Total enthalpy change for freezing of 1 mol of water at $10^{\circ} \mathrm{C}$ to ice at $-10^{\circ} \mathrm{C}$ is ____ (Given : $\Delta_{ ext {fus }} H=x \mathrm{~kJ} / \mathrm{mol}$) $\begin{aligned} & \mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_2 \mathrm{O}(\mathrm{l}) ight]=\mathrm{y}…

Question

Total enthalpy change for freezing of 1 mol of water at $10^{\circ} \mathrm{C}$ to ice at $-10^{\circ} \mathrm{C}$ is ____
(Given : $\Delta_{\text {fus }} H=x \mathrm{~kJ} / \mathrm{mol}$)
$\begin{aligned} & \mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_2 \mathrm{O}(\mathrm{l})\right]=\mathrm{y} \mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1} \\ & \mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_2 \mathrm{O}(\mathrm{s})\right]=\mathrm{z} \mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\end{aligned}$

Aicharya · Accepted Answer

$\begin{aligned} & \Delta H=1 	imes y(0-10)-x 	imes 1000+1 	imes z\left(-10^{\circ}-0^{\circ}ight) \ & \Delta H=-10(100 x+y+z) 	ext { Joule. }\end{aligned}$

Total enthalpy change for freezing of 1 mol of water at $10^{\circ} \mathrm{C}$ to ice at $-10^{\circ}…

Solution