Entropy and the Unavailability of Energy to Do Work

Entropy and the Unavailability of Energy to Do Work

What does a change in entropy mean, and why should we be interested in it? One reason is that entropy is directly related to the fact that not all heat transfer can be converted into work. The next example gives some indication of how an increase in entropy results in less heat transfer into work.

Example: Less Work is Produced by a Given Heat Transfer When Entropy Change is Greater

(a) Calculate the work output of a Carnot engine operating between temperatures of 600 K and 100 K for 4000 J of heat transfer to the engine. (b) Now suppose that the 4000 J of heat transfer occurs first from the 600 K reservoir to a 250 K reservoir (without doing any work, and this produces the increase in entropy calculated above) before transferring into a Carnot engine operating between 250 K and 100 K. What work output is produced? (See the figure below.)

Strategy

In both parts, we must first calculate the Carnot efficiency and then the work output.

Solution (a)

The Carnot efficiency is given by

\({\text{Eff}}_{\text{C}}=1-\cfrac{{T}_{\text{c}}}{{T}_{\text{h}}}\text{.}\)

Substituting the given temperatures yields

\({\text{Eff}}_{\text{C}}=1-\cfrac{\text{100 K}}{\text{600 K}}=0\text{.}\text{833}\text{.}\)

Now the work output can be calculated using the definition of efficiency for any heat engine as given by

\(\text{Eff}=\cfrac{W}{{Q}_{\text{h}}}\text{.}\)

Solving for \(W\) and substituting known terms gives

\(\begin{array}{lll}W& =& {\text{Eff}}_{\text{C}}{Q}_{\text{h}}\\ & =& (0.833)(\text{4000}\phantom{\rule{0.25em}{0ex}}\text{J})=\text{3333 J.}\end{array}\)

Solution (b)

Similarly,

\({\text{Eff}\prime }_{\text{C}}=1-\cfrac{{T}_{\text{c}}}{{\text{T′}}_{\text{c}}}=\text{1}-\cfrac{\text{100 K}}{\text{250 K}}=0.600,\)

so that

\(\begin{array}{lll}W& =& {\text{Eff}\prime }_{\text{C}}{Q}_{h}\\ & =& (0.600)(\text{4000 J})=\text{2400 J.}\end{array}\)

Discussion

There is 933 J less work from the same heat transfer in the second process. This result is important. The same heat transfer into two perfect engines produces different work outputs, because the entropy change differs in the two cases. In the second case, entropy is greater and less work is produced. Entropy is associated with the unavailability of energy to do work.

When entropy increases, a certain amount of energy becomes permanently unavailable to do work. The energy is not lost, but its character is changed, so that some of it can never be converted to doing work—that is, to an organized force acting through a distance. For instance, in the previous example, 933 J less work was done after an increase in entropy of 9.33 J/K occurred in the 4000 J heat transfer from the 600 K reservoir to the 250 K reservoir. It can be shown that the amount of energy that becomes unavailable for work is

\({W}_{\text{unavail}}=\Delta S\cdot {T}_{0}\text{,}\)

where \({T}_{0}\) is the lowest temperature utilized. In the previous example,

\({W}_{\text{unavail}}=(9\text{.}\text{33 J/K})(\text{100 K})=933 J\)

as found.

This lesson is part of:

Thermodynamics

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