1. Break the process into reversible steps for $1\,\text{kg}$ of substance:

   1. Ice at $273\,\text{K}$ $\to$ water at $273\,\text{K}$
   2. Water at $273\,\text{K}$ $\to$ water at $373\,\text{K}$
   3. Water at $373\,\text{K}$ $\to$ vapour at $373\,\text{K}$
   4. Vapour at $373\,\text{K}$ $\to$ vapour at $383\,\text{K}$

   Total entropy change: $\Delta S = \Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4$

2. Step 1: Melting of ice at $273\,\text{K}$

   $\Delta S_1 = \frac{L_f}{T} = \frac{334}{273}$

   $\Delta S_1 \approx 1.223\,\text{kJ kg}^{-1}\text{K}^{-1}$

3. Step 2: Heating liquid water from $273\,\text{K}$ to $373\,\text{K}$

   For heating at constant pressure: $\Delta S_2 = m c_{\ell} \ln\frac{T_2}{T_1}$

   Here $m=1$, $c_{\ell}=4.2\,\text{kJ kg}^{-1}\text{K}^{-1}$.

   $\Delta S_2 = 4.2\ln\frac{373}{273}$

   Using $\ln x = 2.303\log x$, $\ln\frac{373}{273}=2.303(\log 373-\log 273)$

   $=2.303(2.572-2.436)=2.303(0.136)=0.313$

   Therefore, $\Delta S_2 = 4.2\times 0.313 \approx 1.315\,\text{kJ kg}^{-1}\text{K}^{-1}$

4. Step 3: Vaporisation at $373\,\text{K}$

   $\Delta S_3 = \frac{L_v}{T} = \frac{2491}{373}$

   $\Delta S_3 \approx 6.678\,\text{kJ kg}^{-1}\text{K}^{-1}$

5. Step 4: Heating steam from $373\,\text{K}$ to $383\,\text{K}$

   $\Delta S_4 = m c_v \ln\frac{383}{373} = 2.0\ln\frac{383}{373}$

   $\ln\frac{383}{373}=2.303(\log 383-\log 373)$

   $=2.303(2.583-2.572)=2.303(0.011)=0.0253$

   Hence, $\Delta S_4 = 2.0\times 0.0253 \approx 0.0506\,\text{kJ kg}^{-1}\text{K}^{-1}$

6. Add all contributions

   $\Delta S = 1.223 + 1.315 + 6.678 + 0.0506$

   $\Delta S \approx 9.27\,\text{kJ kg}^{-1}\text{K}^{-1}$

   This matches closest with: $\boxed{9.26\,\text{kJ kg}^{-1}\text{K}^{-1}}$

7. Option check

   • A: $7.90$ ❌
   • B: $2.64$ ❌
   • C: $8.49$ ❌
   • D: $9.26$ ✅

Therefore, the correct option is D.