Question

A composite bar of uniform cross-section is made of $25cm$ of copper, $10cm$ of nickel and $15cm$ of aluminium with perfect thermal contacts. The free copper end of the rod is at $100^{\circ }C$ and the free aluminium end is at $0^{\circ }C$. If $K_{Cu}=2K_{Al}$ and $K_{Al}=3K_{Ni}$, then the temperatures of $Cu-Ni$ and $Ni-Al$ junctions are respectively, (Assume no loss of heat occurs from the sides of the rod, $K$ -thermal conductivity).

• A. $82.3{}^{\circ }C,31.3^{\circ }C$
• B. $78.3^{\circ }C,26.1^{\circ }C$
• C. $70^{\circ }C,23.3^{\circ }C$
• D. $90.3^{\circ }C,30.1^{\circ }C$

Answer 1

Answer: (B)

Heat transfer, $\frac{\Delta Q}{\Delta t}=\frac{KA\left(T_1-T_2\right)}{L}$
or $\frac{K_{Cu}(A)\left(100-T_1\right)}{25}=\frac{K_{Ni}(A)\left(T_1-T_2\right)}{10}=\frac{K_{Al}(A)\left(T_2\right)}{15}$
$\because K_{Cu}=2K_{Al}$ and $K_{Al}=3K_{Ni}$
$\therefore \frac{2K_{Al}(A)\left(100-T_1\right)}{25}=\frac{K_{Al}(A)\left(T_2\right)}{15}$
$\Rightarrow 6T_1+5T_2=600...(i)$
$\Rightarrow \frac{K_{Ni}(A)\left(T_1-T_2\right)}{10}=\frac{3K_{Ni}(A)\left(T_2\right)}{15}$
$\Rightarrow T_1=3T_2...(i)$
From Eqs. (i) and (ii), we get
$T_2=26.08^{\circ }C$
$T_1=78.26^{\circ }C$
Hence, the temperature of $Cu-Ni$ and $Ni-Al$ junctions are respectively, $78.26^{\circ }$ and $26.08^{\circ }C$.