Calculate the energy radiated per minute by a black body of surface area `200 cm^(2)` , maintained at `127^(@)`C. `sigma = 5.7xx10^(-8)Wm^(-2)K^(-4)`

Explanation

To solve the problem of calculating the energy radiated per minute by a black body of surface area \(200 \, \text{cm}^2\) maintained at \(127^\circ C\), we will use Stefan-Boltzmann Law, which states that the power radiated by a black body is given by:

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

where:
- \(P\) is the power (energy per second) in watts (J/s),
- \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4\)),
- \(A\) is the surface area in square meters,
- \(T\) is the absolute temperature in Kelvin.

### Step-by-Step Solution:

1. **Convert the temperature from Celsius to Kelvin**:
\[ T(K) = 127 + 273 = 400 \, K \]

2. **Convert the surface area from cm² to m²**:
\[ A = 200 \, \text{cm}^2 = 200 \times 10^{-4} \, \text{m}^2 = 0.02 \, \text{m}^2 \]

3. **Calculate the power radiated using Stefan-Boltzmann Law**:
\[ P = \sigma A T^4 \]
Substituting the values:
\[ P = 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \times 0.02 \, \text{m}^2 \times (400 \, K)^4 \]

4. **Calculate \(T^4\)**:
\[ T^4 = (400)^4 = 256 \times 10^8 \, K^4 \]

5. **Substituting \(T^4\) back into the power equation**:
\[ P = 5.67 \times 10^{-8} \times 0.02 \times 256 \times 10^8 \]

6. **Simplifying the equation**:
\[ P = 5.67 \times 256 \times 0.02 \]
\[ P = 5.67 \times 5.12 = 29.0664 \, \text{W} \]

7. **Calculate the energy radiated per minute**:
\[ \text{Energy per minute} = P \times 60 \, \text{s} \]
\[ \text{Energy per minute} = 29.0664 \, \text{W} \times 60 \, \text{s} = 1743.984 \, \text{J} \]

### Final Answer:
The energy radiated per minute by the black body is approximately \(1744 \, \text{J}\).