The law of entropy increase for bodies in mutual thermal contact may be argued using the fact that the final temperature in the thermal process is higher than the final temperature in a reversible process for work extraction.

When two bodies at unequal temperatures, $T1>T2$, are put in mutual thermal contact, it is observed that the hot body cools down and the cold body warms up. The process continues until the two bodies reach a common temperature ($TF$), which lies somewhere between the two initial temperatures: $T1>TF>T2$. The process is irreversible, implying that the total entropy ($S$) of the two bodies increases in the process. Since entropy is a state function, the standard calculation of the entropy change in each body is performed by choosing any convenient reversible path that connects the initial and final equilibrium states of each body and applying the Clausius formula, $ΔSi=∫dQi/T$, where i = 1, 2. For a reversible heat transfer, $dQi=Ci dT$, where the heat capacity $Ci>0$ is, in general, a function of the temperature of the body and is process dependent. The total change in entropy is the sum of changes for the individual bodies, given by

$ΔS=∫T1TFC1dTT+∫T2TFC2dTT,$
(1)

where $ΔS1=∫T1TFC1dT/T<0$, since $TF, and $ΔS2=∫T2TFC2dT/T>0$, since $TF>T2$. Thus, the second-law inequality, $ΔS>0$, is usually derived for this process by an explicit evaluation of the entropy changes while assuming a specific functional form for $Ci(T)$.1 Even though one term is negative in the above, the sum of the two terms is positive. In this article, we show that $ΔS>0$ may be argued from the fact that TF is greater than the final common temperature Tf that would be reached via a reversible work extraction process—a fact which itself derives from the positivity of the heat capacity of each body.

As a concrete example, consider the case of two bodies with constant volumes and fixed heat capacities C1 and C2, for which we obtain1

$ΔS=C1 ln TFT1+C2 ln TFT2,$
(2)

where the final temperature, obtained from energy conservation, is $TF=αT1+(1−α)T2$, with $α=C1/(C1+C2)$, satisfying $0≤α≤1$. The above expression for the entropy change can be rearranged into the form

$ΔS=(C1+C2) ln (αT1+(1−α)T2T1αT21−α).$
(3)

In this case, the proof of the inequality $ΔS>0$ rests on the inequality between weighted arithmetic and geometric means, given by $αT1+(1−α)T2>T1αT21−α$, for $T1≠T2$. There has been previous discussion around this apparent correspondence between physical laws and mathematical facts such as these inequalities.2–11 However, the fact that one of the means in the above comparison, $T1αT21−α$, is also the final common temperature of the two bodies when subjected to the process of reversible work extraction, seems to have escaped attention in the literature so far. More precisely, consider a reversible process that extracts work from the two bodies initially at temperatures T1 and T2.1,12,13 This may be achieved by introducing a heat engine and running infinitesimal, reversible heat cycles that gradually reduce the temperature difference between the two bodies, until the two bodies obtain a common temperature Tf.14 Now, Tf is determined by the reversibility condition: $C1 ln (Tf/T1)+C2 ln (Tf/T2)=0$, yielding $Tf=T1αT21−α$. So, Eq. (3) may be reexpressed as $ΔS=(C1+C2) ln (TF/Tf)$. Therefore, we can say that the condition TF > Tf directly implies $ΔS>0$. We argue below that TF > Tf holds not only for the case with algebraic means but also in general.

One may wonder, why does Tf figure in the expression for the entropy change in a thermal process? Actually, this suggests a suitable reversible path by which we can bring our initial two-body system to the final state at temperature TF, and which also clarifies that TF > Tf. The alternate path consists of two steps. In the first step, we bring the two bodies to the common temperature Tf by a reversible process, as described above. In the second step, an amount of heat—equal to the total work extracted above—is delivered to the two bodies, bringing them back at the initial total energy, but raising their common temperature to TF. The temperature increases, because the heat capacity of each body is defined to be positive. Since energy in the form of heat is added to the system, so we also expect that the total entropy of the two bodies shows an increase. This line of reasoning is, expectedly, more palatable to the students compared to a plain application of the Clausius formula.15

In conclusion, many previous derivations of the second-law in a thermal equilibration process rely on the use of inequalities between algebraic means (see also Ref. 16). However, as we have seen above, the law of entropy increase follows owing to the reason that the final temperature reached for the thermal process is higher than the final temperature in a reversible process for work extraction. The necessary and sufficient condition for the above proof of the second law is the positivity of the heat capacity of individual bodies, which leads to the condition $TF>Tf$.

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The envisaged reversible step requires the presence of a heat engine, a reversible work source, and a set of auxiliary heat reservoirs. To visualize one such infinitesimal heat cycle—say, the first one—we have prepared the two bodies in initial states using reservoirs at temperatures T1 and T2. Now, assume that we have another pair of reservoirs, which differ from the initial pair by infinitesimally different temperatures, denoted as $T1−δT1$ and $T2+δT2$, where $δTi>0$. Now, an infinitesimal amount of heat dQ1 is reversibly transferred from the hot body to the reservoir at $T1−δT1$, a reversible heat cycle is run, which outputs work dW and rejects heat $dQ2=dQ1−dW$ to the reservoir at $T2+δT2$. This heat is then reversibly transferred to the cold body at T2. In this whole process, the engine and the auxilliary reservoirs undergo a reversible process, so their total change in entropy is zero. As a consequence, the sum of the entropy changes in the bodies is also zero. Furthermore, the temperature of the hot body is reduced to $T1−δT1$ and that of cold body rises to $T2+δT2$, thus decreasing the difference between their temperatures. The sequence of cycles is repeated using heat reservoirs at appropriate temperatures, until the two bodies arrive at a common temperature Tf.
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The procedure is easily extended to the case of n bodies at initial temperatures ${Ti|i=1,2,…,n}$ and heat capacities $Ci(T)>0$. In this case, the net entropy change in the n bodies during the proposed alternate process is expressed as $ΔS=∑i=1n∫TiTf(Ci(T)/T)dT+∑i=1n∫TfTF(Ci(T)/T)dT$. The first sum vanishes due to the reversible work extraction step. Thus, the total entropy change is simply due to heat transfer to the collective system in the second step. Each integral in the second sum contributes positively, since TF > Tf.
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