It can be shown that the change in entropy in heating or cooling a sample is given by the relation $\Delta \mathrm{S}=\mathrm{mc} \ln$ $\left(T_{F} / T_{i}\right)$, where $m$ is the mass of the sample, $c$ is the specific heat of the sample, and $T_{F}$ and $T_{i}$ are the final and initial temperatures, respectively. What is the change in entropy when 20.0 grams of aluminum with a specific heat of $0.900 \mathrm{~J} / \mathrm{g} \mathrm{K}$ is heated from $10.0^{\circ} \mathrm{C}$ to $60.0^{\circ} \mathrm{C}$ ?

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Quizplus · Accepted Answer

The change in entropy ($\Delta S$) can be calculated using the given formula $\Delta S = mc \ln(T_F / T_i)$, where $m = 20.0 \, \text{g}$, $c = 0.900 \, \text{J/gK}$, $T_i = 10.0^\circ C = 283.15 \, \text{K}$, and $T_F = 60.0^\circ C = 333.15 \, \text{K}$. Plugging these values into the formula gives a positive change in entropy, indicating an increase in disorder as the aluminum is heated.

It Can Be Shown That the Change in Entropy in Heating