Ice at $$-5^{\circ}C$$ is heated to become vapor with temperature of $$110^{\circ}C$$ at atmospheric pressure. The entropy change associated with this process can be obtained from

To convert ice at $$-5°C$$ (268 K) to vapor at $$110°C$$ (383 K), the process is divided into heating ice from 268 K to 273 K, melting at 273 K, heating water from 273 K to 373 K, vaporizing at 373 K, and heating steam from 373 K to 383 K.

The change in entropy is given by $$dS = \frac{dq_{rev}}{T}$$.

For heating steps without phase change, the entropy change is calculated using $$\Delta S = \int \frac{C_{p,m}}{T} dT$$.

During a phase change at constant temperature, the entropy change follows $$\Delta S = \frac{\Delta H_m}{T}$$.

Therefore, the total entropy change for the entire process is expressed as $$\Delta S = \int_{268}^{273} \frac{C_{p,m}}{T} dT + \frac{\Delta H_{m,fusion}}{T_f} + \int_{273}^{373} \frac{C_{p,m}}{T} dT + \frac{\Delta H_{m,vaporisation}}{T_b} + \int_{373}^{383} \frac{C_{p,m}}{T} dT$$.

This matches Option 2.

The correct answer is Option 2.