Change in Entropy — Question 9

Question 9

The logarithmic mean is the effective temperature for reversible heat transfer

A body of constant heat capacity $C=100$ J/K warms reversibly from $T_1=300$ K to $T_2=600$ K. Define $T_{\mathrm{eff}}$ by $\Delta S=Q/T_{\mathrm{eff}}$.

1. Derive $T_{\mathrm{eff}}$ and calculate its value.

2. Show that $T_1<T_{\mathrm{eff}}<T_2$ and that the limit as $T_2\to T_1$ is $T_1$. Compare it with the arithmetic mean.

Question 9 - Solution

The logarithmic mean is the effective temperature for reversible heat transfer

(a) Method and calculation

$$Q=C(T_2-T_1),\quad\Delta S=C\ln(T_2/T_1).$$$$T_{\mathrm{eff}}=\frac{T_2-T_1}{\ln(T_2/T_1)}=\frac{300}{\ln2}\approx432.809\,\mathrm K.$$

(b) Interpretation and further calculation

$$\frac1{T_2}<\frac1{T_2-T_1}\int_{T_1}^{T_2}\frac{dT}{T}<\frac1{T_1}\ \Rightarrow\ T_1<T_{\mathrm{eff}}<T_2.$$$$\lim_{T_2\to T_1}\frac{T_2-T_1}{\ln T_2-\ln T_1}=T_1,\quad(T_1+T_2)/2=450\,\mathrm K.$$The effective temperature weights the incremental entropy transfer, so it differs from the arithmetic mean.

Physical check

This effective temperature is an integral summary, not a physical reservoir used throughout the reversible heating.