The temperatures 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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$\frac{1}{2}$

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$\frac{2}{3}$

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$\frac{1}{3}$

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Solution

The correct option is D
$\frac{1}{3}$

Let A be the area of each slab. In the steady state, the rate of heat flow through the composite slab is given by
$\frac{Q}{t} = \frac{T_{2} - T_{1}}{\frac{l_{1}}{K_{1} A} + \frac{l_{2}}{K_{2} A}} = \frac{A (T_{2} - T_{1})}{\frac{l_{1}}{K_{1}} + \frac{l_{2}}{K_{2}}}$ .....(1)
Given that, $l_{1} = x, l_{2} = 4 x, K_{1} = K$ and $K_{2} = 2 K$
Using these values in equation (1), $\frac{Q}{t} = \frac{A (T_{2} - T_{1})}{\frac{x}{K} + \frac{4 x}{2 K}} = [\frac{A (T_{2} - T_{1}) K}{x}] \times \frac{1}{3}$
Comparing this with the given rate of heat transfer, we get $f = \frac{1}{3}$ .
Hence, the correct choice is (d).

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Similar questions

The temperature of the two outer surfaces of a composite slab consisting of two materials having coeffieient of thermal conductivity

K

and

2 K

and thickness

x

and

4 x

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

The temperatures 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