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

Question

Aicharya · Accepted Answer

$\begin{aligned}
& \underset{-5^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{~s})} \rightarrow \underset{0^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{~s})} ; \Delta \mathrm{S}_1=\int_{268 \mathrm{~K}}^{273 \mathrm{~K}} \frac{\mathrm{C}_{\mathrm{p}}, \mathrm{~m} \mathrm{dT}}{\mathrm{~T}} \\
& \underset{0^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{~s})} \rightleftharpoons \underset{0^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{l})} ; \Delta \mathrm{S}_2=\frac{\Delta \mathrm{H}_{\mathrm{m}} \text {, fus }}{273} \\
& \underset{0^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{I})} \rightarrow \underset{100^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{I})} ; \Delta \mathrm{S}_3=\int_{273}^{373} \frac{\mathrm{C}_{\mathrm{p}}, \mathrm{~m} \mathrm{dt}}{\mathrm{~T}} \\
& \underset{100^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{l})} \rightleftharpoons \underset{100^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{~g})} ; \Delta \mathrm{S}_4=\frac{\Delta \mathrm{H}_{\mathrm{m}} \text {, vap }}{373} \\
& \underset{100^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{~g})} \rightarrow \underset{110^{\circ} \mathrm{C}}{\mathrm{H}_2 \mathrm{O}(\mathrm{~g})} ; \Delta \mathrm{S}_5=\int_{373}^{383} \frac{\mathrm{C}_{\mathrm{p}, \mathrm{~m}} \mathrm{dT}}{\mathrm{~T}} \\
& \Delta \mathrm{~S}_{\text {total }}=\Delta \mathrm{S}_1+\Delta \mathrm{S}_2+\Delta \mathrm{S}_3+\Delta \mathrm{S}_4+\Delta \mathrm{S}_5
\end{aligned}$

Ice at $-5^{\circ} \mathrm{C}$ is heated to become vapor with temperature of $110^{\circ} \mathrm{C}$ at…

Solution