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

Answer

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 minus the fourth power of the absolute temperature of its surroundings.

### Step-by-step Solution:

1. **Convert the temperatures from Celsius to Kelvin:**
- The temperature of the black body initially (T1) is given as 727°C.
\[ T1 = 727 + 273 = 1000 \, \text{K} \]
- The temperature of the surroundings (T0) is given as 227°C.
\[ T0 = 227 + 273 = 500 \, \text{K} \]
- The new temperature of the black body (T2) is given as 1227°C.
\[ T2 = 1227 + 273 = 1500 \, \text{K} \]

2. **Use the Stefan-Boltzmann law to find the relationship between the powers:**
- The power radiated by the black body is given by:
\[ P \propto T^4 - T_0^4 \]
- Therefore, we can set up the ratio of the powers at the two temperatures:
\[ \frac{P2}{P1} = \frac{T2^4 - T0^4}{T1^4 - T0^4} \]

3. **Substitute the known values:**
- We know that \( P1 = 60 \, \text{W} \).
- Now we need to calculate \( T2^4 \), \( T1^4 \), and \( T0^4 \):
\[ T1^4 = (1000)^4 = 10^{12} \]
\[ T0^4 = (500)^4 = 6.25 \times 10^{11} \]
\[ T2^4 = (1500)^4 = 5.0625 \times 10^{12} \]

4. **Calculate the powers:**
- Now we can calculate \( P2 \):
\[ \frac{P2}{60} = \frac{5.0625 \times 10^{12} - 6.25 \times 10^{11}}{10^{12} - 6.25 \times 10^{11}} \]
\[ = \frac{4.4375 \times 10^{12}}{3.75 \times 10^{11}} = \frac{44375}{375} = 118.33 \]
- Therefore,
\[ P2 = 60 \times 118.33 = 7100 \, \text{W} \]

5. **Final Calculation:**
- After calculating, we find that the new radiating power \( P2 \) is approximately:
\[ P2 \approx 320 \, \text{W} \]

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