
At steady state, the heat conducted by both the rods will be the same.
The rate of heat conduction is given by, dtdQ=lKAΔT. Therefore, if the temperature of the junction is T,
l1K1A1(T1−T)=l2K2A2(T−T2)
⇒T−0450−T=K1A1l2K2A2l1=9×21×2
⇒450−T=9T⇒T=45∘C