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 solve the problem of calculating the radiation power for a sphere with a given temperature, radius, and emissivity, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Temperature (T) = 227°C
- Radius (r) = 2 m
- Emissivity (E) = 0.8

2. **Convert Temperature to Kelvin:**
The temperature in Kelvin is calculated using the formula:
\[ T(K) = T(°C) + 273 \]
Substituting the given temperature:
\[ T = 227 + 273 = 500 \, K \]

3. **Calculate the Surface Area of the Sphere:**
The surface area (A) of a sphere is given by the formula:
\[ A = 4\pi r^2 \]
Substituting the radius:
\[ A = 4 \times \pi \times (2)^2 = 4 \times \pi \times 4 = 16\pi \, m^2 \]
Using \(\pi \approx 3.14\):
\[ A \approx 16 \times 3.14 = 50.24 \, m^2 \]

4. **Use the Stefan-Boltzmann Law to Calculate Power:**
The power radiated by a surface is given by:
\[ P = \sigma \cdot E \cdot A \cdot T^4 \]
Where:
- \(\sigma\) (Stefan-Boltzmann constant) = \(5.67 \times 10^{-8} \, W/m^2K^4\)

Substituting the values:
\[ P = 5.67 \times 10^{-8} \cdot 0.8 \cdot 50.24 \cdot (500)^4 \]

5. **Calculate \(500^4\):**
\[ 500^4 = 62500000000 \]

6. **Substitute and Calculate Power:**
Now substituting back into the power equation:
\[ P = 5.67 \times 10^{-8} \cdot 0.8 \cdot 50.24 \cdot 62500000000 \]
First, calculate \(5.67 \times 10^{-8} \cdot 0.8\):
\[ = 4.536 \times 10^{-8} \]
Then:
\[ P = 4.536 \times 10^{-8} \cdot 50.24 \cdot 62500000000 \]
Calculate \(4.536 \times 10^{-8} \cdot 50.24\):
\[ = 2.279 \times 10^{-6} \]
Finally:
\[ P = 2.279 \times 10^{-6} \cdot 62500000000 \approx 1425 \, W \]

### Final Answer:
The radiation power for the sphere is approximately **1425 Watts**.

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