Question

The left end of a copper rod (length = 20 cm , area of cross section = 0.20 $c{m}^2$) is maintained at ${20}^{\circ }C$ and the rightend is maintained at ${80}^{\circ }C$. Neglecting any loss of heat through radiation, find (a) the temperature at a point 11 cm from the left end and (b) the heat current through the rod. Thermal conductivity of copper = $385W{m}^{-1\circ }{C}^{-1}$.

Answer 1

l = 20 cm $=20\times {10}^{-2}m,$

A = $0.2c{m}^2=0.2\times {10}^{-4}{m}^2$

$T_1={80}^{\circ }C$ $T_2={20}^{\circ }C$

k = 385

(a) $\frac{Q}{t}=\frac{kA(T_1-T_2)}{l}$

= $\frac{385\times 0.2\times {10}^{-4}(80-20)}{20\times {10}^{-2}}$

= $\frac{385\times {10}^{-5}\times 60}{20\times {10}^{-2}}$

= $385\times 6\times {10}^{-4}\times 10$

=$2310\times {10}^{-3}=2.31$ J/s.

(b) Let the temperature of the 1 point at 11 cm be T.

$\frac{\Delta T}{l}=\frac{Q}{tkA}$

$\Rightarrow \frac{\Delta T}{l}=\frac{2.31}{385\times 0.2\times {10}^{-4}}$

$\Rightarrow \frac{T-20}{11\times {10}^{-2}}=\frac{2.31}{385\times 0.2}\times {10}^4$

$\Rightarrow T-20=\frac{2.31\times {10}^4}{385\times 0.2}\times 11\times {10}^{-2}$

= 33

$\Rightarrow T=33+20={53}^{\circ }C$