Solving 100W Lightbulb Filament Surface Area Problem

[courtneywetts]

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.

 

[vela]

Your answer is a length; the book's answer is an area. Does that give you any ideas?

You need to convert the units.

 

[courtneywetts]

Can someone help me convert these units?

 

[collinsmark]

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.

Can someone help me convert these units?

How many square millimeters are in one square meter?

 

[Nickie]

To solve this problem, we can use the equation P = eσAT^4, where P is power, e is emissivity, σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4), A is the surface area of the filament, and T is the temperature of the filament in Kelvin.

We are given that P = 100 W, e = 1, and T = 3000 K. Plugging these values into the equation, we get:

100 = (1)(5.67 x 10^-8)(A)(3000^4)

Simplifying, we get:

A = 100 / (1)(5.67 x 10^-8)(3000^4)

A = 2.2 x 10^-5 m^2

However, the answer in the book is given in mm^2, so we need to convert our answer to mm^2 by multiplying by 10^6 (since 1 m^2 = 10^6 mm^2). This gives us:

A = (2.2 x 10^-5)(10^6) = 22 mm^2

Therefore, the surface area of the filament is 22 mm^2.