Irreversible Processes and Entropy — Question 4

Question 4

Viscous stirring converts organized work into internal energy

A rigid insulated vessel contains liquid with constant heat capacity $C=1000$ J/K, initially at 300 K. A paddle supplies 6000 J of shaft work to the liquid. The paddle has no net energy storage. Use $\Delta S=C\ln(T_f/T_i)$.

1. Find final temperature and entropy change of the liquid.

2. Compute the heat that a reversible heating path would transfer between the same endpoint temperatures. Explain why work input does not carry entropy into the actual insulated vessel.

Question 4 - Solution

Viscous stirring converts organized work into internal energy

(a) Method and calculation

$$Q=0,\quad W_{\mathrm{by}}=-6000\,\mathrm J,\quad C(T_f-300)=6000\ \Rightarrow\ T_f=306\,\mathrm K.$$$$\Delta S=1000\ln(306/300)\approx19.8026\,\mathrm{J/K}.$$

(b) Interpretation and further calculation

$$Q_{\mathrm{rev}}=\int_{300}^{306}C\,dT=6000\,\mathrm J,\quad\int\frac{\delta Q_{\mathrm{rev}}}T=C\ln(306/300).$$$$S_{\mathrm{generated}}=\Delta S-\int\frac{\delta Q_{\mathrm{actual}}}{T_b}=19.8026\,\mathrm{J/K}.$$Actual heat transfer is zero. Dissipation generates entropy inside the liquid while work transfers energy across the boundary.

Physical check

Equal transferred energies as heat and work need not correspond to equal entropy transfers.