Question

The temperature of the filament of a $100$ W bulb is $4000$${}^o$C in the steady state. The radius of the glass bulb is $400$ m and the thickness of the bulb is $0.4$mm. Assuming no convection, calculate the thermal conductivity of glass taking the temperature outside the bulb to be 27${}^o$C

• A. $2\times 10$${}^{-12}$ J sec${}^{-1}$ m ${}^{-1}$ K${}^{-1}$
• B. $8\times 10$${}^{-12}$ J sec${}^{-1}$ m ${}^{-1}$ K${}^{-1}$
• C. $4\times 10$${}^{-12}$ J sec${}^{-1}$ m ${}^{-1}$ K${}^{-1}$
• D. $5\times 10$${}^{-12}$ J sec${}^{-1}$ m ${}^{-1}$ K${}^{-1}$

Answer 1

The correct option is D $5\times 10$${}^{-12}$ J sec${}^{-1}$ m ${}^{-1}$ K${}^{-1}$

$P=100W$

$A=4\pi r^2=$ Area of speherical bulb

$\Delta T={4000}^{o}C-{27}^{o}C={4273}^{o}K-{300}^{o}K={3973}^{o}K$

$l=0.4mm=4\times {10}^{-4}m$

Thermal conductivity is given as,

$k=\frac{P\times l}{4\pi r^2\Delta T}$

$k=\frac{100\times 4\times {10}^{-4}}{4\pi \times {400}^2\times 3973}$

$\therefore k=5\times {10}^{-12}Jse{c}^{-1}{m}^{-1}{K}^{-1}$