Two rectangular slabs of equal thickness d are mounted in parallel between a hot reservoir… — JAMB UTME

Two rectangular slabs of equal thickness $d$ are mounted in parallel between a hot reservoir at $100^\circ\text{C}$ and a cold reservoir at $20^\circ\text{C}$ . Slab 1 has a thermal conductivity $k_1 = 300\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ and a cross-sectional area $A_1 = 4.0 \times 10^{-4}\text{ m}^2$ . Slab 2 has a thermal conductivity $k_2 = 100\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ and a cross-sectional area $A_2 = 6.0 \times 10^{-4}\text{ m}^2$ . Assuming steady-state heat conduction and no lateral heat loss, what percentage of the total heat transferred per second between the reservoirs conducts through Slab 1?

$66.7\%$ Answer
B
$33.3\%$
C
$50.0\%$
D
$40.0\%$

Answer

The percentage of the total heat transferred per second conducted through Slab 1 is 66.7%.

Fourier's law gives the rate of heat conduction as

P = \frac{k A \Delta T}{d}

. Since the temperature gradient

\frac{\Delta T}{d}

is identical across both parallel slabs, the heat current through each slab is directly proportional to its

k A

product. For Slab 1,

k_1 A_1 = 300 \times 4.0 \times 10^{-4} = 0.12

, while for Slab 2,

k_2 A_2 = 100 \times 6.0 \times 10^{-4} = 0.06

. The total heat current is proportional to

0.12 + 0.06 = 0.18

. Thus, the fraction conducted through Slab 1 is

\frac{0.12}{0.18} = \frac{2}{3}

, which corresponds to

66.7\%

Step-by-Step Solution

Express Fourier's law of heat conduction for each slab in parallel.

Rate of heat flow

P_1 = \frac{k_1 A_1 \Delta T}{d}

and

P_2 = \frac{k_2 A_2 \Delta T}{d}

Both slabs experience the same temperature difference

\Delta T = 100^\circ\text{C} - 20^\circ\text{C} = 80\text{ K}

and have equal length

d

Calculate the effective conductance factors

k A

for both slabs.

k_1 A_1 = 300 \times (4.0 \times 10^{-4}) = 0.12\text{ W}\cdot\text{m}\cdot\text{K}^{-1}

and

k_2 A_2 = 100 \times (6.0 \times 10^{-4}) = 0.06\text{ W}\cdot\text{m}\cdot\text{K}^{-1}

Since

\Delta T / d

is identical for both parallel paths, the heat flow rate is directly proportional to

k A

Determine the total heat flow rate and the percentage carried by Slab 1.

P_{\text{total}} = P_1 + P_2 \propto 0.12 + 0.06 = 0.18

. Percentage through Slab 1

= \frac{0.12}{0.18} \times 100\% = 66.7\%

In parallel conduction, individual heat currents add to form the total heat current.

Key Concept

Parallel Thermal Conduction

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