For a black body at temperature `727^@C`, its radiating power is 60 watt and temperature of surrounding is `227^@C`. If temperature of black body is changed to `1227^@C` then its radiating power will be-

Options

A. 304 W
B. 320 W
C. 240 W
D. 120 W

Explanation

To solve the problem, we will use the Stefan-Boltzmann law, which states that the power radiated by a black body is proportional to the fourth power of its absolute temperature. The formula can be expressed as:

\[ P = \sigma e A (T^4 - T_0^4) \]

Where:
- \( P \) = power radiated
- \( \sigma \) = Stefan-Boltzmann constant
- \( e \) = emissivity (for a black body, \( e = 1 \))
- \( A \) = surface area of the body
- \( T \) = absolute temperature of the black body in Kelvin
- \( T_0 \) = absolute temperature of the surroundings in Kelvin

### Step-by-step Solution:

1. **Convert temperatures from Celsius to Kelvin**:
- For the initial temperature \( T_1 = 727^\circ C \):
\[ T_1 = 727 + 273 = 1000 \, K \]
- For the surrounding temperature \( T_2 = 227^\circ C \):
\[ T_2 = 227 + 273 = 500 \, K \]
- For the new temperature \( T_3 = 1227^\circ C \):
\[ T_3 = 1227 + 273 = 1500 \, K \]

2. **Apply the Stefan-Boltzmann law for the initial state**:
- Given that the radiating power at \( T_1 \) is \( P_1 = 60 \, W \):
\[ P_1 = \sigma A (T_1^4 - T_2^4) \]
- Substitute the values:
\[ 60 = \sigma A (1000^4 - 500^4) \]

3. **Apply the Stefan-Boltzmann law for the new state**:
- For the new temperature \( T_3 \):
\[ P_2 = \sigma A (T_3^4 - T_2^4) \]
- Substitute the values:
\[ P_2 = \sigma A (1500^4 - 500^4) \]

4. **Find the ratio of the powers**:
- From the two equations, we can find the ratio:
\[ \frac{P_1}{P_2} = \frac{T_1^4 - T_2^4}{T_3^4 - T_2^4} \]
- Substitute \( P_1 = 60 \):
\[ \frac{60}{P_2} = \frac{1000^4 - 500^4}{1500^4 - 500^4} \]

5. **Calculate \( 1000^4 - 500^4 \) and \( 1500^4 - 500^4 \)**:
- Factor out \( 500^4 \):
\[ 1000^4 - 500^4 = (2^4 - 1) \cdot 500^4 = 16 - 1 = 15 \cdot 500^4 \]
\[ 1500^4 - 500^4 = (3^4 - 1) \cdot 500^4 = 81 - 1 = 80 \cdot 500^4 \]

6. **Substitute back into the ratio**:
\[ \frac{60}{P_2} = \frac{15 \cdot 500^4}{80 \cdot 500^4} \]
- Cancel \( 500^4 \):
\[ \frac{60}{P_2} = \frac{15}{80} \]

7. **Solve for \( P_2 \)**:
\[ P_2 = 60 \cdot \frac{80}{15} = 60 \cdot \frac{16}{3} = 320 \, W \]

### Final Answer:
The radiating power when the temperature of the black body is changed to \( 1227^\circ C \) will be **320 watts**.