A spherical body of area A and emissivity `e=0.6` is kept inside a perfectly black body. Total heat radiated by the body at temperature T

Options

A. `0.4sigmaAT^4`
B. `0.8sigmaAT^4`
C. `0.6sigmaAT^4`
D. `1.0sigmaAT^4`

Explanation

To solve the problem of finding the total heat radiated by a spherical body with emissivity \( e = 0.6 \) at temperature \( T \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Emissivity**:
- Emissivity (\( e \)) is a measure of how effectively a surface emits thermal radiation compared to a perfect black body. A black body has an emissivity of 1, while our spherical body has an emissivity of 0.6.

2. **Use the Stefan-Boltzmann Law**:
- The Stefan-Boltzmann Law states that the power radiated per unit area of a black body is given by:
\[ P = \sigma T^4 \]
where \( \sigma \) is the Stefan-Boltzmann constant and \( T \) is the absolute temperature in Kelvin.

3. **Calculate the Power Radiated by the Spherical Body**:
- Since our body is not a perfect black body, we need to multiply the power by its emissivity:
\[ P_{\text{body}} = e \cdot \sigma T^4 \]
- Substituting the emissivity value:
\[ P_{\text{body}} = 0.6 \cdot \sigma T^4 \]

4. **Consider the Surface Area**:
- The total power radiated by the spherical body is the power per unit area multiplied by the total area \( A \) of the sphere:
\[ Q = P_{\text{body}} \cdot A = (0.6 \cdot \sigma T^4) \cdot A \]

5. **Final Expression for Total Heat Radiated**:
- Therefore, the total heat radiated \( Q \) by the spherical body at temperature \( T \) is:
\[ Q = 0.6 \sigma A T^4 \]

### Conclusion:
The total heat radiated by the body at temperature \( T \) is given by:
\[ Q = 0.6 \sigma A T^4 \]

### Answer:
The correct option is 3: \( 0.6 \sigma A T^4 \).

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