contestada

the filament of a light bulb has a temperature of 2580oc and radiates 60 w of power. the emissivity of the filament is 0.36. what is the surface area of the filament?

Respuesta :

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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https://brainly.com/question/17485809

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