When $1.000 \mathrm{~kg}$ of ice melts, $33.00 \times 10^{4} \mathrm{~J} / \mathrm{kg}$ of heat are needed. What is the entropy change of the ice in the melting process?

Question

Quizplus · Accepted Answer

The entropy change ($\Delta S$) during the melting process can be calculated using the formula $\Delta S = \frac{Q}{T}$, where $Q$ is the heat added and $T$ is the temperature in Kelvin. Assuming the melting occurs at $0^\circ C$ or $273.15 K$, $\Delta S = \frac{33.00 \times 10^4 \, \text{J}}{273.15 \, \text{K}} \approx 1209 \, \text{J/K}$.

When 1.000 kg1.000 \mathrm{~kg}1.000 kg Of Ice Melts 33.00×104 J/kg33.00 \times 10^{4} \mathrm{~J} / \mathrm{kg}33.00×104 J/kg

When $1.000 \mathrm{~kg}$ Of Ice Melts $33.00 \times 10^{4} \mathrm{~J} / \mathrm{kg}$