Solution:

Given:

- Layer 1: Thermal conductivity \(K_1\)
- Layer 2: Thermal conductivity \(K_2\)
- Thickness of each layer: \(d\)

Formula for Equivalent Thermal Conductivity

The equivalent thermal conductivity for two parallel layers of the same thickness can be given by:

\[
K_{eq} = \frac{K_1 + K_2}{2}
\]

Derivation:

1. Resistance of Each Layer:
The thermal resistance for each layer can be expressed as:
\[
R_1 = \frac{d}{K_1 A}, \quad R_2 = \frac{d}{K_2 A}
\]
where \(A\) is the cross-sectional area.

2. Total Resistance in Parallel:
The total resistance for two resistors in parallel is:
\[
\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}
\]
Substituting the resistances, we get:
\[
\frac{1}{R_{eq}} = \frac{K_1 A}{d} + \frac{K_2 A}{d}
\]
Simplifying:
\[
\frac{1}{R_{eq}} = \frac{A}{d} \left(K_1 + K_2\right)
\]

3. Total Conductivity:
Now, the equivalent conductivity can be expressed as:
\[
K_{eq} = \frac{d}{A} \cdot \frac{1}{R_{eq}} = \frac{d}{A} \cdot \frac{d}{A (K_1 + K_2)} = \frac{K_1 + K_2}{2}
\]

Conclusion:
Thus, for two parallel layers of materials with equal thickness, the correct equivalent thermal conductivity is:

\[
K_{eq} = \frac{K_1 + K_2}{2}
\]

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Solution:

Convection is the process by which heat is transferred through the movement of fluids, including gases like air. In the atmosphere, convection occurs when warm air rises and cool air sinks.

Here's a short explanation of how this works:

1. Heating the Surface: The sun heats the Earth's surface, which in turn warms the air above it.
2. Rising Warm Air: As the air warms, it becomes less dense and rises.
3. Cooling and Sinking: Once the warm air rises, it cools down at higher altitudes, becomes denser, and eventually sinks back down.
4. Cycle Continuation: This cycle of rising warm air and sinking cool air creates convection currents, which distribute heat throughout the atmosphere, influencing weather patterns and temperature distribution.

Overall, convection plays a crucial role in regulating the Earth's climate and weather systems.

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Solution:

The temperature gradient is the rate at which temperature changes with respect to distance. It's given as \( 80^\circ \text{C/m} \), and the length of the rod is \( 0.5 \, \text{m} \).

To find the temperature difference across the rod, we use the formula:

\[
\Delta T = \text{Temperature gradient} \times \text{Length}
\]

Substitute the values:

\[
\Delta T = 80^\circ \text{C/m} \times 0.5 \, \text{m} = 40^\circ \text{C}
\]

Now, the temperature at the hotter end is given as \( 30^\circ \text{C} \), so the temperature at the colder end is:

\[
T_{\text{colder end}} = T_{\text{hotter end}} - \Delta T
\]

Substituting the values:

\[
T_{\text{colder end}} = 30^\circ \text{C} - 40^\circ \text{C} = -10^\circ \text{C}
\]

Therefore, the temperature at the colder end of the rod is \(-10^\circ \text{C}\).

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Solution:

Given that the thermal resistance is the same for both rods, we have the equation:

\[
\frac{L_1}{K_1} = \frac{L_2}{K_2}
\]

So, the ratio of lengths is:

\[
\frac{L_1}{L_2} = \frac{K_1}{K_2}
\]

If the ratio of the thermal conductivities \(K_1:K_2 = 5:4\), the ratio of the lengths will be the same:

\[
\frac{L_1}{L_2} = \frac{5}{4}
\]

Therefore, the correct ratio of the lengths is \(5:4\).

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Solution:

No heat flows through the central rod \(K_5\) if the system is balanced like a Wheatstone bridge. The condition for this is:

\[
\frac{K_1}{K_2} = \frac{K_3}{K_4}
\]

This ensures that the temperature difference between points B and D is zero, preventing any heat flow through \(K_5\).

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