Calculate radiationn powe for sphere whose temperature is `227^(@)C` and radius 2 m and emissivity 0.8.

Options

A. 1425 W
B. 1500W
C. 1255W
D. 1575W

Explanation

To calculate the radiation power of a sphere with a given temperature, radius, and emissivity, we can follow these steps:

### Step-by-Step Solution:

1. **Convert Temperature to Kelvin**:
The temperature given is in Celsius. We need to convert it to Kelvin using the formula:
\[ T(K) = T(°C) + 273 \]
Given \( T = 227°C \):
\[ T = 227 + 273 = 500 \, K \]

2. **Calculate the Surface Area of the Sphere**:
The surface area \( A \) of a sphere is given by the formula:
\[ A = 4\pi r^2 \]
Given the radius \( r = 2 \, m \):
\[ A = 4 \pi (2^2) = 4 \pi (4) = 16\pi \, m^2 \]

3. **Use the Stefan-Boltzmann Law to Calculate Radiation Power**:
The radiation power \( P \) emitted by a body is given by:
\[ P = \sigma A e T^4 \]
where:
- \( \sigma = 5.67 \times 10^{-8} \, W/m^2K^4 \) (Stefan-Boltzmann constant)
- \( e = 0.8 \) (emissivity)
- \( T = 500 \, K \) (absolute temperature)
- \( A = 16\pi \, m^2 \)

Substituting the values:
\[ P = (5.67 \times 10^{-8}) \times (16\pi) \times (0.8) \times (500^4) \]

4. **Calculate \( 500^4 \)**:
First, calculate \( 500^4 \):
\[ 500^4 = 62500000000 \, K^4 \]

5. **Calculate the Power**:
Now substitute \( 500^4 \) back into the power equation:
\[ P = (5.67 \times 10^{-8}) \times (16\pi) \times (0.8) \times (62500000000) \]
- Calculate \( 16\pi \):
\[ 16\pi \approx 50.27 \, m^2 \]
- Now substitute:
\[ P \approx (5.67 \times 10^{-8}) \times (50.27) \times (0.8) \times (62500000000) \]

6. **Final Calculation**:
\[ P \approx (5.67 \times 10^{-8}) \times (50.27) \times (0.8) \times (62500000000) \approx 1425 \, W \]

### Conclusion:
The radiation power of the sphere is approximately \( 1425 \, W \).