Conservation of Energy — Question 4

Question 4

Heat transfer and work in one signed balance

During a process, 80 J enters a system by heat transfer and the system does 50 J of work on its surroundings. Its kinetic energy increases 10 J and potential energy decreases 5 J. Only thermal energy can account for the remaining storage change.

1. Find the net energy transferred into the system.

2. Find its thermal-energy change.

3. Explain why neither heat input nor work output alone equals the thermal-energy change.

Question 4 - Solution

Heat transfer and work in one signed balance

(a) Take heat and work into the system as positive.

Choose transfers into the system as positive: $$Q=+80\ \mathrm J,\qquad W_{\mathrm{on}}=-50\ \mathrm J,$$ $$\boxed{\Delta E=Q+W_{\mathrm{on}}=30\ \mathrm J}.$$

(b) Subtract the other storage changes.

Subtract the other signed storage changes: $$30=\Delta K+\Delta U+\Delta E_{\mathrm{th}}=10-5+\Delta E_{\mathrm{th}},$$ $$\boxed{\Delta E_{\mathrm{th}}=25\ \mathrm J}.$$

(c) Separate boundary processes from state changes.

The general balance is $$\Delta E_{\mathrm{th}}=Q+W_{\mathrm{on}}-\Delta K-\Delta U.$$ Heat and work describe boundary transfers; the incoming energy can change multiple stores or leave through another channel. Thermal energy is stored, while heat labels a transfer process.