Total enthalpy change for freezing of 1 mol of water at 10^ C to ice at -10^ C is ____ (Given : Δ_ fus H=x ~kJ / mol ) aligned & C_ p [H_2 O(l)]=y J mol^ -1 ~K^ -1 & C_ p [H_2 O(s)]=z J mol^ -1 ~K^ -1 aligned

Correct answer: B. -10(100 x+y+z) aligned & Δ H=1 × y(0-10)-x × 1000+1 × z(-10^ -0^ ) & Δ H=-10(100 x+y+z) Joule. aligned

JEE Main 20257 AprilMorning Shift

JEE Main 2025 · 7 April · Morning Shift · Chemistry · Thermodynamics (C)

Total enthalpy change for freezing of 1 mol of water at 10^ C to...

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}

B. $-10(100 x+y+z)$

\begin{aligned} & \Delta H=1 \times y(0-10)-x \times 1000+1 \times z\left(-10^{\circ}-0^{\circ}\right) \\ & \Delta H=-10(100 x+y+z) \text{ Joule. }\end{aligned}