Solving 100W Lightbulb Filament Surface Area Problem

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To find the surface area of a 100W light bulb filament at 3,000 K, the Stefan-Boltzmann law is applied using the equation P = eσAT^4. The correct value for the Stefan-Boltzmann constant σ is 5.670400 x 10^-8 W m^-2 K^-4, not the incorrect value initially referenced. The calculated result of 2.2 x 10^-5 m needs to be converted to square meters to match the book's answer of 22 mm^2. Proper unit conversion is essential, as area is measured in square meters, not meters. Understanding these conversions is crucial for solving the problem accurately.
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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:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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