The temperature of the two outer surfaces of a composite slab, consisting of two materials having co

Q: Why is the heat flow rate $H$ exactly the same in both slabs?

In a steady state, temperature at any point does not change with time. This means no heat is being absorbed or stored by the material. Whatever heat enters the first slab must exit it and enter the second slab, much like water flowing continuously through a pipe.

Q: Why did we assume the interface temperature is $T$?

The two outer surfaces are at $T_2$ and $T_1$. Since heat flows from hot to cold, the junction must be at some intermediate temperature $T$ where $T_2 > T > T_1$. We need this $T$ to calculate the temperature gradient across each individual slab.

Q: Can I solve this using thermal resistance?

Absolutely! This is often the faster method. Treat the slabs as resistors in series. $R_{eq} = R_1 + R_2 = \frac{x}{KA} + \frac{4x}{2KA} = \frac{3x}{KA}$. Then, $H = \frac{\Delta T}{R_{eq}} = \frac{KA(T_2 - T_1)}{3x}$.

Q: What happens to the area $A$ in the equations?

Since the problem describes a composite slab (like a wall), the cross-sectional area $A$ perpendicular to the heat flow is the same for both materials. It cancels out when we equate $H_1$ and $H_2$.

Q: Why is the conductivity of the second slab $2K$ but its thickness is $4x$?

These are just the physical properties given in the problem. A higher conductivity ($2K$) means it conducts heat better, but a greater thickness ($4x$) means heat has to travel further, which increases resistance. The math naturally balances these competing factors.

Question

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity $K$ and $2K$ and thickness $x$ and $4x$ respectively are $T_2$ and $T_1$ ($T_2 > T_1$). The rate of heat transfer through the slab, in a steady state is $\left(\frac{A(T_2 - T_1)K}{x}\right)f$, then $f$ is equal to

Options:

(A) 1

(B) 1/2

(C) 2/3

(D) 1/3

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Slabs

Imagine you are looking at a composite wall made of two entirely different materials fused together. The left section has a thickness of $x$ and a thermal conductivity of $K$. The right section is much thicker at $4x$, but it also conducts heat better, with a conductivity of $2K$.

The left outer surface is baking at a high temperature $T_2$, while the right outer surface is chilling at a lower temperature $T_1$. Because nature abhors an imbalance, heat will naturally flow from the hotter left side to the cooler right side. Our goal is to find the exact rate of this heat transfer.

### The Steady State: A River of Heat

Before we dive into the math, we need to understand the concept of a **steady state**. Think of heat flow like water flowing through a pipe. In a steady state, the pipe isn't expanding or leaking; whatever amount of water enters one end must exit the other end at the exact same rate.

Similarly, in our composite slab, the rate of heat flow ($H$) entering the first slab must perfectly equal the rate of heat flow exiting the second slab. There is no heat accumulating anywhere inside. Let's call the unknown temperature exactly at the junction of these two slabs $T$.

### The Junction: Finding the Middle Ground

Using Fourier's law of heat conduction, we can write the heat current for each slab individually. For the first slab, the heat current $H_1$ is driven by the temperature difference $(T_2 - T)$:

$$ H_1 = \frac{KA(T_2 - T)}{x} $$

For the second slab, the heat current $H_2$ is driven by the temperature difference $(T - T_1)$:

$$ H_2 = \frac{2KA(T - T_1)}{4x} = \frac{KA(T - T_1)}{2x} $$

Because we are in a steady state, we can confidently equate the two:

$$ \frac{KA(T_2 - T)}{x} = \frac{KA(T - T_1)}{2x} $$

Notice how beautifully the area $A$, the conductivity $K$, and the thickness $x$ cancel out from both sides. This leaves us with a simple algebraic equation:

$$ T_2 - T = \frac{T - T_1}{2} $$

Cross-multiplying and rearranging the terms gives us the exact temperature at the interface:

$$ 2T_2 - 2T = T - T_1 \implies 3T = 2T_2 + T_1 \implies T = \frac{2T_2 + T_1}{3} $$

### The Final Flow: Bringing It All Together

Now that we have unmasked the junction temperature $T$, we can substitute it back into our original equation for $H_1$ to find the total steady heat flow rate $H$:

$$ H = \frac{KA}{x} \left( T_2 - \frac{2T_2 + T_1}{3} \right) $$

Taking a common denominator of 3, the expression inside the bracket simplifies elegantly:

$$ H = \frac{KA}{x} \left( \frac{3T_2 - 2T_2 - T_1}{3} \right) = \frac{KA(T_2 - T_1)}{3x} $$

The problem states that the heat transfer rate is given by the expression $\left(\frac{A(T_2 - T_1)K}{x}\right)f$. By comparing our derived result with this given format, it is crystal clear that the mysterious factor $f$ is exactly **$1/3$**.

### The Pro-Move: The Electrical Analogy

While the method above is rigorous, there is a much faster way to solve this using the concept of **Thermal Resistance**. Just like electrical resistors in series, thermal resistances add up!

The thermal resistance of a slab is given by $R = \frac{L}{KA}$. 
For the first slab: $R_1 = \frac{x}{KA}$
For the second slab: $R_2 = \frac{4x}{2KA} = \frac{2x}{KA}$

The equivalent resistance of the entire composite slab is simply their sum:

$$ R_{eq} = R_1 + R_2 = \frac{x}{KA} + \frac{2x}{KA} = \frac{3x}{KA} $$

The total heat flow is then just the total temperature difference divided by the total resistance (analogous to Ohm's Law $I = V/R$):

$$ H = \frac{T_2 - T_1}{R_{eq}} = \frac{T_2 - T_1}{\frac{3x}{KA}} = \frac{KA(T_2 - T_1)}{3x} $$

In just three lines of math, we arrive at the exact same result! Mastering this analogy will save you precious minutes in competitive exams.

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Steady State Condition$

$Fourier’s Law of Heat Conduction$

$Equating Heat Flows$

$Solving for Interface Temperature T$

$Substituting T back into H_{1}$

$Simplifying the Heat Flow Expression$

$Comparing with Given Expression$

$The Way Forward: Thermal Resistance$

The Sigma Insight: Heat Transfer

The Setup

The Steady State

The Junction

The Final Flow

The Pro-Move

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Steady State Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fourier’s Law of Heat Conduction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Heat Flows

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Interface Temperature T

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting T back into H1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Heat Flow Expression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Comparing with Given Expression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward: Thermal Resistance