Q.Two metal rods, $A$ and $B$, of equal length $1\ \text{m}$ and equal cross-sectional area, are joined end to end to form a single composite rod. The thermal conductivity of $A$ is $k_A = 400\ \text{W m}^{-1}\text{K}^{-1}$ and that of $B$ is $k_B = 200\ \text{W m}^{-1}\text{K}^{-1}$. The free end of $A$ is maintained at $100\,^\circ\text{C}$ and the free end of $B$ at $0\,^\circ\text{C}$, and the sides of the composite rod are perfectly lagged. Find the steady-state temperature of the junction between $A$ and $B$.

Given: $k_A = 400\ \text{W m}^{-1}\text{K}^{-1}$, $k_B = 200\ \text{W m}^{-1}\text{K}^{-1}$, $L_A = L_B = 1\ \text{m}$, same cross-sectional area $A$ for both rods, free ends at $100\,^\circ\text{C}$ (rod $A$) and $0\,^\circ\text{C}$ (rod $B$). Let the junction temperature be $T$.

In the steady state, the same heat current must flow through rod $A$ and then on through rod $B$ (none is lost at the junction), so the heat currents through the two rods are equal:

Since $L_A = L_B$ and the common area $A$ cancels:

$40000 - 400T = 200T$ …

Equate the steady-state heat current through each rod (using each rod's own $k$ and the shared, unknown junction temperature …

• Assuming the junction must sit exactly at the midpoint temperature ($50\,^\circ\text{C}$), ignoring that the two rods have different conductivities. …