The gross radiation emitted by a perfectly black body is:

Options

A. Dependent on its temperature
B. Dependent on the area of its surface
C. Dependent on the temperature of the surroundings
D. Independent of the temperature of the surroundings

Explanation

To solve the question regarding the gross radiation emitted by a perfectly black body, we can use the Stefan-Boltzmann Law, which states that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of its absolute temperature (T). The formula is given by:

\[ E = \sigma T^4 \]

where:
- \( E \) is the total energy radiated per unit area,
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( T \) is the absolute temperature of the black body in Kelvin.

### Step-by-Step Solution:

1. **Identify the Formula**: Start with the Stefan-Boltzmann Law, which describes the radiation emitted by a black body.
\[ E = \sigma T^4 \]

2. **Understand the Variables**:
- \( E \) represents the gross radiation emitted,
- \( \sigma \) is a constant (Stefan-Boltzmann constant),
- \( T \) is the absolute temperature in Kelvin.

3. **Consider the Area**: If we are considering a surface area \( A \) of the black body, the total radiation emitted by the body can be expressed as:
\[ Q = \sigma A T^4 \]
where \( Q \) is the total radiation emitted.

4. **Conclusion**: Therefore, the gross radiation emitted by a perfectly black body depends on:
- The temperature \( T \) (as it is raised to the fourth power),
- The area \( A \) of the black body,
- The Stefan-Boltzmann constant \( \sigma \).

### Final Answer:
The gross radiation emitted by a perfectly black body is given by:
\[ Q = \sigma A T^4 \]
This indicates that the radiation depends on the temperature of the body and the area of its surface.

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