Three identical rods $AB$, $CD$ and $PQ$ are joined as shown. $P$ and $Q$ are mid points of $AB$ and $CD$ respectively. Ends $A, B, C$ and $D$ are maintained at $0^o C, 100^o C, 30^o C$ and $60^o C$ respectively. The direction of heat flow in $PQ$ is
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Both the rods $\mathrm{AB}$ and $\mathrm{CD}$ are in steady state.
I.e. the temperature gradient is a constant and the temperature drops linearly with distance.
Temperature at midpoint of $\mathrm{AB}=\frac{T_{A}+T_{B}}{2}=50^{\circ} \mathrm{C}=T_{P}$
Temperature at midpoint of $\mathrm{CD}=\frac{T_{C+T_{D}}}{2}=45^{\circ} C=T_{Q}$
since temperature at $P$ is more than that at $Q,$ heat flows from $P$ to $Q .$
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