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AI-লিখিত অনুমানভিত্তিক উত্তর—ধারণা ও সীমাবদ্ধতা নিচে উল্লেখ করা হয়েছে। এটি উৎসের যাচাইকৃত মূল উত্তর হিসেবে দাবি করা হচ্ছে না।

স্পষ্ট অনুমান: Standard values for specific heat and latent heat in SI units are used: s_ice = 2100 J/(kg·K), L_f = 3.36 × 10^5 J/kg, s_water = 4200 J/(kg·K), L_v = 2.26 × 10^6 J/kg. | The notation 'θ\theta°C' is evaluated both as an arbitrary sub-zero/zero initial temperature θ°C and as a likely typo for 0°C.

Let the initial temperature of ice be $\theta^\circ\text{C} = (273 + \theta)\text{ K}$, where $\theta \le 0$.
Mass of ice, $m = 1\text{ kg}$.

Standard thermodynamic constants:

• Specific heat capacity of ice, $s_{\text{ice}} = 2100\text{ J}/(\text{kg}\cdot\text{K})$
• Specific latent heat of fusion of ice, $L_f = 3.36 \times 10^5\text{ J}/\text{kg}$
• Specific heat capacity of water, $s_{\text{water}} = 4200\text{ J}/(\text{kg}\cdot\text{K})$
• Specific latent heat of vaporization of water, $L_v = 2.26 \times 10^6\text{ J}/\text{kg}$

The conversion takes place in up to four stages:

1. Heating ice from $\theta^\circ\text{C}$ to $0^\circ\text{C}$:
   $\Delta S_1 = \int_{273+\theta}^{273} \frac{m s_{\text{ice}}}{T}\,dT = m s_{\text{ice}} \ln\left(\frac{273}{273 + \theta}\right) = 2100 \ln\left(\frac{273}{273 + \theta}\right)\text{ J/K}$

2. Melting ice at $0^\circ\text{C}$ ($273\text{ K}$) into water at $0^\circ\text{C}$:
   $\Delta S_2 = \frac{m L_f}{T_f} = \frac{1 \times 3.36 \times 10^5}{273} \approx 1230.77\text{ J/K}$

3. Heating water from $0^\circ\text{C}$ ($273\text{ K}$) to $100^\circ\text{C}$ ($373\text{ K}$):
   $\Delta S_3 = m s_{\text{water}} \ln\left(\frac{373}{273}\right) = 1 \times 4200 \times \ln\left(\frac{373}{273}\right) \approx 4200 \times 0.31212 \approx 1310.89\text{ J/K}$

4. Vaporizing water at $100^\circ\text{C}$ ($373\text{ K}$) into steam at $100^\circ\text{C}$:
   $\Delta S_4 = \frac{m L_v}{T_b} = \frac{1 \times 2.26 \times 10^6}{373} \approx 6058.98\text{ J/K}$

Total change in entropy:
$\Delta S = \Delta S_1 + \Delta S_2 + \Delta S_3 + \Delta S_4$
$\Delta S = 2100 \ln\left(\frac{273}{273 + \theta}\right) + 1230.77 + 1310.89 + 6058.98$
$\Delta S = 2100 \ln\left(\frac{273}{273 + \theta}\right) + 8600.64\text{ J/K}$

Special Case (if initial temperature was $0^\circ\text{C}$, i.e., $\theta = 0$):
$\Delta S_1 = 0$
$\Delta S = 1230.77 + 1310.89 + 6058.98 \approx 8600.64\text{ J/K} \approx 8.60 \times 10^3\text{ J/K}$