Question

A composite bar of uniform cross-section is made of $25\, cm$ of copper, $10\, cm$ of nickel and $15\, cm$ 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 $K_{ Cu }=2 K_{ Al }$ and $K_{ Al }=3\, K_{ Ni }$, 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, $\frac{\Delta Q}{\Delta t}=\frac{K A\left(T_{1}-T_{2}\right)}{L}$
or $\frac{K_{ Cu }(A)\left(100-T_{1}\right)}{25}=\frac{K_{ Ni }(A)\left(T_{1}-T_{2}\right)}{10}=\frac{K_{ Al }(A)\left(T_{2}\right)}{15}$
$\because K_{ Cu }=2 K_{ Al }$ and $K_{ Al }=3 K_{ Ni }$
$\therefore \frac{2 K_{ Al }(A)\left(100-T_{1}\right)}{25}=\frac{K_{ Al }(A)\left(T_{2}\right)}{15}$
$\Rightarrow 6 T_{1}+5 T_{2}=600\,\,\,...(i)$
$\Rightarrow \frac{K_{ Ni }(A)\left(T_{1}-T_{2}\right)}{10}=\frac{3 K_{ Ni }(A)\left(T_{2}\right)}{15}$
$\Rightarrow T_{1}=3 T_{2}\,\,\,...(i)$
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 $26.08^{\circ} C$.