Temperature and Heat — Question 6

Question 6

System boundaries distinguish electrical work from heat

An ideal 12.0 V source drives 2.00 A through an immersed resistor for 30.0 s. All supplied energy ultimately remains in a liquid-plus-resistor assembly. Neglect resistor energy storage, external losses and motion. The electrical energy input is $E_{\mathrm{el}}=VI\Delta t$.

1. For the assembly as the system, calculate electrical work input and heat crossing its outer boundary.

2. For the liquid alone as the system, calculate heat received from the resistor and its rise if $C_{\mathrm{liquid}}=720\,\mathrm{J/K}$. Explain the changed classification.

Question 6 - Solution

System boundaries distinguish electrical work from heat

(a) Method and calculation

$$E_{\mathrm{el}}=12.0(2.00)(30.0)=720\,\mathrm J.$$$$W_{\mathrm{el,in}}=720\,\mathrm J,\qquad Q_{\mathrm{outer}}=0.$$The assembly receives energy through the electrical leads, not through an external temperature difference.

(b) Interpretation and further calculation

$$Q_{\mathrm{liquid}}=720\,\mathrm J,\quad\Delta T=Q_{\mathrm{liquid}}/C_{\mathrm{liquid}}=1.00\,\mathrm K.$$$$E_{\mathrm{assembly\ gain}}=E_{\mathrm{liquid\ gain}}=720\,\mathrm J.$$Across the internal resistor–liquid interface the transfer is thermal. Changing the boundary changes the crossing mechanism being counted.

Physical check

Energy is counted once for each chosen boundary; electrical work and the subsequent internal heat transfer must not be added together.