Answer
$83^\circ\;\rm C$
Work Step by Step
We can solve this problem using the principle of heat transfer by conduction. Since the copper and iron rods are welded together, they are in good thermal contact and heat will flow from the hot end of the copper rod to the cold end of the iron rod. The rate of heat transfer $Q$ is given by:
$$Q =\dfrac{ k A (T_1 - T_2) }{ L}\tag 1$$
where $k$ is the thermal conductivity of the material, $A$ is the cross-sectional area, $T_1$ is the temperature at one end, $T_2$ is the temperature at the other end, and $L$ is the length of the rod.
We assume that the heat loss by radiation is negligible.
Since the copper and iron rods have the same dimensions, their cross-sectional areas and lengths are equal. Therefore, the heat rate transfer of both of them is equal and is given by
$$\dfrac{Q_{copper}}{\Delta t} = \dfrac{Q_{iron}}{\Delta t}$$
Using (1);
$$\dfrac{ k_{Cu} A (T_H- T_M) }{L\Delta t} = \dfrac{ k_{Fe} A (T_M- T_C) }{L\Delta t} $$
where $T_H$ is the temperature at the hottest end, $T_M$ is at the middle point, and $T_C$ is at the coldest end.
Thus,
$$k_{Cu} (T_H- T_M)=k_{Fe} (T_M- T_C)$$
Solving for $T_M$
$$k_{Cu} T_H- k_{Cu} T_M =k_{Fe} T_M- k_{Fe}T_C $$
$$k_{Cu} T_H+k_{Fe}T_C =k_{Fe} T_M+ k_{Cu} T_M $$
$$k_{Cu} T_H+k_{Fe}T_C =T_M(k_{Fe} + k_{Cu}) $$
$$T_M=\dfrac{k_{Cu} T_H+k_{Fe}T_C }{k_{Fe} + k_{Cu}}$$
Plugging the known;
$$T_M=\dfrac{(400)(100+273) +(80)(0+273) }{80 + 400}=\bf 356\;\rm K$$
$$T_M=\color{red}{\bf 83}^\circ\rm C$$