Assuming a filament in a 100W light bulb acts like a perfect blackbody, what is the temperature of the hottest portion of the filament if it has a surface area of `6.3 xx 10^(-5) m^(2)` ? The Stefan-Boltzmann constant is `5.67 xx 10^(-8) W //(m^(2).K^(2))`.

Options

A. 130K
B. 2300K
C. 1100K
D. 5800K

Answer

To find the temperature of the hottest portion of the filament in a 100W light bulb, we can use the Stefan-Boltzmann law, which states that the power radiated by a black body is proportional to the fourth power of its temperature.

### Step-by-Step Solution:

1. **Understand the Stefan-Boltzmann Law**:
The power radiated by a black body is given by the formula:
\[ P = \sigma A T^4 \]
where:
- \( P \) is the power (in watts),
- \( \sigma \) is the Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \)),
- \( A \) is the surface area (in m²),
- \( T \) is the absolute temperature (in Kelvin).

2. **Substitute the Known Values**:
Given:
- \( P = 100 \, \text{W} \)
- \( A = 6.3 \times 10^{-5} \, \text{m}^2 \)
- \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \)

Substitute these values into the equation:
\[ 100 = (5.67 \times 10^{-8}) \times (6.3 \times 10^{-5}) \times T^4 \]

3. **Calculate the Product of Constants**:
First, calculate the product of \( \sigma \) and \( A \):
\[ \sigma A = (5.67 \times 10^{-8}) \times (6.3 \times 10^{-5}) = 3.5701 \times 10^{-12} \, \text{W/K}^4 \]

4. **Rearrange the Equation to Solve for \( T^4 \)**:
Rearranging the equation gives:
\[ T^4 = \frac{P}{\sigma A} \]
Substituting the values:
\[ T^4 = \frac{100}{3.5701 \times 10^{-12}} \approx 2.80 \times 10^{13} \, \text{K}^4 \]

5. **Calculate \( T \)**:
Now, take the fourth root to find \( T \):
\[ T = (2.80 \times 10^{13})^{1/4} \]
Using a calculator:
\[ T \approx 2.3 \times 10^{3} \, \text{K} \approx 2300 \, \text{K} \]

### Final Answer:
The temperature of the hottest portion of the filament is approximately **2300 K**.