Calculate the amount of heat radiated per second by a body of surface area `12cm^(2)` kept in thermal equilibrium in a room at temperature `20^(@)C` The emissivity of the surface `=0.80` and `sigma=6.0xx10^(-s)Wm^(-2)K^(-4)` .

Explanation

To calculate the amount of heat radiated per second by a body, 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 with emissivity is given by:

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

Where:
- \( P \) = Power (heat radiated per second) in watts (W)
- \( \epsilon \) = Emissivity of the surface (dimensionless)
- \( \sigma \) = Stefan-Boltzmann constant (\( 6.0 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \))
- \( A \) = Surface area in square meters (m²)
- \( T \) = Absolute temperature in Kelvin (K)

### Step-by-Step Solution:

1. **Convert the Surface Area from cm² to m²**:
\[ A = 12 \, \text{cm}^2 = 12 \times 10^{-4} \, \text{m}^2 = 0.0012 \, \text{m}^2 \]

2. **Convert the Temperature from Celsius to Kelvin**:
\[ T = 20^\circ C = 20 + 273 = 293 \, \text{K} \]

3. **Substitute the Values into the Stefan-Boltzmann Law**:
Using the formula \( P = \epsilon \sigma A T^4 \):
\[ P = 0.80 \times (6.0 \times 10^{-8} \, \text{W/m}^2\text{K}^4) \times (0.0012 \, \text{m}^2) \times (293 \, \text{K})^4 \]

4. **Calculate \( T^4 \)**:
\[ T^4 = (293)^4 = 7.688 \times 10^9 \, \text{K}^4 \]

5. **Calculate the Power**:
\[ P = 0.80 \times (6.0 \times 10^{-8}) \times (0.0012) \times (7.688 \times 10^9) \]
\[ P = 0.80 \times 6.0 \times 10^{-8} \times 0.0012 \times 7.688 \times 10^9 \]
\[ P \approx 0.42 \, \text{W} \]

### Final Answer:
The amount of heat radiated per second by the body is approximately **0.42 Joules**.