Question

A cylindrical copper rod length 1.50m and radius 2.0cm is insulated to prevent heat loss through its curved surface.one end is attached to a thermal reservoir fixed at 300°C and the other is attached to a thermal reservoir fixed at 30°C.What is the rate at which entropy increases for the rod reservoirs system?

Answer 1

Explanations & Calculations

• The rate of heat transfer can be calculated as follows

$\qquad\qquad \begin{aligned} \small R&=\small \frac{Q}{t}=\frac{kA\Delta \theta}{d}\\ &= \small \frac{385WK^{-1}m^{-1}\times\pi (0.02m)^2\times(300-30)K}{1.5m}\\ &=\small 87.085\,W \end{aligned}$

• The entropy change in the 300C resevoir

$\qquad\qquad \begin{aligned} \small \Delta S_{res}&= \small \frac{Q}{300+273}=\frac{Q}{573K} \end{aligned}$

• The entropy change in 30C resevoir

$\qquad\qquad \begin{aligned} \small \Delta S_{sink}&= \small \frac{Q}{30+273K}=\frac{Q}{303K} \end{aligned}$

• The entropy change in the system

$\qquad\qquad \begin{aligned} \small \Delta S_{uni}&= \small \frac{Q}{303}-\frac{Q}{573}\\ &= \small Q\Big[\frac{1}{303}-\frac{1}{273}\Big]\\ \end{aligned}$

• Therefore, the rate of entropy change

$\qquad\qquad \begin{aligned} \small \frac{d\Delta S}{dt}&= \small \frac{Q}{t}\Big[\frac{1}{303}-\frac{1}{573}\Big]\\ &=\small 87.085\Big[1.555\times10^{-3}K^{-1}\Big]\\ &= \small 0.135\,WK^{-1} (>0\therefore \text{spontaneous}) \end{aligned}$