When $2.000 \mathrm{~kg}$ of water evaporates, $22.60 \times 10^{5} \mathrm{~J} / \mathrm{kg}$ of heat are needed. What is the entropy change of the water in the boiling process?

Question

Quizplus · Accepted Answer

The entropy change $\Delta S$ is calculated using $\Delta S = \frac{Q}{T}$, where $Q$ is the heat absorbed and $T$ is the temperature in Kelvin. For water boiling at $100^\circ C$ or $373 \, \mathrm{K}$, $Q = 2.000 \, \mathrm{kg} \times 22.60 \times 10^{5} \, \mathrm{J/kg} = 45.20 \times 10^{5} \, \mathrm{J}$. Thus, $\Delta S = \frac{45.20 \times 10^{5} \, \mathrm{J}}{373 \, \mathrm{K}} = 12,120 \, \mathrm{J/K}$.

When 2.000 kg2.000 \mathrm{~kg}2.000 kg Of Water Evaporates 22.60×105 J/kg22.60 \times 10^{5} \mathrm{~J} / \mathrm{kg}22.60×105 J/kg

When $2.000 \mathrm{~kg}$ Of Water Evaporates $22.60 \times 10^{5} \mathrm{~J} / \mathrm{kg}$