You’ve seen that we can calculate $\text{Δ}S$ for various processes. In addition, entropy and the second law explain why interactions proceed in one direction but not the other. FIGURE 20.21 shows a system that consists of object C, at a colder temperature $T_{\text{C}}$, and object H, at the hotter temperature $T_{\text{H}}$. These objects can interact with each other, but they form an isolated system that does not interact with the rest of the universe. Note that this system is not in thermal equilibrium.

FIGURE 20.21 An interaction of two objects at different temperatures.

The system is the combination of objects C and H. What are the multiplicity and the entropy of this system? For each microstate in C there are ${\text{Ω}}_{\text{H}}$ microstates in H, so the total number of microstates, the multiplicity of the system, is the product ${\text{Ω}}_{\text{tot}}={\text{Ω}}_{\text{C}}{\text{Ω}}_{\text{H}}$. For example, if C has 4 microstates while H has 6, each of the 4 microstates of C can be paired with any one of the 6 microstates of H to give ${\text{Ω}}_{\text{tot}}=6+6+6+6=4\times 6=24$. Thus, using the definition of entropy, we find

We see that the entropy of a combined system is the sum of the entropies of each part. That is, entropy is additive—a useful property for a state variable.

Suppose a small amount of heat $dQ$ is transferred from object H to object C. The heat leaves H and enters C, so $d{Q}_{\text{H}}=-dQ$ and $d{Q}_{\text{C}}=+dQ$. This small energy exchange causes the entropy of the total system to change by

Notice that the expression in parentheses is positive because $T_{\text{C}}<T_{\text{H}}$.

If $dQ>0$, which describes an energy transfer from hot to cold, then $d{S}_{\text{sys}}>0$. The system’s entropy increases, and it will continue to increase with every little energy transfer until $T_{\text{C}}=T_{\text{H}}$, at which time $d{S}_{\text{sys}}=0$. This is the second law of thermodynamics! The total system is an isolated system, and initially it is not in equilibrium because the temperatures of C and H differ. The system’s entropy increases as C and H interact—the ability to disperse energy is increasing—until $T_{\text{C}}=T_{\text{H}}$. At that point, the entropy is maximized and the system is in thermal equilibrium.

Notice that the entropy of object H decreases as heat is withdrawn from it. The second law does not say that entropy can never decrease, only that the entropy of an isolated system cannot decrease. H is not isolated, but its entropy decrease is more than made up for by the entropy increase of C, so the entropy of the combined system, which is isolated, increases.

What would happen if, instead, heat was transferred from cold to hot, making C colder and H hotter? A cold-to-hot heat transfer would have $dQ<0$. From Equation 20.60, you can see that the associated change in the system’s entropy would be $d{S}_{\text{sys}}<0$, which takes the system to a macrostate with a lower multiplicity and a lower entropy. An energy transfer from cold to hot would not violate the first law of thermodynamics—energy would be conserved—but it would violate the second law of thermodynamics by requiring a decrease in the entropy of an isolated system. That’s why a heat transfer from cold to hot never happens.

The key idea here is that a system initially not in thermal equilibrium evolves toward thermal equilibrium because objects within the system interact with each other by exchanging energy. The exchanges conserve energy but increase entropy. As a result, a system evolves toward thermal equilibrium by moving through a series of macrostates of ever-increasing entropy.

We can make this idea more general. FIGURE 20.22 shows a small system interacting both thermally (heat is exchanged) and mechanically (work is done by a moving piston) with a much larger environment. This system is not isolated, so we can’t apply the second law of thermodynamics to the system alone. In principle there’s no limit to the size of the environment, so we can think of the system and its environment as forming the entire universe—that is, $\text{universe}=\text{system}+\text{environment}$. In most cases only the nearby environment is affected by the system, so “universe” is more figurative than literal, but there are situations where a system radiates heat into deep space and truly interacts with the entire universe.

FIGURE 20.22 A system interacting with its environment increases the entropy of the universe.

A system may not be isolated, but the universe as a whole is the ultimate isolated system. Thus the second law of thermodynamics applied to the universe is $\text{Δ}{S}_{\text{universe}}\ge 0\text{;}$ that is, nothing can cause the entropy of the universe to decrease. Entropy is additive, so $S_{\text{universe}}=S_{\text{sys}}+S_{\text{env}}$ and we can write the second law for the universe as

Only idealized reversible processes have $\text{Δ}{S}_{\text{universe}}=0$. Any realistic interaction between a system and its environment has $\text{Δ}{S}_{\text{universe}}>0$ and is irreversible because reversing the process, which would require $\text{Δ}{S}_{\text{universe}}<0$, would violate the second law.

For an isolated system, one that has no interactions with its environment, $\text{Δ}{S}_{\text{sys}}\ge 0$ because the system is the only part of the universe that is changing. The entropy is unchanging only if the system is in equilibrium. Otherwise, the entropy increases as the system evolves toward equilibrium. If the system is not isolated—and most are not—then the second law tells us that any interactions of the system with its environment must increase the entropy of the universe. It’s possible for the system’s entropy to decrease, but only if the environment’s entropy increases by a larger amount.