Q. A composite metal bar of uniform section is made up of length $25\ cm$ of copper, $10\ cm$ of nickel and $15\ cm$ of aluminium. Each part being in perfect thermal contact with the adjoining part. The copper end of the composite rod is maintained at $100℃$ and the aluminium end at $0℃$. The whole rod is covered with belt so that no heat loss occurs at the sides. If $K_{Cu} = 2K_{Al}$ and $K_{Al} = 3K_{Ni}$, then what will be the temperatures of $Cu - Ni$ and $Ni - Al$ junctions respectively
Explanation
(b) If suppose $K_{Ni} = K \Rightarrow K_{Al} = 3K$ and $K_{Cu} = 6K$
Since all metal bars are connected in series
So $\left( \frac{Q}{t} \right)_{Combination} = \left( \frac{Q}{t} \right)_{Cu} = \left( \frac{Q}{t} \right)_{Al} = \left( \frac{Q}{t} \right)_{Ni}$
$and\ \frac{3}{K_{eq}} = \frac{1}{K_{Cu}} + \frac{1}{K_{Al}} + \frac{1}{K_{Ni}} = \frac{1}{6K} + \frac{1}{3K} + \frac{1}{K} = \frac{9}{6K}$
$\Rightarrow K_{eq} = 2K$
Hence, it $\left( \frac{Q}{t} \right)_{Combination} = \left( \frac{Q}{t} \right)_{Cu}$
$\Rightarrow \frac{K_{eq}A(100 - 0)}{l_{Combination}} = \frac{K_{Cu}A(100 - \theta_{1})}{l_{Cu}}$
$\Rightarrow \ \frac{2K\ A(100 - 0)}{(25 + 10 + 15)} = \frac{6K\ A(100 - \theta_{1})}{25} \Rightarrow \theta_{1} = 83.33℃$
Similar if $\left( \frac{Q}{t} \right)_{Combination} = \left( \frac{Q}{t} \right)_{Al}$
$\Rightarrow \frac{2K\ A(100 - 0)}{50} = \frac{3K\ A(\theta_{2} - 0)}{15} \Rightarrow \theta_{2} = 20℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.