Total enthalpy change for freezing of 1 mol water at 10 degrees C to ice at -10 degrees C is ______
(Given : $$\Delta_{fus}H = x$$ kJ/mol, $$C_p[H_2O(l)] = y$$ J mol$$^{-1}$$ K$$^{-1}$$, $$C_p[H_2O(s)] = z$$ J mol$$^{-1}$$ K$$^{-1}$$)

We have to go from liquid water at $$10^{\circ}\text{C}$$ to solid ice at $$-10^{\circ}\text{C}$$. Break the process into three reversible steps:

Step 1 (cooling liquid)
Water (l) $$10^{\circ}\text{C}\rightarrow 0^{\circ}\text{C}$$. For any temperature change, the enthalpy change is $$\Delta H = n\,C_p\,\Delta T$$. Here $$n = 1\;\text{mol}$$, $$C_p(l)=y\;\text{J mol}^{-1}\text{K}^{-1}$$, $$\Delta T = 0-10 = -10\;\text{K}$$. So $$\Delta H_1 = 1 \times y \times (-10)= -10y\;\text{J}$$ $$-(1)$$

Step 2 (freezing at $$0^{\circ}\text{C}$$)
Fusion enthalpy is given for melting, $$\Delta_{fus}H = x\;\text{kJ mol}^{-1}$$. For freezing, enthalpy change has the opposite sign: $$\Delta H_2 = -\,\Delta_{fus}H = -x\;\text{kJ}$$ $$-(2)$$

Step 3 (cooling ice)
Ice (s) $$0^{\circ}\text{C}\rightarrow -10^{\circ}\text{C}$$. Again use $$\Delta H = n\,C_p\,\Delta T$$ with $$C_p(s)=z\;\text{J mol}^{-1}\text{K}^{-1}$$. $$\Delta H_3 = 1 \times z \times (-10)= -10z\;\text{J}$$ $$-(3)$$

Total enthalpy change
Add $$(1),(2),(3)$$, but first convert everything to the same unit. Convert $$\Delta H_2$$ to joules: $$-x\;\text{kJ} = -1000x\;\text{J}$$.

Thus $$\Delta H_{\text{total}} = (-1000x) + (-10y) + (-10z)$$ $$\qquad = -\bigl(1000x + 10y + 10z\bigr)\;\text{J}$$.

Factor out $$-10$$: $$\Delta H_{\text{total}} = -10\bigl(100x + y + z\bigr)\;\text{J}$$.

This matches Option B: $$-10(100x + y + z)$$.

Answer : Option B