Entropy Associated with Disorder

Find the increase in entropy of 1.00 kg of ice originally at $0º C$ that is melted to form water at $0º C$ .

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
As before, the change in entropy can be calculated from the definition of $Δ S$ once we find the energy $Q$ needed to melt the ice.
Solution
The change in entropy is defined as:

$ΔS = \frac{Q}{T} .$
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Here $Q$ is the heat transfer necessary to melt 1.00 kg of ice and is given by
$Q = {mL}_{f,}$
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where $m$ is the mass and $L_{f}$ is the latent heat of fusion. $L_{f} = 334 kJ/kg$ for water, so that
$Q = (1.00 kg) (334 kJ/kg) = 3 . 34 \times 10^{5} J.$
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Now the change in entropy is positive, since heat transfer occurs into the ice to cause the phase change; thus,

$Δ S = \frac{Q}{T} = \frac{3 . 34 \times 10^{5} J}{T} .$
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$T$ is the melting temperature of ice. That is, $T = 0º C=273 K$ . So the change in entropy is
$\begin{array}{l} Δ S & = & \frac{3.34 \times 10^{5} J}{273 K} \\ = & 1.22 \times 10^{3} J/K. \end{array}$
15.65
Discussion
This is a significant increase in entropy accompanying an increase in disorder.

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