Question

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 $f$ equals to

• A. $1$
• B. $1/2$
• C. $2/3$
• D. $1/3$

Answer 1

The correct option is D $1/3$

The thermal resistance of both the blocks will be

$\frac{x}{KA}$ and $\frac{4x}{2KA}=\frac{2x}{KA}$ respectively. Since the two resistors are in series, (as same heat current will flow through both), equivalent thermal resistance will be

$R_{eq.}=\frac{x}{KA}+\frac{2x}{KA}=\frac{3x}{KA}$

Thus, heat current will be

$\frac{dQ}{dt}=\frac{T_2-T_1}{R_{eq.}}=\frac{1}{3}\frac{KA(T_2-T_1)}{x}$

Hence, $f=\frac{1}{3}$