The temperature of a body in increased from `27^(@)C` to `127^(@)C`. By what factor would the radiation emitted by it increase?

Options

A. 2000
B. 1000
C. 100
D. 3.16

Explanation

To solve the problem of how much the radiation emitted by a body increases when its temperature is raised from \(27^\circ C\) to \(127^\circ C\), we can use Stefan-Boltzmann Law, which states that the power radiated by a black body is proportional to the fourth power of its absolute temperature.

### Step-by-Step Solution:

1. **Convert Celsius to Kelvin**:
- The initial temperature \(T_1\) is \(27^\circ C\).
- To convert to Kelvin:
\[ T_1 = 27 + 273 = 300 \, K \]
- The final temperature \(T_2\) is \(127^\circ C\).
- To convert to Kelvin:
\[ T_2 = 127 + 273 = 400 \, K \]

2. **Apply Stefan-Boltzmann Law**:
- According to the Stefan-Boltzmann Law, the energy emitted (\(E\)) is proportional to the fourth power of the temperature:
\[ E \propto T^4 \]
- Thus, the ratio of the energies emitted at the two temperatures can be expressed as:
\[ \frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4 \]

3. **Substitute the Temperatures**:
- Substitute \(T_1\) and \(T_2\) into the equation:
\[ \frac{E_2}{E_1} = \left(\frac{400}{300}\right)^4 \]

4. **Calculate the Ratio**:
- Simplifying the fraction:
\[ \frac{400}{300} = \frac{4}{3} \]
- Now raise this to the power of 4:
\[ \left(\frac{4}{3}\right)^4 = \frac{256}{81} \]

5. **Calculate the Numerical Value**:
- Now calculate \(\frac{256}{81}\):
\[ \frac{256}{81} \approx 3.16 \]

6. **Conclusion**:
- Therefore, the radiation emitted by the body increases by a factor of approximately \(3.16\).

### Final Answer:
The radiation emitted by the body increases by a factor of \(3.16\).

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