Problem 170 - Master Resource Book in JEE Main Physics

Question

A slab consists of two portions of different materials of same thickness and having the conductivities \(K_{1}\) and \(K_{2}\). The equivalent thermal conductivity of the slab is [Karnataka CET 2008] (a) \(K_{1}+K_{2}\) (b) \(\frac{K_{1} K_{2}}{K_{1}+K_{2}}\) (c) \(\frac{2 K_{1} K_{2}}{K_{1}+K_{2}}\) (d) \(\sqrt{K_{1}+K_{2}}\)

Step-by-Step Solution

Answer

The equivalent thermal conductivity is \(\frac{2 K_{1} K_{2}}{K_{1}+K_{2}}\).

Step 1: Understanding the Setup

The slab consists of two portions with different thermal conductivities, each with a thickness of 'd'. We want to find the equivalent thermal conductivity for the whole slab.

Step 2: Recognize Parallel Arrangement

Since the two materials have the same thickness, the equivalent thermal conductivity for the slab is calculated by assuming the portions are in parallel. The formula for the equivalent thermal conductivity in parallel is used here.

Step 3: Apply the Formula for Equivalent Conductivity

The formula for the equivalent thermal conductivity when two conductors are in parallel is given by: \[ K_{eq} = \frac{2 K_{1} K_{2}}{K_{1} + K_{2}} \] This is because both materials are contributing equally along the same path of heat transfer, similar to parallel resistances in electricity.

Step 4: Identify the Correct Option

From the given options, option (c) matches the derived formula: \[ \frac{2 K_{1} K_{2}}{K_{1} + K_{2}} \] Therefore, option (c) is the correct answer for the equivalent thermal conductivity of the slab.

Key Concepts

Heat Transfer

Heat transfer is the movement of heat from one place to another through conduction, convection, or radiation. In materials like a slab composed of two different materials, conduction is the primary mode of heat transfer. Conduction occurs when heat is directly transmitted through a material due to a difference in temperature between its parts. The rate of heat transfer depends on the material's thermal conductivity. A material with high thermal conductivity allows heat to pass through more easily, allowing for efficient heat transfer.

• Conduction: Direct heat transfer through materials.
• Thermal Conductivity: Indicates how well a material conducts heat.

Understanding how heat moves through materials helps in determining how well the slab will transfer heat. This becomes particularly important when dealing with materials having different rates of conductivity.

Parallel Arrangement

In the context of thermal conductivity, materials can be arranged either in series or parallel. In a parallel arrangement, two or more pathways allow heat to flow simultaneously through different materials. This is akin to multiple lanes of traffic rather than a single file lane. When materials are in parallel, each with the same thickness, we assume they contribute to heat transfer equally. Much like parallel resistors in electrical circuits, the materials' resistances to heat flow combine in a way that requires an equivalent conductivity formula for accurate calculation.

• Parallel Arrangement: Multiple paths for heat flow.
• Results in combined effect of all material conductivities.

When you have materials with different conductivities, the parallel setup allows for a sort of balance, creating an averaged effect with their thermal contributions.

Thermal Resistance

Thermal resistance is an important concept when understanding heat transfer in materials. It refers to a material's resistance to heat flow, akin to resistance in an electrical circuit. The greater the thermal resistance, the less heat is transferred through the material. In simpler terms, thermal resistance tells us how much a material will impede the transfer of heat. In the context of the slab with two materials, thermal resistance helps to understand how each material affects the total heat transfer.

• Thermal Resistance: A measure of how well a material resists heat flow.
• Higher resistance means less heat transfer.

In a parallel arrangement, thermal resistances blend together to form a new resistance that factors into the overall heat transfer through the slab.

Conductivity Formula

The conductivity formula is crucial for calculating the equivalent conductivity when materials are arranged in parallel. For the problem of the slab with two materials of different conductivities, the formula used is derived from electrical principles where parallel resistances are combined.The formula for equivalent thermal conductivity in a parallel arrangement is expressed as:\[ K_{eq} = \frac{2 K_{1} K_{2}}{K_{1} + K_{2}} \]This formula considers both materials' contributions to the entire slab while accounting for the effect of their specific conductivities.

• Formula reflects shared contribution to heat transfer.
• Makes use of both materials' conductive properties in calculation.

This understanding helps design materials with optimal thermal properties and compute how well a composite slab will transfer heat under specified conditions.