Conservation of Energy — Question 8

Question 8

Thermal redistribution inside an insulated pair

Two bodies are enclosed together with no external work or heat transfer. Their constant thermal storage coefficients are $C_A=40$ J/K and $C_B=60$ J/K. Initially A is at $80^\circ$C and B at $20^\circ$C. Use the supplied relation $\Delta E_{\mathrm{th}}=C\Delta T$ and assume a common final temperature, with no other energy changes.

1. Find the final temperature.

2. Find the thermal-energy change of each body.

3. Explain why averaging the two initial temperatures without weights is generally incorrect.

Question 8 - Solution

Thermal redistribution inside an insulated pair

(a) Conserve the sum of thermal energies.

Use the supplied thermal storage law for each body: $$40(T_f-80)+60(T_f-20)=0\Rightarrow100T_f=4400.$$ $$\boxed{T_f=44^\circ\mathrm C}.$$

(b) Evaluate each signed change.

Evaluate the signed energy changes: $$\boxed{\Delta E_A=40(44-80)=-1440\ \mathrm J},$$ $$\boxed{\Delta E_B=60(44-20)=+1440\ \mathrm J},\qquad\Delta E_A+\Delta E_B=0.$$

(c) Weight by storage per degree.

The general final temperature is weighted by storage capacity: $$\boxed{T_f=\frac{C_AT_A+C_BT_B}{C_A+C_B}}.$$ The arithmetic mean follows only when $C_A=C_B$. Here the larger coefficient of B pulls the final temperature closer to its initial 20 degrees.