Question

As per the given figure, two plates A and B of thermal conductivity K and 2 K are joined together to form a compound plate. The thickness of plates are 4.0 cm and 2.5 cm, respectively and the area of cross-section is 120 cm2 for each plate. The equivalent thermal conductivity of the compound plate is $(1+\frac{5}{α})K,$ then the value of α will be ___.

Answer 1

The correct answer is: 21

$\frac{L_1}{K_1A_1}+\frac{L_1}{K_1A_1}=\frac{L_1+L_2}{K_{eff}A_eff}$

$⇒K_{eff}=\frac{13K}{10.5}=(1+\frac{5}{21})K$

⇒ α = 21