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 $100$ K and that of the copper rod at $0$ K. Assuming that the rods are perfectly insulated from the surrounding, the temperature at the junction of the two rods is :

• A. $20$ K
• B. $30$ K
• C. $40$ K
• D. $50$ K

Answer 1

Answer: (A)

$20$ K

Given : $K_s=50W{m}^{-1}{K}^{-1}$ $K_c=400W{m}^{-1}{K}^{-1}$

As the two rods are connected in series with each other, thus heat flowing through both the rods must be same i.e. $Q_s=Q_c$

$\therefore$ $\frac{K_{s}A\Delta T_{s}t}{l_s}=\frac{K_c\Delta T_{c}t}{l_c}$

where $A$ is the area of each rod.

$\therefore$ $\frac{50\times A(100-T)t}{0.1}=\frac{400\times (T-0)t}{0.2}$

Or $100-T=4T$

$⟹$ $T=20$