The energy radiated per hour from the surface of a filament 0.5 cm long and of radius 0.32 cm of an incandescent lamp at a certain temperature is `2.625xx10^(5)` J. If the relative emittance of the surface is 0.8 calculate the temperature of the filament.

Explanation

To solve the problem step by step, let's follow the outlined approach:

### Step 1: Understand the Problem
We need to find the temperature of a filament given the energy it radiates per hour, its dimensions, and its relative emittance.

### Step 2: Write Down the Formula
The energy radiated by a surface is given by the Stefan-Boltzmann law:
\[ E = \epsilon \cdot \sigma \cdot A \cdot T^4 \]
Where:
- \( E \) = energy radiated (in watts)
- \( \epsilon \) = relative emittance (given as 0.8)
- \( \sigma \) = Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \))
- \( A \) = surface area of the filament (in m²)
- \( T \) = temperature (in Kelvin)

### Step 3: Convert Energy from Joules per Hour to Watts
The energy radiated is given as \( 2.625 \times 10^5 \) J/hour. To convert this to watts (since 1 watt = 1 joule/second):
\[ E = \frac{2.625 \times 10^5 \, \text{J}}{3600 \, \text{s}} \approx 73 \, \text{W} \]

### Step 4: Calculate the Surface Area of the Filament
The filament is cylindrical, so the surface area \( A \) can be calculated using the formula:
\[ A = 2 \pi r h \]
Where:
- \( r = 0.32 \, \text{cm} = 0.32 \times 10^{-2} \, \text{m} \)
- \( h = 0.5 \, \text{cm} = 0.5 \times 10^{-2} \, \text{m} \)

Calculating the area:
\[ A = 2 \pi (0.32 \times 10^{-2}) (0.5 \times 10^{-2}) = 2 \pi (0.32 \times 0.5) \times 10^{-4} \approx 1.005 \times 10^{-4} \, \text{m}^2 \]

### Step 5: Substitute Values into the Formula
Now we can substitute the known values into the Stefan-Boltzmann equation:
\[ 73 = 0.8 \cdot (5.67 \times 10^{-8}) \cdot (1.005 \times 10^{-4}) \cdot T^4 \]

### Step 6: Solve for Temperature \( T \)
First, calculate the right side:
\[ 0.8 \cdot (5.67 \times 10^{-8}) \cdot (1.005 \times 10^{-4}) \approx 4.56 \times 10^{-11} \]

Now, rearranging the equation:
\[ T^4 = \frac{73}{4.56 \times 10^{-11}} \approx 1.60 \times 10^{12} \]

Taking the fourth root to find \( T \):
\[ T \approx (1.60 \times 10^{12})^{1/4} \approx 1972 \, \text{K} \]

### Final Answer
The temperature of the filament is approximately **1972 K**.

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