Entropy and the Second Law of Thermodynamics

Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy

Photograph shows a glass of a beverage with ice cubes and a straw, placed on a paper napkin on the table. There is a piece of sliced lemon at the edge of the glass. There is condensate around the outside surface of the glass, giving the appearance that the ice is melting.

The ice in this drink is slowly melting. Eventually the liquid will reach thermal equilibrium, as predicted by the second law of thermodynamics. (credit: Jon Sullivan, PDPhoto.org)

There is yet another way of expressing the second law of thermodynamics. This version relates to a concept called entropy. By examining it, we shall see that the directions associated with the second law—heat transfer from hot to cold, for example—are related to the tendency in nature for systems to become disordered and for less energy to be available for use as work. The entropy of a system can in fact be shown to be a measure of its disorder and of the unavailability of energy to do work.

Making Connections: Entropy, Energy, and Work

Recall that the simple definition of energy is the ability to do work. Entropy is a measure of how much energy is not available to do work. Although all forms of energy are interconvertible, and all can be used to do work, it is not always possible, even in principle, to convert the entire available energy into work. That unavailable energy is of interest in thermodynamics, because the field of thermodynamics arose from efforts to convert heat to work.

We can see how entropy is defined by recalling our discussion of the Carnot engine. We noted that for a Carnot cycle, and hence for any reversible processes, \({Q}_{\text{c}}/{Q}_{\text{h}}={T}_{\text{c}}/{T}_{\text{h}}\). Rearranging terms yields

\(\cfrac{{Q}_{\text{c}}}{{T}_{\text{c}}}=\cfrac{{Q}_{\text{h}}}{{T}_{\text{h}}}\)

for any reversible process. \({Q}_{\text{c}}\) and \({Q}_{\text{h}}\) are absolute values of the heat transfer at temperatures \({T}_{\text{c}}\) and \({T}_{\text{h}}\), respectively. This ratio of \(Q/T\) is defined to be the change in entropy\(\Delta S\) for a reversible process,

\(\Delta S={(\cfrac{Q}{T})}_{\text{rev}}\text{,}\)

where \(Q\) is the heat transfer, which is positive for heat transfer into and negative for heat transfer out of, and \(T\) is the absolute temperature at which the reversible process takes place. The SI unit for entropy is joules per kelvin (J/K). If temperature changes during the process, then it is usually a good approximation (for small changes in temperature) to take \(T\) to be the average temperature, avoiding the need to use integral calculus to find \(\Delta S\).

The definition of \(\Delta S\) is strictly valid only for reversible processes, such as used in a Carnot engine. However, we can find \(\Delta S\) precisely even for real, irreversible processes. The reason is that the entropy \(S\) of a system, like internal energy \(U\), depends only on the state of the system and not how it reached that condition. Entropy is a property of state. Thus the change in entropy \(\Delta S\) of a system between state 1 and state 2 is the same no matter how the change occurs. We just need to find or imagine a reversible process that takes us from state 1 to state 2 and calculate \(\Delta S\) for that process. That will be the change in entropy for any process going from state 1 to state 2. (See the figure below.)

The diagram shows a schematic representation of a system that goes from state one with entropy S sub one to state two with entropy S sub two. The two states are shown as two circles drawn a distance apart. Two arrows represent two different processes to take the system from state one to state two. A straight arrow pointing from state one to state two shows a reversible process. The change in entropy for this process is given by delta S equals Q divided by T. The second process is marked as a curving arrow from state one to state two, showing an irreversible process. The change in entropy for this process is given by delta S sub irreversible equals delta S sub reversible equals S sub two minus S sub one.

When a system goes from state 1 to state 2, its entropy changes by the same amount \(\Delta S\), whether a hypothetical reversible path is followed or a real irreversible path is taken.

Now let us take a look at the change in entropy of a Carnot engine and its heat reservoirs for one full cycle. The hot reservoir has a loss of entropy \(\Delta {S}_{\text{h}}=-{Q}_{\text{h}}/{T}_{\text{h}}\), because heat transfer occurs out of it (remember that when heat transfers out, then \(Q\) has a negative sign). The cold reservoir has a gain of entropy \(\Delta {S}_{\text{c}}={Q}_{\text{c}}/{T}_{\text{c}}\), because heat transfer occurs into it. (We assume the reservoirs are sufficiently large that their temperatures are constant.) So the total change in entropy is

\(\Delta {S}_{\text{tot}}=\Delta {S}_{\text{h}}+\Delta {S}_{\text{c}}\text{.}\)

Thus, since we know that \({Q}_{\text{h}}/{T}_{\text{h}}={Q}_{\text{c}}/{T}_{\text{c}}\) for a Carnot engine,

\(\Delta {S}_{\text{tot}}\text{=–}\cfrac{{Q}_{\text{h}}}{{T}_{\text{h}}}+\cfrac{{Q}_{\text{c}}}{{T}_{\text{c}}}=0\text{.}\)

This result, which has general validity, means that the total change in entropy for a system in any reversible process is zero.

The entropy of various parts of the system may change, but the total change is zero. Furthermore, the system does not affect the entropy of its surroundings, since heat transfer between them does not occur. Thus the reversible process changes neither the total entropy of the system nor the entropy of its surroundings. Sometimes this is stated as follows: Reversible processes do not affect the total entropy of the universe. Real processes are not reversible, though, and they do change total entropy. We can, however, use hypothetical reversible processes to determine the value of entropy in real, irreversible processes. The following example illustrates this point.

Example: Entropy Increases in an Irreversible (Real) Process

Spontaneous heat transfer from hot to cold is an irreversible process. Calculate the total change in entropy if 4000 J of heat transfer occurs from a hot reservoir at C\({T}_{\text{h}}=\text{600 K}(\text{327º C})\) to a cold reservoir at C\({T}_{\text{c}}=\text{250 K}(-\text{23º C})\), assuming there is no temperature change in either reservoir. (See the figure below.)

Strategy

How can we calculate the change in entropy for an irreversible process when \(\Delta {S}_{\text{tot}}=\Delta {S}_{\text{h}}+\Delta {S}_{\text{c}}\) is valid only for reversible processes? Remember that the total change in entropy of the hot and cold reservoirs will be the same whether a reversible or irreversible process is involved in heat transfer from hot to cold. So we can calculate the change in entropy of the hot reservoir for a hypothetical reversible process in which 4000 J of heat transfer occurs from it; then we do the same for a hypothetical reversible process in which 4000 J of heat transfer occurs to the cold reservoir. This produces the same changes in the hot and cold reservoirs that would occur if the heat transfer were allowed to occur irreversibly between them, and so it also produces the same changes in entropy.

Solution

We now calculate the two changes in entropy using \(\Delta {S}_{\text{tot}}=\Delta {S}_{\text{h}}+\Delta {S}_{\text{c}}\). First, for the heat transfer from the hot reservoir,

\(\Delta {S}_{\text{h}}=\cfrac{-{Q}_{\text{h}}}{{T}_{\text{h}}}=\cfrac{-\text{4000 J}}{\text{600 K}}=–6\text{.}\text{67 J/K}\text{.}\)

And for the cold reservoir,

\(\Delta {S}_{\text{c}}=\cfrac{{Q}_{\text{c}}}{{T}_{\text{c}}}=\cfrac{\text{4000 J}}{\text{250 K}}=\text{16}\text{.}0 J/K\text{.}\)

Thus the total is

\(\begin{array}{ccc}\Delta {S}_{\text{tot}}& =& \Delta {S}_{\text{h}}+\Delta {S}_{\text{c}}\\ & =& (–6\text{.}\text{67 +16}\text{.}0)\phantom{\rule{0.25em}{0ex}}\text{J/K}\\ & =& \text{9.33 J/K.}\end{array}\)

Discussion

There is an increase in entropy for the system of two heat reservoirs undergoing this irreversible heat transfer. We will see that this means there is a loss of ability to do work with this transferred energy. Entropy has increased, and energy has become unavailable to do work.

Part a of the figure shows the irreversible heat transfer from the hot system to the cold system. The hot reservoir at temperature T sub h is represented by a rectangular section in the top and the cold reservoir at temperature T sub c is shown as a rectangular section at the bottom. Heat Q is shown to flow from hot reservoir to cold reservoir as shown by a continuous bold arrow pointing downward. The heat is a direct transfer from T sub h to T sub c. The entropy change delta S for an irreversible process is shown equal to entropy change delta S for a reversible process. Part b of the figure shows two reversible heat transfers from the hot system to the cold system. The hot reservoir at temperature T sub h is represented by a rectangular section in the top and the cold reservoir at temperature T sub c is shown as a rectangular section at the bottom. Heat Q is shown to flow out of the hot reservoir, and an equal amount of heat Q is shown to flow into the cold reservoir as shown by two arrows representing two reversible processes and not a direct transfer from T sub h to T sub c. The entropy change delta S for an irreversible process is shown equal to entropy change delta S for a reversible process.

(a) Heat transfer from a hot object to a cold one is an irreversible process that produces an overall increase in entropy. (b) The same final state and, thus, the same change in entropy is achieved for the objects if reversible heat transfer processes occur between the two objects whose temperatures are the same as the temperatures of the corresponding objects in the irreversible process.

It is reasonable that entropy increases for heat transfer from hot to cold. Since the change in entropy is \(Q/T\), there is a larger change at lower temperatures. The decrease in entropy of the hot object is therefore less than the increase in entropy of the cold object, producing an overall increase, just as in the previous example. This result is very general:

There is an increase in entropy for any system undergoing an irreversible process.

With respect to entropy, there are only two possibilities: entropy is constant for a reversible process, and it increases for an irreversible process. There is a fourth version of the second law of thermodynamics stated in terms of entropy:

The total entropy of a system either increases or remains constant in any process; it never decreases.

For example, heat transfer cannot occur spontaneously from cold to hot, because entropy would decrease.

Entropy is very different from energy. Entropy is not conserved but increases in all real processes. Reversible processes (such as in Carnot engines) are the processes in which the most heat transfer to work takes place and are also the ones that keep entropy constant. Thus we are led to make a connection between entropy and the availability of energy to do work.

Entropy and the Second Law of Thermodynamics