Thermal Properties of Matter - Result Question 33
34. Consider a compound slab consisting of two different materials having equal thicknesses and thermal conductivities $K$ and $2 K$, respectively. The equivalent thermal conductivity of the slab is
[2003]
(a) $\frac{4}{3} K$
(b) $\frac{2}{3} K$
(c) $\sqrt{3} K$
(d) $3 K$
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Answer:
Correct Answer: 34. (a)
Solution:
- (a) In series, equivalent thermal conductivity
$K _{\text{eq }}=\frac{2 K_1 K_2}{K_1+K_2}$
or, $K _{\text{eq }}=\frac{2 \times K \times 2 K}{K+2 K}=\frac{4}{3} K$
In electricity equivalent resistance in series grouping $R_s=R_1+R_2+R_3 \ldots . .+R_n$
In heat-thermal resistance in series $R_s^{n}=R_1+R_2+$ $R_3 \ldots . .+R_n$
$ \frac{\ell_1+\ell_2+\ldots . . \ell_n}{K s}=\frac{\ell_1}{K_1 A}+\frac{\ell_2}{K_2 A}+\ldots . .+\frac{\ell_n}{K_n A} $
For $n$ no. of rods of equal length
$ K_s=\frac{n}{\frac{1}{K_1}+\frac{1}{K_2} \ldots . .+\frac{1}{K_n}} $
