A rod CD of thermal resistance 10.0 kW^-1 is joined at the middle of an identical rod AB as shown in

Q: Why is the resistance of AC and BC taken as 5?

Thermal resistance is directly proportional to the length of the conductor. Since C is the midpoint of AB, the length of AC and BC is half of the total length, making their resistance half of 10.

Q: Can we assume heat flows in any direction initially?

Yes! Just like in electrical circuits, you can assume any direction for the heat current. If your assumption is wrong, the final value will simply come out negative, but the magnitude will be correct.

Q: What does 'steady state' mean in this context?

Steady state means that the temperature at any given point in the rods is no longer changing with time. Because of this, the heat entering the junction must exactly equal the heat leaving it.

Q: Why do we use Kirchhoff's Current Law for heat?

Heat current is mathematically analogous to electrical current. Temperature difference acts like voltage (potential difference), and thermal resistance acts like electrical resistance. Hence, KCL applies perfectly.

Q: What if the rods had different cross-sectional areas?

Thermal resistance is given by $R = \frac{L}{KA}$. If the cross-sectional area $A$ changes, the resistance would change inversely. You would need to calculate the new resistances before applying KCL.

Question

A rod $CD$ of thermal resistance $10.0 	ext{ kW}^{-1}$ is joined at the middle of an identical rod $AB$ as shown in figure. The end $A, B$ and $D$ are maintained at $200^\circ	ext{C}, 100^\circ	ext{C}$ and $125^\circ	ext{C}$ respectively. The heat current in $CD$ is $P$ watt. The value of $P$ is ....... .

Sigma Percentile · Accepted Answer

### The Electrical Analogy of Heat

Imagine heat as a flowing river of energy. Just as water flows from high elevation to low elevation, heat flows from a region of high temperature to a region of low temperature. This beautiful symmetry allows us to borrow powerful tools from electrical circuits to solve complex thermal problems.

In this problem, we are given a T-shaped metallic structure. Instead of electrical resistance, we have **thermal resistance**, and instead of electrical current, we have **heat current**.

### Decoding the Thermal Resistances

We are given that the thermal resistance of rod $CD$ is $10 	ext{ kW}^{-1}$. The problem states that rod $AB$ is identical to rod $CD$. This implies that the total thermal resistance of rod $AB$ is also $10 	ext{ kW}^{-1}$.

However, rod $CD$ is joined at the exact middle of rod $AB$ at a junction we'll call $C$. Because thermal resistance is directly proportional to the length of the conductor ($R \propto L$), splitting rod $AB$ into two equal halves means the resistance is also halved.

Therefore, the resistance of segment $AC$ is:
$$R_{AC} = \frac{10}{2} = 5 	ext{ kW}^{-1}$$

Similarly, the resistance of segment $BC$ is:
$$R_{BC} = \frac{10}{2} = 5 	ext{ kW}^{-1}$$

### Applying Kirchhoff's Law for Heat

Now, let's focus on the junction $C$. Let the steady-state temperature at this junction be $T$. In a steady state, thermal energy does not accumulate at any point. This means the total heat entering the junction must perfectly balance the total heat leaving it. This is the thermal equivalent of **Kirchhoff's Current Law (KCL)**.

Let's assume heat flows from the hottest end $A$ ($200^\circ	ext{C}$) into junction $C$, and then splits, flowing outwards to ends $B$ ($100^\circ	ext{C}$) and $D$ ($125^\circ	ext{C}$).

We can write the heat current equation as:
$$I_{AC} = I_{CB} + I_{CD}$$

Using the formula for heat current $I = \frac{\Delta T}{R}$, we substitute our values:
$$\frac{200 - T}{5} = \frac{T - 100}{5} + \frac{T - 125}{10}$$

### The Final Calculation

To make the algebra cleaner, let's multiply the entire equation by $10$ to eliminate the denominators:
$$2(200 - T) = 2(T - 100) + (T - 125)$$

Expanding the brackets, we get:
$$400 - 2T = 2T - 200 + T - 125$$

Grouping the $T$ terms on one side and the constants on the other:
$$400 + 200 + 125 = 2T + 2T + T$$
$$725 = 5T$$

Solving for $T$, we find the junction temperature:
$$T = 145^\circ	ext{C}$$

Finally, the question asks for the heat current $P$ flowing through rod $CD$. We simply apply the heat current formula to segment $CD$:
$$P = I_{CD} = \frac{T - T_D}{R_{CD}}$$

Substituting our known values:
$$P = \frac{145 - 125}{10} = \frac{20}{10} = 2 	ext{ W}$$

The heat current in rod $CD$ is exactly **$2 	ext{ W}$**.

Enter Numerical Value:

Visualized Solution

System Setup

Thermal Resistance

Junction Temperature

Heat Current Equation

Substituting Values

Solving for $T$

Calculating Junction Temperature

Heat Current in $C D$

What If?

The Sigma Insight: Heat Transfer

The Electrical Analogy of Heat

Decoding the Thermal Resistances

Applying Kirchhoff's Law for Heat

The Final Calculation

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