A 0.42-kg quantity of ethanol, in the liquid state at its melting point of $- 114.4 ^ { \circ } \mathrm { C }$, is frozen at atmospheric pressure. The heat of fusion of ethanol is $1.04 \times 10 ^ { 5 } \mathrm {~J} / \mathrm { kg }$, and its molecular mass is $46.1 \mathrm {~g} / \mathrm { mol }$. What is the change in the entropy of the ethanol as it freezes?
A) $280 \mathrm {~J} / \mathrm { K }$
B) $250 \mathrm {~J} / \mathrm { K }$
C) $- 280 \mathrm {~J} / \mathrm { K }$
D) $- 250 \mathrm {~J} / \mathrm { K }$
E) $- 310 \mathrm {~J} / \mathrm { K }$

Question

Quizplus · Accepted Answer

The change in entropy ($\Delta S$) during the freezing process can be calculated using the formula $\Delta S = \frac{Q}{T}$, where $Q$ is the heat absorbed or released (in this case, released since it's freezing) and $T$ is the temperature in Kelvin. First, convert the mass of ethanol to kilograms (it's already in kg), calculate the heat released using the heat of fusion ($Q = m \cdot \text{heat of fusion}$), and convert the temperature from Celsius to Kelvin ($T_{K} = T_{C} + 273.15$). The negative sign in the answer indicates that the system is releasing heat, which is expected during freezing.

A 042-Kg Quantity of Ethanol, in the Liquid State at Its