Question

In the process:
$H_{2}O(s,-{10}^{\circ }C,1atm)⟶H_{2}O(l,{10}^{\circ }C,1atm)$
$C_{P}forice=9calde{g}^{-1}mo{l}^{-1},C_{P}for{H}_{2}O=18calde{g}^{-1}mo{l}^{-1}$.
Latent heat of fusion of ice $=1440calmo{l}^{-1}at{0}^{\circ }C$.
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.

• A. 1
• B. 2
• C. 3
• D. 3rd Law is not used at all

Answer 1

The correct option is B 2

Step 1. (using the third law of thermodynamics):
$\left(Forchanging{H}_{2}O\right(s\left)\right(-{10}^{\circ }C,1atm)⟶H_{2}O(s,0^{\circ }C1atm)$
$\Delta S_1=\stackrel{0}{\underset{-10}{\int }}n\frac{C_P}{T}dT=1\times 9\times 2.3\times log\frac{273}{263}$

$=0.336calde{g}^{-1}mo{l}^{-1}$

Step 2 (using the second law of thermodynamics):
$H_{2}O\left(s\right)(0^{\circ }C,1atm)⟶H_{2}O\left(l\right)(0^{\circ }C,1atm)$
$\Delta S_2=\frac{q_{rev}}{T}=\frac{1440}{273}=5.258calde{g}^{-1}mo{l}^{-1}$
Step 3 (using the third law of thermodynamics):
$H_{2}O\left(l\right)(0^{\circ }C,1atm)⟶H_{2}O\left(l\right)({10}^{\circ }C,1atm)$
$\Delta S_3=\stackrel{10}{\underset{0}{\int }}n\frac{C_P}{T}dT=1\times 18\times 2.3\times log\frac{283}{273}$

$=0.647calde{g}^{-1}mo{l}^{-1}$

$\Delta S=\Delta S_1+\Delta S_2+\Delta S_3=0.336+5.258+0.647$

$=6.258calde{g}^{-1}mo{l}^{-1}$