Order to Disorder

Order to Disorder

Entropy is related not only to the unavailability of energy to do work—it is also a measure of disorder. This notion was initially postulated by Ludwig Boltzmann in the 1800s. For example, melting a block of ice means taking a highly structured and orderly system of water molecules and converting it into a disorderly liquid in which molecules have no fixed positions. (See the figure below.) There is a large increase in entropy in the process, as seen in the following example.

Example: 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 \(0º C\).

Strategy

As before, the change in entropy can be calculated from the definition of \(\Delta S\) once we find the energy \(Q\) needed to melt the ice.

Solution

The change in entropy is defined as:

\(\mathrm{\Delta S}=\cfrac{Q}{T}\text{.}\)

Here \(Q\) is the heat transfer necessary to melt 1.00 kg of ice and is given by

\(Q={\text{mL}}_{\text{f}\text{,}}\)

where \(m\) is the mass and \({L}_{\text{f}}\) is the latent heat of fusion. \({L}_{\text{f}}=\text{334}\phantom{\rule{0.25em}{0ex}}\text{kJ/kg}\) for water, so that

\(Q=(1.00 kg)(334 kJ/kg)=3\text{.}\text{34}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{J.}\)

Now the change in entropy is positive, since heat transfer occurs into the ice to cause the phase change; thus,

\(\Delta S=\cfrac{Q}{T}=\cfrac{3\text{.}\text{34}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{J}}{T}\text{.}\)

\(T\) is the melting temperature of ice. That is, \(T=0º\text{C=273 K}\). So the change in entropy is

\(\begin{array}{lll}\Delta S& =& \cfrac{3\text{.}\text{34}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{J}}{\text{273 K}}\\ & =& \text{1.22}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{J/K.}\end{array}\)

Discussion

This is a significant increase in entropy accompanying an increase in disorder.

In another easily imagined example, suppose we mix equal masses of water originally at two different temperatures, say \(\text{20.0º C}\) and \(\text{40.0º C}\). The result is water at an intermediate temperature of \(\text{30.0º C}\). Three outcomes have resulted: entropy has increased, some energy has become unavailable to do work, and the system has become less orderly. Let us think about each of these results.

First, entropy has increased for the same reason that it did in the example above. Mixing the two bodies of water has the same effect as heat transfer from the hot one and the same heat transfer into the cold one. The mixing decreases the entropy of the hot water but increases the entropy of the cold water by a greater amount, producing an overall increase in entropy.

Second, once the two masses of water are mixed, there is only one temperature—you cannot run a heat engine with them. The energy that could have been used to run a heat engine is now unavailable to do work.

Third, the mixture is less orderly, or to use another term, less structured. Rather than having two masses at different temperatures and with different distributions of molecular speeds, we now have a single mass with a uniform temperature.

These three results—entropy, unavailability of energy, and disorder—are not only related but are in fact essentially equivalent.

This lesson is part of:

Thermodynamics

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