Q. 1.56$\times 10^{5}$ J of heat is conducted through is 2 $m^{2}$ wall of 12 cm thick in one hour. Temperature difference between the two sides of the wall is 20$℃$. The thermal conductivity of the material of the wall is (in ${Wm}^{- 1\ }K^{- 1}$)
A) 0.11
B) 0.13
C) 0.15
D) 1.2
Explanation
(b) $\frac{d\mathcal{Q}}{dt} = \frac{KA\mathrm{\Delta}T}{x}$
$\frac{1.56 \times 10^{5}}{3600} = \frac{K \times 2 \times 20}{12 \times 10^{- 2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ K = \frac{1.56 \times 10^{5} \times 12 \times 10^{- 2}}{3600 \times 2 \times 10}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1.56}{12} = 0.13$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.