Assuming a filament in a 100 W light bulb acts like a perfect blackbody, what is

Question

Assuming a filament in a 100 W light bulb acts like a perfect blackbody, what is the temperature of the hottest portion of the filament if it has a surface area of 6.3x10-5m2?The Stefan-Boltzmann constant is 5.67x10-8W/(m2? K2). (1.6pts) 5800 K 1100 K 2300 K 130 K
Assuming a filament in a 100 W light bulb acts like a perfect blackbody, what is the temperature of the hottest portion of the filament if it has a surface area of 6.3x10-5m2?The Stefan-Boltzmann constant is 5.67x10-8W/(m2? K2). (1.6pts) 5800 K 1100 K 2300 K 130 K
Assuming a filament in a 100 W light bulb acts like a perfect blackbody, what is the temperature of the hottest portion of the filament if it has a surface area of 6.3x10-5m2?The Stefan-Boltzmann constant is 5.67x10-8W/(m2? K2). (1.6pts) 5800 K 1100 K 2300 K 130 K
Assuming a filament in a 100 W light bulb acts like a perfect blackbody, what is the temperature of the hottest portion of the filament if it has a surface area of 6.3x10-5m2?The Stefan-Boltzmann constant is 5.67x10-8W/(m2? K2). Assuming a filament in a 100 W light bulb acts like a perfect blackbody, what is the temperature of the hottest portion of the filament if it has a surface area of 6.3x10-5m2?The Stefan-Boltzmann constant is 5.67x10-8W/(m2? K2). (1.6pts) 5800 K 1100 K 2300 K 130 K
5800 K 5800 K 1100 K 1100 K 2300 K 2300 K 130 K
130 K
130 K

Explanation / Answer

TheStefan-Boltzmannformula is:

intensity = (S-B)*T^4

So the power emitted by a blackbody over its area A should be:
P = A*(S-B)*T^4

or T = 4th_root(P/((S-B)*A)
= 4th_root(100/((5.67e(-8)*(6.3e(-5)))
= 4th_root(1e15/((6.3)(5.67)))
= 1e3 * 4th_root(1000/((6.3)(5.67))
= 1e3* 4th_root(27.9947)
= 1e3*(2.30)
= 2300 degrees Kelvin.

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