Three rods of identical cross-sectional area and made from the same metal form the sides of an isosc

Q: Why is the net heat flow at point C zero?

In a steady state, the temperature of any point in the system does not change with time. If more heat entered than left, the temperature would rise. Thus, incoming heat must equal outgoing heat.

Q: How do we know the direction of heat flow?

Heat always flows from a region of higher temperature to a region of lower temperature. Since $T_B = \sqrt{2}T$ is greater than $T_A = T$, heat flows from B to C and then from C to A.

Q: Why did we assume the lengths of AB and BC to be $a$?

The problem states it's an isosceles right-angled triangle. The two sides forming the right angle must be equal. Assigning them a variable $a$ helps in calculating the hypotenuse length using the Pythagorean theorem.

Q: Does the cross-sectional area or thermal conductivity affect the final ratio?

In this specific problem, since all rods are made of the same metal and have identical cross-sectional areas, $K$ and $A$ are constant and cancel out from the steady-state equation.

Q: What if the triangle was not right-angled?

If the triangle wasn't right-angled, the length of the third side would be different, which would change the thermal resistance of that rod and consequently alter the steady-state temperature $T_c$.

Question

Three rods of identical cross-sectional area and made from the same metal form the sides of an isosceles triangle $ABC$, right angled at $B$. The points $A$ and $B$ are maintained at temperatures $T$ and $\sqrt{2}T$ respectively. In the steady state, the temperature of the point $C$ is $T_c$. Assuming that only heat conduction takes place, $T_c/T$ is

Options:

(A) $\frac{1}{2(\sqrt{2}-1)}$

(B) $\frac{3}{\sqrt{2}+1}$

(C) $\frac{1}{\sqrt{3}(\sqrt{2}-1)}$

(D) $\frac{1}{\sqrt{2}+1}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a triangular frame made of three identical metal rods. The problem tells us that this frame forms an isosceles right-angled triangle, with the right angle at vertex $B$.

Since it is an isosceles right-angled triangle, the two perpendicular sides must be equal in length. Let's assign them a length of $a$. So, $AB = BC = a$.

Using the Pythagorean theorem, we can easily find the length of the hypotenuse $AC$. It will be $\sqrt{a^2 + a^2} = \sqrt{2}a$. We are also given the temperatures at two vertices: $T_A = T$ and $T_B = \sqrt{2}T$. Our goal is to find the steady-state temperature at vertex $C$, which we will call $T_c$.

### The Principle of Steady State

The key to solving this problem lies in the phrase **"steady state"**. In a steady state, the temperature at any point in the system does not change with time.

For the temperature at junction $C$ to remain constant, the net heat flowing into the junction must be exactly equal to the net heat flowing out of it. This is essentially the principle of conservation of energy applied to heat transfer.

We know that heat always flows from a region of higher temperature to a region of lower temperature. Since $T_B = \sqrt{2}T$ is greater than $T_A = T$, heat will naturally flow from $B$ to $C$, and then continue from $C$ to $A$. Therefore, the heat current entering $C$ from $B$ ($\dot{Q}_{BC}$) must equal the heat current leaving $C$ towards $A$ ($\dot{Q}_{CA}$).

### Setting Up the Master Equation

The rate of heat flow, or heat current, through a rod is given by the formula $\dot{Q} = \frac{KA}{l}\Delta T$, where $K$ is the thermal conductivity, $A$ is the cross-sectional area, $l$ is the length, and $\Delta T$ is the temperature difference.

Let's write the expressions for the heat currents in our two rods. For rod $BC$, the length is $a$ and the temperature difference is $(\sqrt{2}T - T_c)$. So, $\dot{Q}_{BC} = \frac{KA}{a}(\sqrt{2}T - T_c)$.

For rod $CA$, the length is $\sqrt{2}a$ and the temperature difference is $(T_c - T)$. So, $\dot{Q}_{CA} = \frac{KA}{\sqrt{2}a}(T_c - T)$.

Equating these two expressions gives us our master equation:

$$ \frac{KA}{a}(\sqrt{2}T - T_c) = \frac{KA}{\sqrt{2}a}(T_c - T) $$

### Algebraic Simplification

This equation might look a bit messy, but it simplifies beautifully. Notice that the term $\frac{KA}{a}$ is present on both sides. Since the rods are identical, $K$ and $A$ are the same, and we can cancel this entire term out!

This leaves us with a much cleaner equation:

$$ \sqrt{2}T - T_c = \frac{T_c - T}{\sqrt{2}} $$

To get rid of the fraction, let's multiply the entire equation by $\sqrt{2}$. This gives:

$$ \sqrt{2}(\sqrt{2}T - T_c) = T_c - T $$

Expanding the left side, we get:

$$ 2T - \sqrt{2}T_c = T_c - T $$

### Final Calculation

Now, it's just a matter of simple algebra to isolate $T_c$. Let's group all the terms containing $T$ on the left side and all the terms containing $T_c$ on the right side.

Adding $T$ to both sides and adding $\sqrt{2}T_c$ to both sides, we get:

$$ 2T + T = T_c + \sqrt{2}T_c $$

$$ 3T = T_c(1 + \sqrt{2}) $$

Finally, dividing by $(1 + \sqrt{2})$, we find the steady-state temperature at $C$:

$$ T_c = \frac{3T}{\sqrt{2} + 1} $$

The question asks for the ratio $T_c/T$, which is simply:

$$ \frac{T_c}{T} = \frac{3}{\sqrt{2} + 1} $$

This perfectly matches option (b). The elegance of this problem lies in how the physical properties of the rods cancel out, leaving a purely geometric and algebraic relationship!

Select Answer:

Visualized Solution

$A B = B C = a, A C = 2 a$

$\dot{Q}_{in} = \dot{Q}_{out}$

$\dot{Q} = \frac{K A}{l} Δ T$

$\frac{K A}{a} (2 T - T_{c}) = \frac{K A}{2 a} (T_{c} - T)$

$2 (2 T - T_{c}) = T_{c} - T$

$3 T = T_{c} (1 + 2)$

$\frac{T _{c}}{T} = \frac{3}{2 + 1}$

$What if K_{1} \neq = K_{2} \neq = K_{3} ?$

The Sigma Insight: Heat Transfer

Analyzing the Setup

The Principle of Steady State

Setting Up the Master Equation

Algebraic Simplification

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB=BC=a,AC=2​a

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q˙​in​=Q˙​out​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q˙​=lKA​ΔT

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } aKA​(2​T−Tc​)=2​aKA​(Tc​−T)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2​(2​T−Tc​)=Tc​−T

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3T=Tc​(1+2​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TTc​​=2​+13​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if K1​=K2​=K3​?