The ratio of energy of emitted radiation of black body at ` 27^(@)C and 927^(@) c` is

Options

A. `1:256`
B. `1:64`
C. `1:16 `
D. `1:4`

Explanation

To find the ratio of the energy of emitted radiation of a black body at temperatures of 27°C and 927°C, we can use the Stefan-Boltzmann law, which states that the power (energy per unit time) emitted by a black body is proportional to the fourth power of its absolute temperature (in Kelvin).

### Step-by-Step Solution:

1. **Convert Celsius to Kelvin**:
- The temperature in Kelvin (K) is given by the formula:
\[ T(K) = T(°C) + 273 \]
- For the first temperature (27°C):
\[ T_1 = 27 + 273 = 300 \, K \]
- For the second temperature (927°C):
\[ T_2 = 927 + 273 = 1200 \, K \]

2. **Apply the Stefan-Boltzmann Law**:
- The power emitted by a black body is given by:
\[ P \propto T^4 \]
- Therefore, the ratio of the powers emitted at the two temperatures can be expressed as:
\[ \frac{P_1}{P_2} = \left(\frac{T_1}{T_2}\right)^4 \]

3. **Substitute the Temperatures**:
- Substitute \(T_1 = 300 \, K\) and \(T_2 = 1200 \, K\):
\[ \frac{P_1}{P_2} = \left(\frac{300}{1200}\right)^4 \]

4. **Simplify the Ratio**:
- Simplifying the fraction:
\[ \frac{300}{1200} = \frac{1}{4} \]
- Now raise this to the power of 4:
\[ \left(\frac{1}{4}\right)^4 = \frac{1}{256} \]

5. **Final Result**:
- Thus, the ratio of the energy emitted at 27°C to that at 927°C is:
\[ \frac{P_1}{P_2} = \frac{1}{256} \]

### Conclusion:
The ratio of energy of emitted radiation of a black body at 27°C and 927°C is \( \frac{1}{256} \).