Q. A metal plate $4\ mm$ thick has a temperature difference of $32℃$ between its faces. It transmits $200\ kcal/h$ through an area of $5cm^{2}$. Thermal conductivity of the material is
A) $58.33\ W/m$-$℃$
B) $33.58\ W/m$-$℃$
C) $5 \times 10^{- 4}\ W/m$-$℃$
D) None of these
Explanation
(a) Here, $\mathrm{\Delta}x = 4\ mm = 4 \times 10^{- 3}m$
$\mathrm{\Delta}T = 32℃$
Transmit heat per hours
$\frac{\mathrm{\Delta}Q}{\mathrm{\Delta}T} = 200\ kcal/h = \frac{200 \times 1000 \times 4.2}{60 \times 60}J/s = 233.33J/s$
$A = 5cm^{2} = 5 \times 10^{- 4}m^{2}$
We know that, $\frac{\mathrm{\Delta}Q}{\mathrm{\Delta}T} = KA\left( \frac{\mathrm{\Delta}T}{\mathrm{\Delta}x} \right)$
$\therefore$ Thermal conductivity of material, $K = \frac{\mathrm{\Delta}Q/\mathrm{\Delta}T}{A(\mathrm{\Delta}T/\mathrm{\Delta}x)}$
Or $K = \frac{233.33 \times 4 \times 10^{- 3}}{5 \times 10^{- 4} \times 32} = 58.33W/m$-$℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.