Solving 100W Lightbulb Filament Surface Area Problem

• courtneywetts
In summary, the problem asks for the surface area of a filament in a 100 W light bulb with a temperature of 3,000 K and emissivity of 1. Using the Stefan-Boltzmann law, we can solve for the surface area (A) by rearranging the formula P = e(sigma)(A)(T^4). The constant sigma is 5.670400 x 10^-8 in SI units, and the units for the equation are in Joules/second/meter squared/kelvin to the fourth power. After converting the units and solving for A, we get a value of 2.2 x 10^-5 meters squared. To convert this to the answer given in the book of
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.

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?

Last edited:

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.

1. How do you calculate the surface area of a 100W lightbulb filament?

The surface area of a 100W lightbulb filament can be calculated by first finding the length of the filament and then using the formula for the surface area of a cylinder: 2πrh + 2πr^2, where r is the radius of the filament and h is the length.

2. What is the purpose of calculating the surface area of a lightbulb filament?

Calculating the surface area of a lightbulb filament is important for understanding the efficiency and heat dissipation of the bulb. It can also be used in designing more efficient lightbulbs and predicting their lifespan.

3. How does the wattage of a lightbulb affect the surface area of its filament?

The wattage of a lightbulb does not directly affect the surface area of its filament. However, a higher wattage bulb will produce more heat, which can lead to a larger surface area due to thermal expansion of the filament.

4. Are there any other factors that can affect the surface area of a lightbulb filament?

Yes, the material of the filament and its shape can also affect the surface area. For example, a coiled filament will have a larger surface area compared to a straight filament of the same length.

5. How can the surface area of a lightbulb filament be reduced?

The surface area of a lightbulb filament can be reduced by using a more efficient material, such as tungsten, and by designing the filament to be shorter and thicker. This will decrease the amount of heat produced and therefore reduce the surface area.

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