A black body is at `727^(@)C`. It emits energy at a rate which is proportional to

Options

A. `(727)^(2)`
B. `(1000)^(4)`
C. `(1000)^(2)`
D. `(727)^(4)`

Explanation

To solve the problem, we need to determine how the rate of energy emission from a black body is related to its temperature.

### Step-by-Step Solution:

1. **Understand the Concept of a Black Body**:
A black body is an idealized physical object that absorbs all incoming radiation and re-emits energy based on its temperature.

2. **Use Stefan-Boltzmann Law**:
The rate of energy emission (E) from a black body is given by the Stefan-Boltzmann law, which states:
\[ E = \sigma e A T^4 \]
where:
- \(E\) = energy emitted per unit time (rate of emission)
- \(\sigma\) = Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4\))
- \(e\) = emissivity of the body (for a perfect black body, \(e = 1\))
- \(A\) = surface area of the body
- \(T\) = absolute temperature in Kelvin

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

4. **Apply the Stefan-Boltzmann Law**:
Since we are interested in the rate of energy emission being proportional to temperature, we can simplify the equation:
\[ E \propto T^4 \]
Thus, substituting the temperature:
\[ E \propto (1000)^4 \]

5. **Conclusion**:
The rate of energy emission is proportional to the fourth power of the absolute temperature. Therefore, the answer to the question is that the rate of energy emission from the black body at \(727^\circ C\) is proportional to \(T^4\).

### Final Answer:
The rate of energy emission is proportional to \(T^4\), where \(T\) is the absolute temperature in Kelvin.

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