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

Chapter contents: Chapter 8: Thermal Physics

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. — **Figure 8.13.1.** 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)

Entropy is a property of a system’s state that helps us describe the direction of spontaneous processes. For reversible heat transfer at a constant absolute temperature, the entropy change is the heat transferred into the system divided by that temperature. The total entropy of an isolated system cannot decrease: it remains constant for an ideal reversible process and increases for an irreversible process.

Making Connections: Entropy, Energy, and Work

A heat engine converts part of the energy transferred to it as heat into mechanical work. It is not always possible, even in principle, to convert all of a system’s energy into work. Entropy helps determine how much energy is unavailable to do 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,

Q_{c} / Q_{h} = T_{c} / T_{h}

. Rearranging terms yields

\frac{Q_{c}}{T_{c}} = \frac{Q_{h}}{T_{h}}

(8.13.1)

for a reversible Carnot cycle.

Q_{c}

and

Q_{h}

are absolute values of the heat transfer at temperatures

T_{c}

and

T_{h}

, respectively. This ratio of

Q / T

is defined to be the change in entropy

Δ S

for reversible heat transfer at constant temperature,

Δ S = {(\frac{Q}{T})}_{rev}

(8.13.2)

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

Δ S

The expression above applies to reversible heat transfer at constant temperature. However, we can find

Δ 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

Δ 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

Δ S

for that process. That will be the change in entropy for any process going from state 1 to state 2. (See Figure 8.13.2.)

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. — **Figure 8.13.2.** When a system goes from state 1 to state 2, its entropy changes by the same amount $Δ S$ , 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 reversible Carnot engine and its heat reservoirs for one full cycle. The hot reservoir has a loss of entropy

Δ S_{h} = - Q_{h} / T_{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

Δ S_{c} = Q_{c} / T_{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

Δ S_{tot} = Δ S_{h} + Δ S_{c}

Thus, since we know that

Q_{h} / T_{h} = Q_{c} / T_{c}

for a reversible Carnot engine,

Δ S_{tot} = - \frac{Q_{h}}{T_{h}} + \frac{Q_{c}}{T_{c}} = 0

This result means the total change in entropy of the surrounding for one full cycle of a reversible Carnot engine is zero. The engine itself returns to its initial state after one full cycle, so its entropy is unchanged. Thus, the total entropy change of the engine and its surroundings is zero.

The entropy of the system and that of its surroundings may each change, but their changes sum to zero for a reversible process. 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 8.13.1

Entropy Increases in an Irreversible (Real) Process

A hot reservoir at 600 K loses 4000 J through heat transfer to a cold reservoir at 250 K. The reservoirs are sufficiently large that their temperature changes are negligible. Calculate their total change in entropy. (See Figure 8.13.3.)

Strategy

The hot reservoir’s internal energy decreases by 4000 J, while the cold reservoir’s internal energy increases by the same amount. These changes take each reservoir to a new state, even though its temperature remains nearly constant.

To calculate the entropy changes, imagine bringing each reservoir separately to its final state by reversible heat transfer. The hot reservoir releases 4000 J to another body infinitesimally cooler than itself; the cold reservoir receives 4000 J from another body infinitesimally warmer than itself. We can then use

Δ S = \frac{Q}{T}

for each reservoir.

Solution

For the hot reservoir,

Δ S_{h} = \frac{- Q_{h}}{T_{h}} = \frac{- 4000 J}{600 K} = - 6 . 67 J / K

For the cold reservoir,

Δ S_{c} = \frac{Q_{c}}{T_{c}} = \frac{4000 J}{250 K} = 16 . 0 J / K

Thus the total is

\begin{array}{rcl} Δ S_{tot} & = & Δ S_{h} + Δ S_{c} \\ = & (- 6 . 67 + 16.0) J/K \\ = & 9.33 J/K \end{array}

Discussion

The cold reservoir gains more entropy than the hot reservoir loses, so the total entropy increases. The actual heat transfer conserves energy, but reduces the energy available 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 shows separate reversible heat transfers: the hot reservoir releases heat to an auxiliary body and the cold reservoir receives heat from another auxiliary body. 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. — **Figure 8.13.3.** (a) Heat transfer from a hot reservoir to a cold one is irreversible and produces an overall increase in entropy. (b) Each reservoir can reach the same final state through a separate reversible heat transfer with an auxiliary body at an infinitesimally different temperature. These separate transfers give the same entropy changes for the two reservoirs as the actual irreversible transfer.

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 the system and its surroundings together during an irreversible process.

With respect to entropy, there are only two possibilities: the combined entropy of the system and its surroundings is constant for a reversible process and 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 an isolated 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. The total entropy of a system and its surroundings increases in irreversible processes. For engines operating between the same two heat reservoirs, reversible processes yield the most work for a given heat input and keep the combined entropy of the system and its surroundings constant. Thus we are led to make a connection between entropy and the availability of energy to do work.

Order to Disorder

The microscopic interpretation of entropy relates it to the number of arrangements compatible with a system’s macroscopic state. For example, when ice melts, its molecules gain access to many more arrangements in the liquid. The entropy increases, as the following example shows. The statistical interpretation is developed in the following section.

Example 8.13.2

Entropy Associated with Disorder

Find the increase in entropy of 1.00 kg of ice originally at

0° C

that is melted to form water at

0° C

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

For melting at constant temperature, the change in entropy is:

Δ S = \frac{Q}{T}

Here

Q

is the heat transfer necessary to melt 1.00 kg of ice and is given by

Q = m L_{f}

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

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}

T

is the melting temperature of ice. That is,

T = 0° C = 273 K

. So the change in entropy is

\begin{array}{rcl} Δ S & = & \frac{3.34 \times 10^{5} J}{273 K} \\ = & 1.22 \times 10^{3} J/K \end{array}

Discussion

This is a significant increase in entropy accompanying the change from solid to liquid.

The diagram has two images. The first image shows molecules of ice. They are represented as tiny spheres joined to form a floral pattern. The system is shown as ordered. The second image shows what happens when ice melts. The change in entropy delta S is marked between the two images shown by an arrow pointing from first image toward the second image with change in entropy delta S shown greater than zero. The second image represents water shown as tiny spheres moving in a random state. The system is marked as disordered. — **Figure 8.13.4.** When ice melts, its molecules gain access to more microscopic arrangements in the liquid. Its entropy increases during this phase transition.

In another easily imagined example, suppose we mix equal masses of water originally at two different temperatures, say

20.0° C

and

40.0° C

. The result is water at an intermediate temperature of

30.0° 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.

In this example, the increase in entropy accompanies a decrease in the energy available to do work.

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