Question

The operating temperature of an incandescent bulb (with a tungsten filament) of power $60\text{W}$ is $3000\text{K}$. If the surface area of the filament be $25{\text{mm}}^2$. Find its emissivity $e$.$(\sigma =5.67\times {10}^{-8}\text{W}/{\text{m}}^2-{\text{K}}^4)$

• A. $0.52$
• B. $0.48$
• C. $0.64$
• D. $0.36$

Answer 1

The correct option is A $0.52$

Given $T=3000\text{K}$; Power $=\frac{dQ}{dt}=60\text{W}$
Area $A=25\times {10}^{-6}{\text{m}}^2,e=?$
Total radiation lost per second $=\frac{dQ}{dt}=Ae\sigma T^4$,
$60\text{W}=(25\times {10}^{-6}{\text{m}}^2)e(5.67\times {10}^{-8}{\text{m}}^{-2}-{\text{K}}^{-4})(3000{)}^4{\text{K}}^4$
$60=114.82e$
$ϵ=0.52$