The temperatures of two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivities K and 2K and thicknesses, x and 4x respectively are $T_{2} a n d T_{1} w i t h T_{2} > T_{1}$ . The rate of heat transfer through the slab, in a steady state is $\frac{k A (T_{2} - T_{1})}{x} α$ What is the value of $α$ ?

1

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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 $\frac{1}{3}$
$\frac{Q}{Δ} = \frac{T_{2} - T_{1}}{\frac{l_{1}}{K_{l} A} + \frac{T_{2}}{K_{2} A}} = \frac{A (T_{2} - T_{1})}{\frac{l_{1}}{K_{1}} + \frac{l_{2}}{K_{2}}} G i v e n l_{1} = x, l_{2} = 4 x, 4 x, K_{1} = K a n d K_{2} = 2 K .$
Using these values, we get
$\frac{Q}{Δ t} = \frac{A (T_{2} - T_{1})}{\frac{x}{K} + \frac{4 x}{2 k}} = [\frac{K A (T_{2} - T_{1})}{x}] \times \frac{1}{3} ∴ α = \frac{1}{3}$

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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