An electric stove burner of diameter 20cm is at a temperature of `250^(@)C`. If `sigma = 5.67 xx 10^(-8) W//m^(2) . K^(4)`, at what rate is the burner radiating energy ? Assume the emissivity `epsilon= 0.6`.

Options

A. 4W
B. 80 W
C. 320 W
D. 1600 W

Answer

To find the rate at which the electric stove burner is radiating energy, 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 absolute temperature. The formula for the power radiated by a body is given by:

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

Where:
- \( P \) = power radiated (in watts)
- \( \epsilon \) = emissivity of the surface (dimensionless)
- \( \sigma \) = Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} \, \text{W/m}^2 \cdot \text{K}^4 \))
- \( A \) = surface area of the burner (in m²)
- \( T \) = absolute temperature (in Kelvin)

### Step 1: Convert the temperature from Celsius to Kelvin
The temperature in Celsius is given as \( 250^\circ C \). To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

Calculating:

\[ T = 250 + 273.15 = 523.15 \, K \]

### Step 2: Calculate the area of the burner
The burner is circular with a diameter of \( 20 \, cm \). First, we convert the diameter to meters:

\[ d = 20 \, cm = 0.2 \, m \]

The radius \( r \) is half of the diameter:

\[ r = \frac{d}{2} = \frac{0.2}{2} = 0.1 \, m \]

The area \( A \) of the circular burner is given by the formula:

\[ A = \pi r^2 \]

Calculating:

\[ A = \pi (0.1)^2 = \pi (0.01) \approx 0.0314 \, m^2 \]

### Step 3: Substitute the values into the Stefan-Boltzmann equation
Now we can substitute the values into the formula:

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

Substituting the known values:

- \( \epsilon = 0.6 \)
- \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2 \cdot \text{K}^4 \)
- \( A \approx 0.0314 \, m^2 \)
- \( T = 523.15 \, K \)

Calculating \( T^4 \):

\[ T^4 = (523.15)^4 \approx 7.55 \times 10^9 \, K^4 \]

Now substituting these values into the equation:

\[ P = 0.6 \times (5.67 \times 10^{-8}) \times (0.0314) \times (7.55 \times 10^9) \]

Calculating \( P \):

\[ P \approx 0.6 \times 5.67 \times 10^{-8} \times 0.0314 \times 7.55 \times 10^9 \]

Calculating step by step:

1. \( 5.67 \times 10^{-8} \times 0.0314 \approx 1.78 \times 10^{-9} \)
2. \( 1.78 \times 10^{-9} \times 7.55 \times 10^9 \approx 13.43 \)
3. \( P \approx 0.6 \times 13.43 \approx 8.06 \, W \)

### Final Answer
The rate at which the burner is radiating energy is approximately \( 8.06 \, W \).

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