A rod $$CD$$ of thermal resistance $$10.0$$ KW$$^{-1}$$ is joined at the middle of an identical rod $$AB$$ as shown in figure. The ends $$A$$, $$B$$ and $$D$$ are maintained at 200°C, 100°C and 125°C respectively. The heat current in $$CD$$ is _________ P W. The value of $$P$$ is _________.

We have three points whose temperatures are fixed: $$A \;(200^{\circ}\text{C}),\; B \;(100^{\circ}\text{C}),\; D \;(125^{\circ}\text{C})$$. An identical rod $$AB$$ has total thermal resistance $$10\;\text{K W}^{-1}$$, and rod $$CD$$ also has thermal resistance $$10\;\text{K W}^{-1}$$. Because the connection to rod $$CD$$ is made at the mid-point $$C$$ of $$AB$$, rod $$AB$$ is effectively split into two equal halves: $$AC$$ and $$CB$$.

The resistance of a portion is directly proportional to its length. Since each half is half the full length, each half has half the resistance: $$R_{AC}=R_{CB}=\dfrac{10}{2}=5\;\text{K W}^{-1}.$$

Let the steady-state temperature of the junction $$C$$ be $$T\;({}^{\circ}\text{C}).$$ For any rod the heat current $$Q$$ is given by the basic conduction relation $$Q=\dfrac{\Delta T}{R},$$ where $$\Delta T$$ is the temperature difference across the rod and $$R$$ is its thermal resistance.

The heat current from $$A$$ to $$C$$ is therefore $$Q_{AC}=\dfrac{200-T}{5}.$$

The heat current from $$C$$ to $$B$$ is $$Q_{CB}=\dfrac{T-100}{5}.$$

The heat current from $$C$$ to $$D$$ is $$Q_{CD}=\dfrac{T-125}{10}.$$

At junction $$C$$ the system is in steady state, so the heat arriving at $$C$$ from $$A$$ must equal the heat leaving $$C$$ towards $$B$$ and $$D$$: $$Q_{AC}=Q_{CB}+Q_{CD}.$$ Substituting the expressions, we get $$\dfrac{200-T}{5}=\dfrac{T-100}{5}+\dfrac{T-125}{10}.$$

To clear denominators, multiply the entire equation by $$10$$: $$2(200-T)=2(T-100)+(T-125).$$

Simplifying each side, we obtain $$400-2T=2T-200+T-125.$$ Combining like terms on the right, $$400-2T=3T-325.$$

Now move all the $$T$$ terms to one side and the constants to the other: $$400+325=3T+2T.$$ $$725=5T.$$ So, $$T=\dfrac{725}{5}=145^{\circ}\text{C}.$$

The required heat current in rod $$CD$$ is $$Q_{CD}=\dfrac{T-125}{10}=\dfrac{145-125}{10}=\dfrac{20}{10}=2\;\text{W}.$$

Hence, the heat current in $$CD$$ is $$2\;\text{W}$$, so the value of $$P$$ is $$2$$.

So, the answer is $$2$$.