Energy is being emitted from the surface of a black body at `127^(@)C` temperature at the rate of `1.0xx10^(6)J//sec-m^(2)`. Temperature of the black at which the rate of energy emission is `16.0xx10^(6)J//sec-m^(2)` will be

Answer

To solve the problem, we will use the Stefan-Boltzmann law, which states that the energy emitted per unit area per unit time (intensity) from a black body is proportional to the fourth power of its absolute temperature. The formula can be expressed as:

\[ I = \sigma T^4 \]

where:
- \( I \) is the intensity (energy emitted per unit area per unit time),
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( T \) is the absolute temperature in Kelvin.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Initial temperature \( T_0 = 127^\circ C \)
- Initial intensity \( I_0 = 1.0 \times 10^6 \, J/s \cdot m^2 \)

2. **Convert the Initial Temperature to Kelvin:**
\[ T_0 = 127 + 273 = 400 \, K \]

3. **Identify the Final Conditions:**
- Final intensity \( I_f = 16.0 \times 10^6 \, J/s \cdot m^2 \)

4. **Set Up the Proportionality Relation:**
Since intensity is proportional to the fourth power of temperature, we can write:
\[ \frac{I_f}{I_0} = \left(\frac{T_f}{T_0}\right)^4 \]
Substituting the known values:
\[ \frac{16.0 \times 10^6}{1.0 \times 10^6} = \left(\frac{T_f}{400}\right)^4 \]
Simplifying the left side:
\[ 16 = \left(\frac{T_f}{400}\right)^4 \]

5. **Take the Fourth Root:**
To find \( T_f \), we take the fourth root of both sides:
\[ \frac{T_f}{400} = 16^{1/4} \]
Since \( 16 = 2^4 \), we have:
\[ 16^{1/4} = 2 \]
Thus:
\[ \frac{T_f}{400} = 2 \]

6. **Solve for \( T_f \):**
\[ T_f = 2 \times 400 = 800 \, K \]

7. **Convert the Final Temperature to Celsius:**
\[ T_f = 800 - 273 = 527^\circ C \]

### Final Answer:
The temperature of the black body at which the rate of energy emission is \( 16.0 \times 10^6 \, J/s \cdot m^2 \) will be \( 527^\circ C \).