PhysicsThe 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 T2 and T1(T2>T1).The rate of heat transfer through the slab, in a steady state is A(T2−T1)Kxf, with f equal to

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

A
$\frac{2}{3}$
B
$\frac{1}{2}$
C
1
D
$\frac{1}{3}$

Solution:

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,
$\frac{d Q}{d t} = \frac{Δ T}{\frac{3 x}{K A}} = \frac{(T_{2} - T_{1}) K A}{3 x}$
$= \frac{1}{3} \{\frac{A (T_{2} - T_{1}) K}{x}\} ∴ f = \frac{1}{3}$

Solution:

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