circ} C , 1\, atm \right) \rightarrow H _2 O (l)\left(10^{\circ} C , 1 \,atm \right)$
$\Delta S_3=\int\limits_0^{10} n \frac{C_P}{T} d T=1 \times 18 \times 2.3 \times \log \frac{283}{273} $
$=0.647 \,cal \,deg ^{-1} \,mol ^{-1} $
$\Delta S=\Delta S_1+\Delta S_2+\Delta S_3$
$=0.336+5.258+0.647 $
$=6.258\, cal\, deg ^{-1} \,mol ^{-1}$