A 300 W lamp loses all its energy by emission of radiation from the surface of its filament. If the area of surface of filament is `2.4cm^(2)` and is of emissivity `0.4,` estimate its temperature. Given that `sigma = 1.36 xx 10^(-12) cal cm^(-2) s^(-1) K^(-4) J=4.2 J cal^(-1)`. Neglect the absorption from the surroundings.

Answer

To estimate the temperature of the filament in the lamp, we will use the Stefan-Boltzmann law, which relates the power emitted by a black body to its temperature. The formula we will use is:

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

Where:
- \( P \) is the power (in watts),
- \( \varepsilon \) is the emissivity (dimensionless),
- \( \sigma \) is the Stefan-Boltzmann constant (in \( \text{W/m}^2\text{K}^4 \)),
- \( A \) is the surface area (in \( \text{m}^2 \)),
- \( T \) is the absolute temperature (in Kelvin).

### Step 1: Convert the area from cm² to m²
Given the area of the filament is \( 2.4 \, \text{cm}^2 \), we convert it to square meters:

\[ A = 2.4 \, \text{cm}^2 \times \frac{1 \, \text{m}^2}{10^4 \, \text{cm}^2} = 2.4 \times 10^{-4} \, \text{m}^2 \]

### Step 2: Convert the Stefan-Boltzmann constant to the correct units
The given value of \( \sigma \) is \( 1.36 \times 10^{-12} \, \text{cal/cm}^2\text{s} \text{K}^4 \). We need to convert this to \( \text{W/m}^2\text{K}^4 \):

1. Convert calories to joules:
\[ 1 \, \text{cal} = 4.2 \, \text{J} \]
Therefore,
\[ \sigma = 1.36 \times 10^{-12} \, \text{cal/cm}^2\text{s} \text{K}^4 \times 4.2 \, \text{J/cal} = 5.712 \times 10^{-12} \, \text{J/cm}^2\text{s} \text{K}^4 \]

2. Convert \( \text{cm}^2 \) to \( \text{m}^2 \):
\[ \sigma = 5.712 \times 10^{-12} \, \text{J/cm}^2\text{s} \text{K}^4 \times 10^4 \, \text{cm}^2/\text{m}^2 = 5.712 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \]

### Step 3: Rearrange the formula to solve for temperature \( T \)
From the formula \( P = \varepsilon \sigma A T^4 \), we can rearrange it to find \( T \):

\[ T^4 = \frac{P}{\varepsilon \sigma A} \]

### Step 4: Substitute the known values
Substituting the values into the equation:
- \( P = 300 \, \text{W} \)
- \( \varepsilon = 0.4 \)
- \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \)
- \( A = 2.4 \times 10^{-4} \, \text{m}^2 \)

\[ T^4 = \frac{300}{0.4 \times 5.67 \times 10^{-8} \times 2.4 \times 10^{-4}} \]

### Step 5: Calculate \( T^4 \)
Calculating the denominator:
\[ 0.4 \times 5.67 \times 10^{-8} \times 2.4 \times 10^{-4} = 5.44 \times 10^{-11} \]

Now substituting back:
\[ T^4 = \frac{300}{5.44 \times 10^{-11}} \approx 5.51 \times 10^{12} \]

### Step 6: Calculate \( T \)
Taking the fourth root:
\[ T = (5.51 \times 10^{12})^{1/4} \approx 1.37 \times 10^3 \, \text{K} = 1370 \, \text{K} \]

### Final Answer
The estimated temperature of the filament is approximately **1370 K**.

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