Question

A composite block is made of slabs $A,B,C,D$ and $E$ of different thermal conductivities (given in terms of a constant $K$) and sizes (given in terms of length $L$ and area of cross-section $S$) as shown in the figure. All slabs are of the same width. Heat $Q$ flows only from left to right through the blocks. Then in the steady state:

• A. heat flow through slabs $A$ and $E$ are same
• B. heat flow through slab $B$ is minimum
• C. temperature difference across slab $E$ is smallest
• D. heat flow through $C=$ heat flow through $B+$ heat flow through $D$

Answer 1

Answer: (A), (B), (C)

temperature difference across slab $E$ is smallest


The given system comprises three components shown separately in the figure, whose thermal resistances are
$R_A=\frac{L}{2KS},R_B=\frac{4L}{KS}$,
$R_C=\frac{3L}{KS},R_D=\frac{12L}{5KS}$
and $R_E=\frac{L}{6KS}$
Since they are in series, thermal current (heat flow) through $A$ and $E$ will be the same (say $H$)
And since $R_B:R_C:R_D=4:3:\frac{12}{5}$, currents through them will be
$I_B=\frac{H}{4},I_C=\frac{H}{3},I_D=\frac{5H}{12}$
So, $I_C\ne I_B+I_D$ and heat flow through slab $B$ is minimum ($H/4$)

Temperature differences across slabs are given by:
$\Delta {\theta }_A=H{R}_A=\frac{HL}{2KS}$,
$\Delta {\theta }_B=\frac{H}{4}{R}_B=\frac{HL}{KS}$,
$\Delta {\theta }_C=\frac{H}{3}{R}_C=\frac{HL}{KS}$,
$\Delta {\theta }_D=\frac{5H}{12}{R}_D=\frac{HL}{KS}$ and $\Delta {\theta }_E=H{R}_E=\frac{HL}{6KS}$,
$\Rightarrow \Delta {\theta }_E$ is the smallest.

Hence, the correct options are (a), (b) and (c).