Problem 175 - Master Resource Book in JEE Main Physics

Question

The temperature of the two outer surface of a composite slab, consisting of two materials having coefficients of thermal conductivity \(K\) and \(2 K\) and thickness \(x\) and \(4 x\), respectively are \(T_{2}\) and \(T_{1}\left(T_{2}>T_{1}\right)\). The rate of heat transfer through the slab, in a steady state is \(\left(\frac{A\left(T_{2}-T_{1}\right) K}{x}\right) f\), with \(f\) equal to \(\quad\) [UP SEE 2007] (a) 1 (b) \(\frac{1}{2}\) (c) \(\frac{2}{3}\) (d) \(\frac{1}{3}\)

Step-by-Step Solution

Answer

f = \(\frac{1}{3}\) (option d).

Step 1: Understand the Problem

We are given a composite slab with two materials, each with different thermal conductivities and thicknesses. We need to calculate the rate of heat transfer through the slab and find the value of the factor \( f \) in the expression \( \left(\frac{A(T_{2}-T_{1}) K}{x}\right) f \).

Step 2: Identify Thermal Resistance

The thermal resistance for each layer can be calculated using the formula \( R = \frac{L}{A K} \), where \( L \) is the thickness and \( K \) is the thermal conductivity. For the first layer, \( R_1 = \frac{x}{A K} \) and for the second layer, \( R_2 = \frac{4x}{A (2K)} = \frac{2x}{A K} \).

Step 3: Compute Total Thermal Resistance

The total thermal resistance \( R_{total} \) of the slab is the sum of the resistances of the two layers: \[ R_{total} = R_1 + R_2 = \frac{x}{A K} + \frac{2x}{A K} = \frac{3x}{A K} \].

Step 4: Calculate the Rate of Heat Transfer

The rate of heat transfer \( Q \) in steady state is given by the formula \( Q = \frac{T_2 - T_1}{R_{total}} \). Substituting \( R_{total} \), we get: \[ Q = \frac{A K (T_2 - T_1)}{3x} \].

Step 5: Match with Given Expression

The given rate of heat transfer is \( \left( \frac{A(T_{2}-T_{1}) K}{x} \right) f \). Setting this equal to the calculated \( Q \), we have: \[ \frac{A K (T_2 - T_1)}{3x} = \left( \frac{A (T_2 - T_1) K}{x} \right) f \].

Step 6: Solve for Factor f

To find \( f \), compare the two expressions: \[ \frac{1}{3} = f \]. So, \( f = \frac{1}{3} \).

Key Concepts

Thermal Conductivity

Thermal conductivity is a property that describes how well a material can conduct heat. It is denoted by the symbol \( K \). High thermal conductivity means that heat passes through the material quickly, while low thermal conductivity indicates that heat passes through slowly.
In the context of composite slabs, thermal conductivity plays a crucial role in determining the rate at which heat transfers through each layer of the slab.

• The first material in our problem has a thermal conductivity of \( K \).
• The second material has a thermal conductivity of \( 2K \), meaning it conducts heat twice as effectively.

It's important to understand that even if one material has a higher thermal conductivity, the cumulative effect of both materials must be assessed to determine the overall heat conduction through the composite slab.

Thermal Resistance

Thermal resistance is a measure of a material's resistance to heat flow. The higher the thermal resistance, the less heat can pass through the material. It is influenced by both the thickness of the material (\( L \)) and its thermal conductivity (\( K \)). Thermal resistance is calculated using the formula \( R = \frac{L}{A K} \) where \( A \) is the cross-sectional area.

• In the first layer of our composite slab, the thermal resistance is \( R_1 = \frac{x}{A K} \).
• The second layer has a thermal resistance \( R_2 = \frac{2x}{A K} \) due to its higher thickness and different conductivity.

To find the total thermal resistance, you simply add the resistances of both layers:\[ R_{total} = R_1 + R_2 = \frac{x}{A K} + \frac{2x}{A K} = \frac{3x}{A K} \]This total thermal resistance helps us to understand how difficult it is for heat to transfer through the entire composite slab.

Rate of Heat Transfer

The rate of heat transfer is how quickly heat energy passes through a material from a higher temperature area to a lower temperature area. In a composite slab, where multiple materials are involved, understanding this rate involves both thermal conductivity and thermal resistance.
The formula to calculate the rate of heat transfer is:\[ Q = \frac{T_2 - T_1}{R_{total}} \]In this problem:

• \( T_2 \) is the temperature of the hotter surface.
• \( T_1 \) is the temperature of the cooler surface.

By substituting the total thermal resistance into the formula, we get:\[ Q = \frac{A K (T_2 - T_1)}{3x} \]This tells us that the rate of heat transfer depends inversely on the total thermal resistance. As the resistance increases, the rate of heat transfer decreases. In our formula, \( f \) was found to be \( \frac{1}{3} \), helping us understand the relationship between all parameters involved.