A composite block is made of slabs A, B, C, D and E of different thermal conductivities (given in te

Q: Why are slabs A and E considered to be in series?

Because heat flows only from left to right, all the heat entering the system must pass entirely through slab A. It then splits into the parallel branches B, C, and D. Finally, all those branches merge back together, and 100% of the heat must pass through slab E to exit. Components that carry the exact same total flow are in series.

Q: How did we determine the heights of slabs B, C, and D?

Look at the arrows on the left side of the diagram. They indicate the y-coordinates of the boundaries between the slabs. The top is 0, the boundary between B and C is at 1L, the boundary between C and D is at 3L, and the bottom is at 4L. This means B has a height of 1L, C has a height of 2L, and D has a height of 1L.

Q: Why is the heat flow through E not considered the maximum?

While the heat flow through E is the total heat flow of the system, the heat flow through A is also the total heat flow. Since they are equal, E does not have a strictly 'maximum' heat flow compared to A.

Q: How do we calculate the heat flow in the parallel branches?

In a parallel circuit, the temperature difference (voltage drop) across all branches is the same. The heat flow (current) through any single branch is simply that shared temperature difference divided by the branch's individual thermal resistance.

Q: Why is the temperature difference smallest for slab E?

For components in series, the temperature difference is directly proportional to their thermal resistance ($\Delta T = H \cdot R$). Since slab E has the largest cross-sectional area and highest thermal conductivity among the series components, it has the smallest thermal resistance, and thus the smallest temperature drop.

Question

A composite block is made of slabs $A, B, C, D$ and $E$ of different thermal conductivities (given in terms of a constant $K$) and sizes (given in terms of length, $L$) as shown in the figure. All slabs are of same width. Heat $Q$ flows only from left to right through the blocks. Then, in steady state

Options:

(A) heat flow through $A$ and $E$ slabs are same

(B) heat flow through slab $E$ is maximum

(C) temperature difference across slab $E$ is smallest

(D) heat flow through $C$ = heat flow through $B$ + heat flow through $D$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine this composite block as a giant electrical circuit, but instead of electrical current, we have heat flowing steadily from left to right. The slabs act as thermal resistors.

Notice the structure carefully: Slab $A$ is on the far left, taking the full brunt of the incoming heat. Then, the heat splits into three parallel paths through slabs $B$, $C$, and $D$. Finally, all the heat recombines and exits through slab $E$ on the far right. This means slab $A$, the parallel combination of $B, C, D$, and slab $E$ are all in **series**.

### The Master Equation

To solve this, we need the thermal resistance formula:

$$R = \frac{L}{KA}$$

where $L$ is the length of the heat path, $K$ is the thermal conductivity, and $A$ is the cross-sectional area perpendicular to the heat flow. Let's assume the uniform width of all slabs is $w$.

Let's calculate the resistance for each slab by carefully reading the dimensions from the diagram:

**Slab A:** Length is $1L$. It spans the full height of $4L$, so its area is $4Lw$. 
$$R_A = \frac{L}{2K(4Lw)} = \frac{1}{8Kw}$$

**Slab B:** Length is $4L$. Its height is $1L$ (from $3L$ to $4L$ on the y-axis), so area is $Lw$.
$$R_B = \frac{4L}{3K(Lw)} = \frac{4}{3Kw}$$

**Slab C:** Length is $4L$. Its height is $2L$ (from $1L$ to $3L$), so area is $2Lw$.
$$R_C = \frac{4L}{4K(2Lw)} = \frac{1}{2Kw}$$

**Slab D:** Length is $4L$. Its height is $1L$ (from $0$ to $1L$), so area is $Lw$.
$$R_D = \frac{4L}{5K(Lw)} = \frac{4}{5Kw}$$

**Slab E:** Length is $1L$. It spans the full height of $4L$, so area is $4Lw$.
$$R_E = \frac{L}{6K(4Lw)} = \frac{1}{24Kw}$$

### Equivalent Resistance of the Parallel Section

Slabs $B$, $C$, and $D$ are in parallel. We find their equivalent resistance $R_{BCD}$ by adding their reciprocals:

$$\frac{1}{R_{BCD}} = \frac{1}{R_B} + \frac{1}{R_C} + \frac{1}{R_D}$$

$$\frac{1}{R_{BCD}} = \left(\frac{3}{4} + 2 + \frac{5}{4}\right) Kw = \left(\frac{8}{4} + 2\right) Kw = 4Kw$$

$$R_{BCD} = \frac{1}{4Kw}$$

### Evaluating the Options

**Option (a):** Since slabs $A$ and $E$ are in the main series line of the circuit, the total heat current $H$ must pass through both of them. Therefore, the heat flow through $A$ and $E$ is exactly the same. **(Correct)**

**Option (b):** The heat flow through $E$ is equal to the heat flow through $A$. It is the total heat flow, but it is not uniquely the "maximum" since $A$ shares this value.

**Option (c):** The temperature difference across any series component is given by Ohm's law for heat: $\Delta T = H \cdot R$. Since the heat current $H$ is the same for $A$, the $BCD$ section, and $E$, the component with the smallest resistance will have the smallest temperature drop. 
Comparing the resistances: $R_E (1/24) < R_A (1/8) < R_{BCD} (1/4)$. 
Clearly, $R_E$ is the smallest, so the temperature difference across slab $E$ is the smallest. **(Correct)**

**Option (d):** In the parallel section, the temperature difference $\Delta T_{BCD}$ is the same for slabs $B$, $C$, and $D$. The heat flow through each is $H_i = \frac{\Delta T_{BCD}}{R_i}$.

$$H_B = \frac{\Delta T_{BCD}}{4/(3Kw)} = 0.75 Kw \Delta T_{BCD}$$
$$H_C = \frac{\Delta T_{BCD}}{1/(2Kw)} = 2.0 Kw \Delta T_{BCD}$$
$$H_D = \frac{\Delta T_{BCD}}{4/(5Kw)} = 1.25 Kw \Delta T_{BCD}$$

Let's check the given relation: $H_B + H_D = 0.75 + 1.25 = 2.0$. This is exactly equal to $H_C$! **(Correct)**

By systematically breaking down the physical geometry into a thermal circuit, we've elegantly proven that options (a), (c), and (d) are correct.

Select Answer:

Visualized Solution

Visualizing the Thermal Circuit

Thermal Resistance Formula

Resistance of Slab A

Resistances of Slabs B, C, and D

Resistance of Slab E

Equivalent Resistance of Parallel Section

Evaluating Option A

Evaluating Option C

Evaluating Option D

Conclusion

The Sigma Insight: Heat Transfer

Analyzing the Setup

The Master Equation

Equivalent Resistance of the Parallel Section

Evaluating the Options

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