A partition wall has two layers of different materials A and B in contact with each other. They have the same thickness but the thermal conductivity of layer A is twice that of layer B. At steady state the temperature difference across the layer B is 50 K, then the corresponding difference across the layer A is

Options

A. 50K
B. 12.5K
C. 25K
D. 60 K

Answer

To solve the problem, we need to apply the concept of heat conduction through two layers of materials with different thermal conductivities. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- We have two layers: Layer A and Layer B.
- The thickness of both layers is the same, denoted as \( x \).
- The thermal conductivity of Layer A is twice that of Layer B. If we denote the thermal conductivity of Layer B as \( k \), then the thermal conductivity of Layer A is \( 2k \).
- The temperature difference across Layer B (\( \Delta T_B \)) is given as 50 K.

2. **Use Fourier's Law of Heat Conduction:**
- According to Fourier's law, the heat transfer \( Q \) through a material is given by:
\[ Q = \frac{k \cdot A \cdot \Delta T}{x} \]
- Here, \( k \) is the thermal conductivity, \( A \) is the cross-sectional area, \( \Delta T \) is the temperature difference, and \( x \) is the thickness.

3. **Set Up the Equations for Heat Transfer:**
- For Layer A:
\[ Q_A = \frac{2k \cdot A \cdot \Delta T_A}{x} \]
- For Layer B:
\[ Q_B = \frac{k \cdot A \cdot \Delta T_B}{x} \]

4. **Equate the Heat Transfer:**
- At steady state, the heat transfer through both layers must be equal:
\[ Q_A = Q_B \]
- Substituting the expressions for \( Q_A \) and \( Q_B \):
\[ \frac{2k \cdot A \cdot \Delta T_A}{x} = \frac{k \cdot A \cdot \Delta T_B}{x} \]

5. **Cancel Common Terms:**
- The terms \( k \), \( A \), and \( x \) can be canceled from both sides:
\[ 2 \Delta T_A = \Delta T_B \]

6. **Substitute the Known Temperature Difference:**
- We know \( \Delta T_B = 50 \, \text{K} \):
\[ 2 \Delta T_A = 50 \, \text{K} \]

7. **Solve for \( \Delta T_A \):**
- Divide both sides by 2:
\[ \Delta T_A = \frac{50 \, \text{K}}{2} = 25 \, \text{K} \]

### Final Answer:
The corresponding temperature difference across Layer A is \( \Delta T_A = 25 \, \text{K} \).

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