A body at a temperature of `727^(@)C` and having surface area `5 cm^(2)`, radiations `300 J` of energy each minute. The emissivity is(Given Boltzmann constant `=5.67xx10^(-8) Wm^(-2)K^(-4)`

Options

A. `e= 0.18`
B. `e= 0.02`
C. `e= 0.2`
D. `e= 0.15`

Explanation

To find the emissivity of the body, we can use the Stefan-Boltzmann law, which states:

\[ Q = e \sigma A T^4 t \]

Where:
- \( Q \) = total energy radiated (in joules)
- \( e \) = emissivity (dimensionless)
- \( \sigma \) = Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \))
- \( A \) = surface area (in square meters)
- \( T \) = absolute temperature (in Kelvin)
- \( t \) = time (in seconds)

### Step-by-Step Solution:

1. **Convert Temperature to Kelvin**:
The given temperature is \( 727^\circ C \). To convert it to Kelvin:
\[ T = 727 + 273 = 1000 \, K \]

2. **Convert Surface Area to Square Meters**:
The surface area is given as \( 5 \, \text{cm}^2 \). To convert it to square meters:
\[ A = 5 \, \text{cm}^2 = 5 \times 10^{-4} \, \text{m}^2 \]

3. **Convert Time to Seconds**:
The energy is radiated per minute, so we need to convert minutes to seconds:
\[ t = 1 \, \text{minute} = 60 \, \text{seconds} \]

4. **Substitute Values into the Stefan-Boltzmann Law**:
We know that \( Q = 300 \, J \), \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \), \( T = 1000 \, K \), \( A = 5 \times 10^{-4} \, \text{m}^2 \), and \( t = 60 \, s \). Plugging these values into the formula:
\[ 300 = e \cdot (5.67 \times 10^{-8}) \cdot (5 \times 10^{-4}) \cdot (1000)^4 \cdot 60 \]

5. **Calculate \( T^4 \)**:
\[ T^4 = (1000)^4 = 10^{12} \, K^4 \]

6. **Calculate the Right Side**:
\[ \text{Right Side} = (5.67 \times 10^{-8}) \cdot (5 \times 10^{-4}) \cdot (10^{12}) \cdot 60 \]
\[ = (5.67 \times 5 \times 60) \times 10^{-8 -4 + 12} \]
\[ = (1701) \times 10^{0} = 1701 \, W \]

7. **Solve for Emissivity \( e \)**:
Rearranging the equation to find \( e \):
\[ e = \frac{Q}{\text{Right Side}} = \frac{300}{1701} \]
\[ e \approx 0.176 \]

8. **Final Result**:
The emissivity \( e \) is approximately \( 0.18 \).

### Conclusion:
The emissivity of the body is approximately \( 0.18 \).