Emissive power of an ideal black body at `127^(@)C` is E. the temperature at which it increases to 102% is

Options

A. 400 K
B. 100 K
C. 402 K
D. 502 K

Answer

To solve the problem, we need to determine the temperature at which the emissive power of an ideal black body increases to 102% of its initial value at 127°C. We will use Stefan-Boltzmann Law, which states that the emissive power (E) of a black body is proportional to the fourth power of its absolute temperature (T).

### Step-by-Step Solution:

1. **Convert the initial temperature to Kelvin**:
\[ T_1 = 127^\circ C + 273 = 400 \, K \]

2. **Write the expression for emissive power at the initial temperature**:
According to Stefan-Boltzmann Law, the emissive power \( E \) at temperature \( T_1 \) is given by:
\[ E = \sigma T_1^4 \]
where \( \sigma \) is the Stefan-Boltzmann constant.

3. **Calculate the emissive power at 102% of the initial value**:
The new emissive power \( E_2 \) when it increases to 102% is:
\[ E_2 = 1.02 E = 1.02 \sigma T_1^4 \]

4. **Set up the equation for the new temperature \( T_2 \)**:
The emissive power at the new temperature \( T_2 \) can also be expressed using the Stefan-Boltzmann Law:
\[ E_2 = \sigma T_2^4 \]

5. **Equate the two expressions for emissive power**:
\[ 1.02 \sigma T_1^4 = \sigma T_2^4 \]
Here, we can cancel \( \sigma \) from both sides:
\[ 1.02 T_1^4 = T_2^4 \]

6. **Substitute the value of \( T_1 \)**:
\[ 1.02 (400)^4 = T_2^4 \]

7. **Calculate \( T_2^4 \)**:
\[ T_2^4 = 1.02 \times 400^4 \]

8. **Take the fourth root to find \( T_2 \)**:
\[ T_2 = (1.02)^{1/4} \times 400 \]

9. **Calculate \( (1.02)^{1/4} \)**:
Using a calculator or approximation, we find:
\[ (1.02)^{1/4} \approx 1.00499 \]

10. **Final calculation for \( T_2 \)**:
\[ T_2 \approx 1.00499 \times 400 \approx 402 \, K \]

11. **Convert \( T_2 \) back to Celsius**:
\[ T_2 = 402 - 273 \approx 129^\circ C \]

### Final Answer:
The temperature at which the emissive power increases to 102% is approximately \( 129^\circ C \).