Question

Calculate the enthalpy change of fusion of $1$ mol of water at $5^{o}C$ to ice at $-5^{o}C$.
Given : ${\Delta }_{fus}H=6.03kJmo{l}^{-1}$ at $0^{o}C$
$C_p\left[H_{2}O\right(l\left)\right]=75.3Jmo{l}^{-1}{K}^{-1}$
$C_p\left[H_{2}O\right(s\left)\right]=36.8Jmo{l}^{-1}{K}^{-1}$

• A. $-6.59kJmo{l}^{-1}$
• B. $6.59kJmo{l}^{-1}$
• C. $-3.29kJmo{l}^{-1}$
• D. $3.29kJmo{l}^{-1}$

Answer 1

Answer: (A)

$-6.59kJmo{l}^{-1}$

The change of $5℃$ water to $-5℃$ ice takes place in three steps.

1) $5℃$ water to $0℃$ water.

2) $0℃$ water to $0℃$ ice.

3) $0℃$ ice to $-5℃$ ice.

Enthalpy change for step 1 is

$H_1=C_p\left[H_{2}O\right(l\left)\right]\times \Delta T{H}_1=75.3\times (-5)H_1=-376.5{Jmol}^{-1}=-0.376kJ{mol}^{-1}$

Enthalpy change for step 2

$H_2=-{\Delta }_{fus}H{H}_2=-6.03kJ{mol}^{-1}$

Enthalpy change for step 3 is

$H_3=C_p\left[H_{2}O\right(s\left)\right]\times \Delta T{H}_3=36.8\times (-5)H_3=-184{Jmol}^{-1}=-0.184kJ{mol}^{-1}$

So, total enthalpy change

$=H_1+H_2+H_3=-0.376-6.03-0.184=-6.59kJ{mol}^{-1}$

So, correct answer is option A.