A 1.00-kg sample of steam at 100.0 °C condenses to water at 100.0 °C.What is the entropy change of the sample if the heat of vaporization of water is 2.26 × 10⁶ J/kg?

Question

Quizplus · Accepted Answer

The entropy change ($\Delta S$) for the condensation process can be calculated using the formula $\Delta S = -\frac{Q}{T}$, where $Q$ is the heat released during condensation and $T$ is the absolute temperature in Kelvin. Given that the heat of vaporization ($Q$) is $2.26 \times 10^6$ J/kg for a 1.00-kg sample, and the temperature ($T$) is 100.0 °C or 373.15 K, the entropy change is $\Delta S = -\frac{2.26 \times 10^6 \, \text{J/kg}}{373.15 \, \text{K}} = -6.06 \times 10^3 \, \text{J/K}$.

A 100-Kg Sample of Steam at 100