Solving 100W Lightbulb Filament Surface Area Problem

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Homework Help Overview

The problem involves calculating the surface area of a light bulb filament operating at a temperature of approximately 3,000 K, given a power output of 100 W and an emissivity of 1. The original poster attempts to apply the Stefan-Boltzmann law to find the surface area but encounters confusion regarding unit conversions and the correct value of the Stefan-Boltzmann constant.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the Stefan-Boltzmann law and question the original poster's use of the constant's value and its units. There are inquiries about unit conversions and the relationship between length and area in the context of the problem.

Discussion Status

Some participants have provided guidance on unit conversions and clarified the correct value of the Stefan-Boltzmann constant. There is an ongoing exploration of the implications of these conversions and the need for clarity on the units involved.

Contextual Notes

There is a noted discrepancy between the original poster's calculated value and the expected answer from the textbook, prompting questions about unit conversion and the correct application of the formula. The discussion reflects a lack of consensus on the correct approach but indicates productive engagement with the problem's requirements.

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Homework Statement



The filament of typical light bulb is at a temperature of about 3,000 K. For a 100 W bulb, what is the surface area of the filament. You can assume that the entire 100 W goes into heating the filament and the filament has an emissivity of 1.

Homework Equations


The equation is P= e(sigma)(A)(T^4)
P is power
e is the emissivity
sigma is the constant 5.67 x 10 ^-5

The Attempt at a Solution



I have gotten to the point where i got 2.2 x 10^-5 m.
The answer in the back of the book is 22 mm^2.
I can't figure out how this was gotten.
 
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Your answer is a length; the book's answer is an area. Does that give you any ideas?

You need to convert the units.
 
Can someone help me convert these units?
 
courtneywetts said:

Homework Statement



The filament of typical light bulb is at a temperature of about 3,000 K. For a 100 W bulb, what is the surface area of the filament. You can assume that the entire 100 W goes into heating the filament and the filament has an emissivity of 1.


Homework Equations


The equation is P= e(sigma)(A)(T^4)
P is power
e is the emissivity
sigma is the constant 5.67 x 10 ^-5
Where did you get this sigma number, and what are the units of the number? Are you sure the exponent on the 10 is correct?

I think your thinking of the Stefan–Boltzmann law.

P = \varepsilon \sigma A T^4
In SI units, \sigma = 5.670400 \times 10^{-8} \ [\mathrm{\ J \ s^{-1} \ m^{-2} \ K^{-4}} \ ].

Or if you'd rather uses Watts instead of Joules/sec, you could say, \sigma = 5.670400 \times 10^{-8} \ [\mathrm{\ W \ m^{-2} \ K^{-4}} \ ]

The Attempt at a Solution



I have gotten to the point where i got 2.2 x 10^-5 m.
Well, it seems that you used the correct "-8" in your \sigma to get that value, instead of the "-5" that you had listed in the relevant equations section. :thumbs:

But you need to look at the units you ended up with. Is area measured in meters or square meters?

The answer in the back of the book is 22 mm^2.
I can't figure out how this was gotten.

courtneywetts said:
Can someone help me convert these units?

How many square millimeters are in one square meter? :wink:
 
Last edited:

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