Temperature — Question 2

Question 2

A hotter sample can store less modeled energy

Two equilibrium samples obey the supplied relation $U=CT$ with the same zero-energy convention. Sample A has $C_A=100\,\mathrm{J/K}$ and $T_A=400\,\mathrm K$; B has $C_B=1000\,\mathrm{J/K}$ and $T_B=300\,\mathrm K$. They are initially isolated.

1. Compare their temperatures and energies. Does the energy ranking identify the hotter sample?

2. If each receives $1000\,\mathrm J$ while $C$ remains constant, find the two temperature increases and final temperature difference.

Question 2 - Solution

A hotter sample can store less modeled energy

(a) Method and calculation

Temperature ranks A above B, but$$U_A=100(400)=40000\,\mathrm J,\qquad U_B=1000(300)=300000\,\mathrm J.$$$$T_A>T_B,\qquad U_A<U_B.$$Energy is extensive; it cannot replace temperature as the thermal-state coordinate.

(b) Interpretation and further calculation

From the stipulated model, $\Delta U=C\Delta T$:$$\Delta T_A=1000/100=10\,\mathrm K,\quad\Delta T_B=1000/1000=1\,\mathrm K.$$$$T_{A,f}-T_{B,f}=410-301=109\,\mathrm K.$$Equal energy increments need not produce equal temperature changes.

Physical check

The conclusion follows within the same energy reference; arbitrary energy zero shifts make cross-system energy rankings even less useful as thermometers.