The energy emitted per second by a black body at `1227^(@)C` is E. If the temperature of the black body is increased to `2727^(@)C` the energy emitted per second in the terms of E in the second case.

Options

A. 12E
B. 16E
C. 8E
D. 9E

Answer

To solve the problem, we will use Stefan-Boltzmann Law, which states that the energy emitted per second (E) by a black body is proportional to the fourth power of its absolute temperature (T) in Kelvin. The formula is given by:

\[ E = \sigma A T^4 \]

where:
- \( E \) is the energy emitted per second,
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( A \) is the surface area of the black body,
- \( T \) is the absolute temperature in Kelvin.

### Step 1: Convert temperatures from Celsius to Kelvin
1. For the first temperature \( T_1 = 1227^\circ C \):
\[ T_1 = 1227 + 273 = 1500 \, K \]

2. For the second temperature \( T_2 = 2727^\circ C \):
\[ T_2 = 2727 + 273 = 3000 \, K \]

### Step 2: Write the equations for energy emitted
1. The energy emitted at \( T_1 \):
\[ E_1 = \sigma A (1500)^4 \]

2. The energy emitted at \( T_2 \):
\[ E_2 = \sigma A (3000)^4 \]

### Step 3: Find the ratio of the energies
To find how \( E_2 \) relates to \( E_1 \), we take the ratio:
\[ \frac{E_2}{E_1} = \frac{\sigma A (3000)^4}{\sigma A (1500)^4} \]

The \( \sigma A \) terms cancel out:
\[ \frac{E_2}{E_1} = \frac{(3000)^4}{(1500)^4} \]

### Step 4: Simplify the ratio
This can be simplified further:
\[ \frac{E_2}{E_1} = \left(\frac{3000}{1500}\right)^4 = (2)^4 = 16 \]

### Conclusion
Thus, the energy emitted per second when the temperature is increased to \( 2727^\circ C \) is:
\[ E_2 = 16E \]

### Final Answer
The energy emitted per second in the second case is \( 16E \).

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