What is the change in entropy when $15.0 \mathrm {~g}$ of water at $100 ^ { \circ } \mathrm { C }$ are turned into steam at $100 ^ { \circ } \mathrm { C }$ ? The latent heat of vaporization of water is $22.6 \times 10 ^ { 5 } \mathrm {~J} / \mathrm { kg }$.
A) $0 \mathrm {~J} / \mathrm { K }$
B) $- 90.8 \mathrm {~J} / \mathrm { K }$
C) $339 \mathrm {~J} / \mathrm { K }$
D) $90.8 \mathrm {~J} / \mathrm { K }$
E) $- 339 \mathrm {~J} / \mathrm { K }$

Question

Quizplus · Accepted Answer

The change in entropy is calculated using the formula $\Delta S = \frac{Q}{T}$, where $Q$ is the heat absorbed and $T$ is the temperature in Kelvin. Here, $Q = mL = 0.015 \, \text{kg} \times 22.6 \times 10^5 \, \text{J/kg} = 3390 \, \text{J}$. The temperature $T = 100^\circ C = 373 \, \text{K}$. Thus, $\Delta S = \frac{3390}{373} \approx 9.08 \, \text{J/K}$, which is approximately $90.8 \, \text{J/K}$.

What Is the Change in Entropy When 15.0 g15.0 \mathrm {~g}15.0 g Of Water At

What Is the Change in Entropy When $15.0 \mathrm {~g}$ Of Water At