Equilibrium temperature from entropy

Problem

Two bodies of equal heat capacity $C=800\;\text{J/K}$, one at $300\;\text{K}$ and one at $500\;\text{K}$, are brought into contact. Calculate the final equilibrium temperature (constant heat capacity) and verify that $\Delta S>0$.

Solution

The heat capacities are equal, so the equilibrium temperature is the arithmetic mean: $T_e=\frac{500+300}{2}=400\;\text{K}$ The total entropy change is the sum of the two contributions, each of the form $\Delta S=C\ln\frac{T_e}{T_\text{init}}$: $\Delta S=C\ln\frac{T_e}{T_2}+C\ln\frac{T_e}{T_1}=800\left[\ln\frac{400}{500}+\ln\frac{400}{300}\right]=800\left[-0{,}2231+0{,}2877\right]=800\cdot 0{,}0645=+51{,}6\;\text{J/K}>0$ The result is positive: the process (heat exchange between bodies at different temperatures) is irreversible. $\ev{T_e=400\;\text{K},\ \Delta S\approx 51{,}6\;\text{J/K}>0}$

Links

Topics: Entropy and the second law Concepts: Entropy · Temperature · Second law of thermodynamics Skills: Entropy balance · Calorimetric equation