The surface area of the filament of a light bulb is 0.159×10⁻⁴m².Surface area is inversely to emissivity and directly proportional to power.
Every body in this universe emits some radiations whether it will be large or small.
We know that surface area of filament of bulb=Power radiated by the body/Emissivity of the body, or
=>A=P/I
Now, Power radiated by the body is given(P)=σAT⁴,where
σ is Stefan constant having value 5.67040×10⁻⁸ w/m²-K⁴.
A is the surface area
and T is the temperature of body.
Now, temperature is given in degree celsius.
So, on converting temperature into kelvin, we get
K=C+273
=>K=2580+273
=>K=2853Kelvin
Value of T⁴ is =(2853)⁴ =(8139609×8139609)K⁴
On putting values of power, area and temperature, we get
=>A=P/σT⁴
=>A=60 / [(5.67040×10⁻⁸)×(8139609×8139609)
=>A=60/(5.67040×10⁻⁸)×(66.21×10¹²)
=>A=60/375.48×10⁴
=>A=0.159×10⁻⁴m²
Hence, surface area is 0.159×10⁻⁴m².
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