Change in Entropy — Question 3

Question 3

Integrating a variable heat capacity distinguishes heat from entropy

A rigid body has total heat capacity $C(T)=a+bT$, where $a=20.0$ J/K and $b=.0500$ J/K$^2$. It warms reversibly from 300 K to 600 K, with no work.

1. Find heat input and entropy change by separate integrals.

2. Compare the exact entropy with $Q/450\,\mathrm K$. Explain why an arithmetic mean temperature generally cannot replace the integral.

Question 3 - Solution

Integrating a variable heat capacity distinguishes heat from entropy

(a) Method and calculation

$$Q=\int_{300}^{600}(a+bT)dT=a(300)+\frac b2(600^2-300^2)=12750\,\mathrm J.$$$$\Delta S=\int_{300}^{600}\left(\frac aT+b\right)dT=a\ln2+b(300)\approx28.8629\,\mathrm{J/K}.$$

(b) Interpretation and further calculation

$$Q/(450\,\mathrm K)=28.3333\,\mathrm{J/K}\ne\Delta S.$$$$T_{\mathrm{effective}}=Q/\Delta S\approx441.743\,\mathrm K.$$The entropy integral weights each incremental heat input by its own reciprocal temperature, rather than assigning all heat a single arithmetic-average temperature.

Physical check

An energy integral uses $C(T)$; an entropy integral uses $C(T)/T$.