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Q.

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 $(\frac{A (T_{2} - T_{1}) K}{x}) f,$ with $f$ equal to

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a

1

b

$\frac{2}{3}$

c

$\frac{1}{2}$

d

$\frac{1}{3}$

answer is D.

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Detailed Solution

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The thermal resistance is given by
$\frac{x}{K A} + \frac{4 x}{2 K A} = \frac{x}{K A} + \frac{2 x}{K A} = \frac{3 x}{K A}$
Amount of heat flow per second,
$\frac{d Q}{d t} = \frac{Δ T}{\frac{3 x}{K A}} = \frac{(T_{2} - T_{1}) K A}{3 x}$
$= \frac{1}{3} \{\frac{A (T_{2} - T_{1}) K}{x}\} ∴ f = \frac{1}{3}$

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