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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 $\left( {\frac{{A({T_2} - {T_1})K}}{x}} \right)f$, with $f $ which equal to
$1$
$\frac{1}{2}$
$\frac{2}{3}$
$\frac{1}{3}$
Solution
(d) Equation of thermal conductivity of the given combination ${K_{eq}} = \frac{{{l_1} + {l_2}}}{{\frac{{{l_1}}}{{{K_1}}} + \frac{{{l_2}}}{{{K_2}}}}} = \frac{{x + 4x}}{{\frac{x}{K} + \frac{{4x}}{{2K}}}} = \frac{5}{3}K$.
Hence rate of flow of heat through the given combination is $\frac{Q}{t} = \frac{{{K_{eq}}.A({T_2} – {T_1})}}{{(x + 4x)}} = \frac{{\frac{5}{3}K\,A\,({T_2} – {T_1})}}{{5x}}$=$\frac{{\frac{1}{3}K\,A\,({T_2} – {T_1})}}{x}$
On comparing it with given equation we get $f = \frac{1}{3}$
