A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount…
A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat $Q$ in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod when placed in thermal contact with the two reservoirs in time $\mathrm{t}$ ?
$Q / 4$
$Q / 16$
$2 \mathrm{Q}$
$Q / 2$
Solution
In steady state the amount of heat flowing from one face to the other face in time $t$ is given by $Q=\frac{K A\left(\theta_1-\theta_2\right) t}{l}$ where $K$ is coefficient of thermal conductivity of material of rod $\Rightarrow \quad \frac{\mathrm{Q}}{\mathrm{t}} \propto \frac{\mathrm{A}}{\mathrm{l}} \propto \frac{\mathrm{r}^2}{\mathrm{l}}$
As the metallic rod is melted and the material is formed into a rod of half the radius
$\begin{aligned}
\mathrm{V}_1 & =\mathrm{V}_2 \\
\Rightarrow \quad \pi \mathrm{r}_1^2 \mathrm{l}_1 & =\pi \mathrm{r}_2^2 \mathrm{l}_2 \\
\quad \quad \quad \mathrm{l}_1 & =\frac{\mathrm{l}_2}{4}
\end{aligned}$
Now, from Eqs. (i) and (ii)
$\begin{aligned}
\Rightarrow \quad & \frac{Q_1}{Q_2}=\frac{r_1^2}{l_1} \times \frac{l_2}{r_2^2}=\frac{r_1^2}{l_1} \times \frac{4 l_1}{\left(r_1 / 2\right)^2} \\
\Rightarrow \quad Q_1 & =16 Q_2
\end{aligned}$
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