1. Entropy change for a multi-step heating process

For entropy, the general expression is

When a substance is heated at constant pressure within a single phase,

so

For a phase change occurring reversibly at constant temperature,

2. Break the given process into steps

Ice at $-5^\circ\text{C}$ means initial temperature is

and final state is steam at $110^\circ\text{C}$, i.e.

At atmospheric pressure, the process occurs in these reversible steps:

1. Heat ice from $268\,\text{K}$ to $273\,\text{K}$:

   $$\Delta S_1 = \int_{268}^{273} \frac{C_{p,m}(\text{ice})}{T}\, dT$$

2. Melt ice at $273\,\text{K}$:

   $$\Delta S_2 = \frac{\Delta H_{m,\text{fusion}}}{T_f}$$

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

   $$\Delta S_3 = \int_{273}^{373} \frac{C_{p,m}(\text{liquid})}{T}\, dT$$

4. Vaporize water at $373\,\text{K}$:

   $$\Delta S_4 = \frac{\Delta H_{m,\text{vaporisation}}}{T_b}$$

5. Heat steam from $373\,\text{K}$ to $383\,\text{K}$:

   $$\Delta S_5 = \int_{373}^{383} \frac{C_{p,m}(\text{vapour})}{T}\, dT$$

Hence total entropy change is

3. Check the options

Option A

It uses

instead of

So this is an expression for enthalpy-like heating contributions, not entropy. Incorrect.

Option B

Again uses

without dividing by $T$. Incorrect.

Option C

This is too vague and also not the correct decomposition for the full process. Entropy requires

not simply $\int C_p dT + q_{\text{rev}}/T$. Incorrect.

Option D

This includes:

• heating contributions as $\int \frac{C_{p,m}}{T} dT$
• fusion entropy as $\frac{\Delta H_{m,\text{fusion}}}{T_f}$
• vaporization entropy as $\frac{\Delta H_{m,\text{vaporisation}}}{T_b}$

This is exactly the correct thermodynamic expression. Correct.

4. Final answer

The correct option is: