A 100W bulb has tungsten filament of total length `1.0m` and raidius `4xx10^(-5)m` . The emiswsivity of the filament is `0.8` and `sigma=6.0xx10^(-8)Wm^(-2)K^(-4)` . Calculate the temperature of the filament when the bulb is oprating at correct wattage.

Explanation

To find the temperature of the tungsten filament in the 100W bulb, we can follow these steps:

### Step 1: Calculate the Surface Area of the Filament
The filament is modeled as a cylinder, and the surface area \( A \) of a cylinder is given by the formula:
\[ A = 2 \pi r l \]
where \( r \) is the radius and \( l \) is the length of the filament.

Given:
- Length \( l = 1.0 \, m \)
- Radius \( r = 4 \times 10^{-5} \, m \)

Substituting the values:
\[ A = 2 \pi (4 \times 10^{-5}) (1.0) = 8 \pi \times 10^{-5} \, m^2 \]

### Step 2: Use the Stefan-Boltzmann Law
The power radiated by the filament can be calculated using the Stefan-Boltzmann law:
\[ P = \epsilon \sigma A T^4 \]
where:
- \( P \) is the power (100 W),
- \( \epsilon \) is the emissivity (0.8),
- \( \sigma \) is the Stefan-Boltzmann constant (\( 6.0 \times 10^{-8} \, W/m^2K^4 \)),
- \( A \) is the area we calculated in Step 1,
- \( T \) is the temperature in Kelvin.

### Step 3: Rearranging the Formula to Solve for Temperature
Rearranging the formula to solve for \( T \):
\[ T^4 = \frac{P}{\epsilon \sigma A} \]
\[ T = \left( \frac{P}{\epsilon \sigma A} \right)^{1/4} \]

### Step 4: Substitute the Values
Now we substitute the known values into the equation:
\[ T = \left( \frac{100}{0.8 \times 6.0 \times 10^{-8} \times 8 \pi \times 10^{-5}} \right)^{1/4} \]

### Step 5: Calculate the Area
First, calculate the area:
\[ A = 8 \pi \times 10^{-5} \approx 2.51 \times 10^{-4} \, m^2 \]

### Step 6: Substitute and Calculate Temperature
Now substituting the area back into the temperature equation:
\[ T = \left( \frac{100}{0.8 \times 6.0 \times 10^{-8} \times 2.51 \times 10^{-4}} \right)^{1/4} \]

Calculating the denominator:
\[ 0.8 \times 6.0 \times 10^{-8} \times 2.51 \times 10^{-4} \approx 1.21 \times 10^{-11} \]

Now substituting this value back:
\[ T = \left( \frac{100}{1.21 \times 10^{-11}} \right)^{1/4} \]
\[ T = \left( 8.26 \times 10^{12} \right)^{1/4} \]
\[ T \approx 1700 \, K \]

### Final Answer
The temperature of the filament when the bulb is operating at correct wattage is approximately \( 1700 \, K \).

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