Question

Two slabs are of the thicknesses $ d_1 $ and $ d_2 $. Their thermal conductivities are $ K_1 $ and $ K_2 $, respectively. They are in series. The free ends of the combination of these two slabs are kept at temperatures $ \theta_1 $ and $ \theta_2 $. Assume $ \theta_1>\theta_2 $. The temperature $ \theta $ of their common junction is

• A. \( \frac{K_1 d_1}{K_2 d_2} \)
• B. \( \frac{K_1 d_1}{K_1 d_2} \)
• C. \( \frac{K_2 d_2 \theta_1 + K_1 d_1 \theta_2}{K_1 d_2 + K_2 d_1} \)
• D. \( \frac{K_1 d_2 \theta_1 + K_2 d_1 \theta_2}{K_1 d_1 + K_2 d_2} \)

Hint:

When two slabs with different thermal conductivities are in series, use the formula \( \theta = \frac{K_2 d_2 \theta_1 + K_1 d_1 \theta_2}{K_1 d_2 + K_2 d_1} \) to calculate the temperature at the junction.

Answer 1

The Correct Option is C

The problem involves calculating the temperature at the common junction of two slabs in series with thicknesses \( d_1 \) and \( d_2 \), and thermal conductivities \( K_1 \) and \( K_2 \) respectively. The end temperatures of the slabs are \( \theta_1 \) and \( \theta_2 \), and we want to find the temperature \( \theta \) at their common junction.

Given that the slabs are in series and heat flow is steady, the rate of heat transfer through both slabs is the same. Using the formula for heat flow through a slab, we have:

For slab 1: \[ \frac{K_1 \cdot A \cdot (\theta_1 - \theta)}{d_1} \]

For slab 2: \[ \frac{K_2 \cdot A \cdot (\theta - \theta_2)}{d_2} \]

Since the heat flow rate is the same, equate the two expressions:

\[ \frac{K_1 \cdot A \cdot (\theta_1 - \theta)}{d_1} = \frac{K_2 \cdot A \cdot (\theta - \theta_2)}{d_2} \]

Cancel out the area \( A \) and rearrange the equation to solve for \( \theta \):

\[ K_1 d_2 (\theta_1 - \theta) = K_2 d_1 (\theta - \theta_2) \]

Expanding both sides gives:

\[ K_1 d_2 \theta_1 - K_1 d_2 \theta = K_2 d_1 \theta - K_2 d_1 \theta_2 \]

Rearrange to find \( \theta \):

\[ K_1 d_2 \theta_1 + K_2 d_1 \theta_2 = \theta (K_1 d_2 + K_2 d_1) \]

Solve for \( \theta \):

\[ \theta = \frac{K_2 d_1 \theta_2 + K_1 d_2 \theta_1}{K_1 d_2 + K_2 d_1} \]

This corresponds to the option: \(\frac{K_2 d_2 \theta_1 + K_1 d_1 \theta_2}{K_1 d_2 + K_2 d_1}\).

Answer 2

For two slabs in series with different thermal conductivities and thicknesses, the heat transfer rate is the same for both slabs. 
The temperature gradient is different across each slab, but the rate of heat transfer \( Q \) through each slab remains the same. 
The relation for steady-state heat conduction is: \[ Q = \frac{K_1 A (\theta_1 - \theta)}{d_1} = \frac{K_2 A (\theta - \theta_2)}{d_2} \] Here: 
- \( K_1 \) and \( K_2 \) are the thermal conductivities, 
- \( d_1 \) and \( d_2 \) are the thicknesses of the slabs, 
- \( \theta_1 \) and \( \theta_2 \) are the temperatures at the free ends, 
- \( \theta \) is the temperature at the common junction. Equating the two expressions for \( Q \), we get: \[ \frac{K_1 (\theta_1 - \theta)}{d_1} = \frac{K_2 (\theta - \theta_2)}{d_2} \] Multiplying out and solving for \( \theta \), we get: \[ K_1 d_2 (\theta_1 - \theta) = K_2 d_1 (\theta - \theta_2) \] \[ K_1 d_2 \theta_1 - K_1 d_2 \theta = K_2 d_1 \theta - K_2 d_1 \theta_2 \] Rearranging for \( \theta \), we get: \[ \theta (K_1 d_2 + K_2 d_1) = K_1 d_2 \theta_1 + K_2 d_1 \theta_2 \] \[ \theta = \frac{K_2 d_2 \theta_1 + K_1 d_1 \theta_2}{K_1 d_2 + K_2 d_1} \] 
Thus, the correct answer is: \[ \text{(3) } \frac{K_2 d_2 \theta_1 + K_1 d_1 \theta_2}{K_1 d_2 + K_2 d_1} \]