Question

As per the given figure, two plates $A$ and $B$ of thermal conductivity $K$ and $2K$ are joined together to form a compound plate. The thickness of plates are $4.0cm$ and $2.5cm$ respectively and the area of cross-section is $120c{m}^2$ for each plate. The equivalent thermal conductivity of the compound plate is $\left(1+\frac{5}{\alpha }\right)K$, then the value of $\alpha$ will be _______.

Answer 1

Answer: 21

$\frac{\Delta Q}{\Delta t}=\left(\frac{1}{R}\right)\Delta T$
$R$ : Thermal resistivity
$\therefore R_1=\frac{L_1}{K_{1}A}=\frac{L_1}{K(120)}$
$L_1=4cm$
$A=120c{m}^2$
$R_2=\frac{2.5}{(2K)(120)}$
Now, $R_{eq}$ of this series combination
$R_{eq}=R_1+R_2$
where $L_{eq}=4+2.5=6.5$
$\frac{L_{eq}}{K_{eq}(A)}=\frac{4}{K(120)}+\frac{5}{\frac{2}{2K(120)}}$
$\frac{6.5}{K_{eq}(120)}=\frac{4}{K(120)}+\frac{5}{4K(120)}$
$\frac{6.5}{K_{eq}}=\frac{21}{4K}$
$K_{cq}=\frac{26}{21}K=\left(1+\frac{5}{21}\right)K$
$\therefore a=21$