An experimental tungsten light bulb filament has a length of 7 cm and a diameter of 0.074 cm. The filament is basically just a wire that heats due to resistance to electrical current. The ends of the wire are not exposed. The total filament emissivity is  0.323 and its temperature is 2,378 K. Calculate the power emitted by the lamp in watts.

To calculate the power emitted by the tungsten filament, we can use the Stefan-Boltzmann law. This law relates the power radiated per unit surface area of a blackbody to its temperature:

\[
P = \epsilon \sigma A T^4
\]

Where:
- \( P \) is the total power radiated (in watts),
- \( \epsilon \) is the emissivity of the material (dimensionless),
- \( \sigma \) is the Stefan-Boltzmann constant, which is \( 5.67 \times 10^{-8} \, \text{W} \, \text{m}^{-2} \, \text{K}^{-4} \),
- \( A \) is the surface area of the filament (in square meters),
- \( T \) is the temperature of the filament (in Kelvin).

### Step 1: Calculate the surface area of the filament

The filament is a cylindrical wire, so the surface area \( A \) can be calculated using the formula for the lateral surface area of a cylinder:

\[
A = \pi d L
\]

Where:
- \( d \) is the diameter of the filament,
- \( L \) is the length of the filament.

First, let's convert the diameter and length from centimeters to meters:

\[
d = 0.074 \, \text{cm} = 0.074 \times 10^{-2} \, \text{m} = 7.4 \times 10^{-4} \, \text{m}
\]
\[
L = 7 \, \text{cm} = 7 \times 10^{-2} \, \text{m} = 0.07 \, \text{m}
\]

Now, calculate the surface area:

\[
A = \pi \times 7.4 \times 10^{-4} \, \text{m} \times 0.07 \, \text{m}
\]

### Step 2: Calculate the power emitted by the filament

Substitute the values into the Stefan-Boltzmann law equation to calculate the power:

\[
P = 0.323 \times 5.67 \times 10^{-8} \, \text{W} \, \text{m}^{-2} \, \text{K}^{-4} \times A \times (2378 \, \text{K})^4
\]

Let's perform the calculations step by step.The power emitted by the tungsten filament is approximately **95.3 watts**.

Would you like more details on this calculation, or do you have any further questions?

Here are five related questions:
1. How does the emissivity of a material affect the power it emits?
2. What is the impact of temperature on the power emitted according to the Stefan-Boltzmann law?
3. How would the power emitted change if the filament length were doubled?
4. What are the assumptions behind using the Stefan-Boltzmann law for this calculation?
5. How does the surface area of the filament influence the total emitted power?

**Tip:** The Stefan-Boltzmann law assumes that the object behaves like a blackbody, which means it radiates energy efficiently at all wavelengths. If the object has an emissivity less than 1, it radiates less energy.

Learn how to calculate the power emitted by a tungsten filament using the Stefan-Boltzmann law. This physics problem involves thermal radiation, emissivity, filament dimensions, and temperature. Find the solution step-by-step with detailed calculations.

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