Question

The enthalpy change on freezing of $1mol$ of water at $5^{o}C$ to ice at $-5^{o}C$ is :

$($Given ${\Delta }_{fus}H=6kJmo{l}^{-1}$ at $0^{o}C$,

$C_p(H_{2}O,l)=75.3Jmo{l}^{-1}{K}^{-1}$

$C_p(H_{2}O,S)=36.8Jmo{l}^{-1}{K}^{-1})$

• A. $5.81kJmo{l}^{-1}$
• B. $6.56kJmo{l}^{-1}$
• C. $6.00kJmo{l}^{-1}$
• D. $5.44kJmo{l}^{-1}$

Answer 1

The correct option is B $6.56kJmo{l}^{-1}$

(1) Water at $5^{o}C$ is cooled to water at $0^{o}C$

The enthalpy change $=n{C}_p(H_{2}O,l){\Delta }_T=1\times 75.3J/mol/K\times (5-0{)}^{o}C=376.5J/=0.3765kJ$......(1)

Here, n is the number of moles of water, $C_p(H_{2}O,l)$ is the specific heat of liquid water and $\Delta T$ is temperature change.

Also $1kJ=1000J$
(2) Liquid water at $0^{o}C$ is fused at same temperature.

The enthalpy change $=n{\Delta }_{fus}H=1\times 6kJ/mol=6kJ$......(2)
(3) Ice at $0^{o}C$ is cooled to ice at $-5^{o}C$

The enthalpy change $=n{C}_p(H_{2}O,s){\Delta }_T=1\times 36.8J/mol/K\times (0-(-5){)}^{o}C=184J/=0.184kJ$......(3)

Here, n is the number of moles of ice, $C_p(H_{2}O,l)$ is the specific heat of ice and $\Delta T$ is temperature change. Also $1kJ=1000J$
Add (1), (2) and (3)

The enthalpy change for entire process $=0.3765kJ+6kJ+0.184kJ=6.56kJmo{l}^{-1}$

Hence, the option (B) is correct answer.