The Question is from neet-physics thermal-properties-of-matter thermal-conductivity, and it was created by the author Mr. ANIL KUMAR.G

Question

Two rods of identical dimensions, but made of different materials A and B, are joined end to end. The thermal conductivity of A is twice that of B. If the free end of A is maintained at $100^\circ C$ and the free end of B is at $0^\circ C$, what is the temperature at the interface when steady state is reached?

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
 **Primary Concept**: Heat conduction and steady-state heat flow.
 **Related Topics**: Thermal conductivity, Fourier's law of heat conduction.
 **Prerequisites**: Understanding of heat transfer mechanisms and solving simultaneous equations.
## SOLUTION PATHWAY
**Given Information**:
  $k_A = 2k_B$ (Thermal conductivity of A is twice that of B)
  $T_A = 100^\circ C$ (Temperature at the free end of rod A)
  $T_B = 0^\circ C$ (Temperature at the free end of rod B)
  Identical dimensions (length $L$ and cross-sectional area $A$ are the same for both rods)
  Steady state is reached (heat flow is constant)
**Method & Steps**:
 1. Apply Fourier's law of heat conduction.  At steady state, the rate of heat flow through rod A ($Q_A$) equals the rate of heat flow through rod B ($Q_B$).
  Reasoning:  In steady state, no heat is accumulating at the interface; the heat flowing into the interface must equal the heat flowing out.
  Formula/Rule: $Q = kA\frac{ΔT}{L}$, where $Q$ is the rate of heat flow, $k$ is the thermal conductivity, $A$ is the cross-sectional area, $ΔT$ is the temperature difference, and $L$ is the length.
 2. Set up equations for heat flow in rods A and B. Let $T_i$ be the temperature at the interface.
  Working: For rod A, $Q_A = k_A A \frac{100 - T_i}{L}$. For rod B, $Q_B = k_B A \frac{T_i - 0}{L}$. Since $Q_A = Q_B$, we have $k_A A \frac{100 - T_i}{L} = k_B A \frac{T_i}{L}$.
  Key Point: The areas and lengths cancel out because the rods have identical dimensions.
 3. Solve for $T_i$. Substitute $k_A = 2k_B$ into the equation:
  $2k_B (100 - T_i) = k_B T_i$
  $200 - 2T_i = T_i$
  $3T_i = 200$
  $T_i = \frac{200}{3} \approx 66.7^\circ C$
  Result: The temperature at the interface is approximately $66.7^\circ C$.
  Verification: Substitute $T_i$ back into the equations for $Q_A$ and $Q_B$ to verify that they are equal.
 ## OPTION ANALYSIS
A :x: Incorrect calculation or misunderstanding of the concept of thermal conductivity.
B :x: Incorrect calculation or misunderstanding of the concept of thermal conductivity.
C ✓: Correct application of Fourier's law and solving the resulting equation.
D :x: Incorrect calculation or misunderstanding of the concept of thermal conductivity.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Ignoring the different thermal conductivities of the materials: This leads to an incorrect assumption that the interface temperature is the average of the end temperatures.
  How to avoid:  Remember to use Fourier's law correctly and account for the different $k$ values.
  2.Incorrectly setting up the equations for heat flow: Errors in setting up the temperature differences can lead to incorrect results.
  How to avoid: Carefully consider the direction of heat flow and the temperature differences across each rod.
 **Quick Tips**:
  1.Remember that in steady state, the heat flow rate is constant throughout the system.
  2.The interface temperature will be closer to the temperature of the rod with higher thermal conductivity.

Mr. ANIL KUMAR.G · Answer

$50^\circ C$

Mr. ANIL KUMAR.G · Answer

$33.3^\circ C$

Mr. ANIL KUMAR.G · Answer

The correct Option is "$66.7^\circ C$"

Mr. ANIL KUMAR.G · Answer

$75^\circ C$

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