Question

A cylindrical steel rod of length $0.10m$ and thermal conductivity $50W{m}^{-1}{K}^{-1}$ is welded end to end to copper rod of thermal conductivity $400W{m}^{-1}{K}^{-1}$ and of the same area of cross-section but $0.20m$ long. The free end of the steel rod is maintained at $100^{\circ }C$ and that of the copper rod at $0^{\circ }C$. Assuming that the rods are perfectly insulated from the surrounding, the temperature at the junction of the two rods is

• A. $20^{\circ }C$
• B. $30^{\circ }C$
• C. $40^{\circ }C$
• D. $50^{\circ }C$

Answer 1

Answer: (A)

In steady state, rate of heat flow through both rods is equal.
So ,${\left(\frac{kA\left(T_1-T\right)}{l}\right)}_{\text{steel }}={\left(\frac{kA\left(T-T_2\right)}{l}\right)}_{\text{copper }}$

Substituting given values, we have
$\frac{50.A\left(100-T\right)}{0.1}=\frac{400.A\left(T-0\right)}{0.2}$
$\Rightarrow 100\left(100-T\right)=400\left(T\right)$
$\Rightarrow T=20^{\circ }C$