The rate of radiation of a black body at 0° C is E . The rate of radiation of this black body at `273^(@)C` will be

Options

A. 16E
B. 8E
C. 4E
D. E

Explanation

To solve the problem, we need to determine the rate of radiation of a black body at 273°C, given that the rate of radiation at 0°C is E. We will use the Stefan-Boltzmann law, which states that the rate of radiation (E) emitted by a black body is proportional to the fourth power of its absolute temperature (T).

### 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 0°C:
\[ T_1 = 0 + 273 = 273 \, K \]
- For 273°C:
\[ T_2 = 273 + 273 = 546 \, K \]

2. **Apply the Stefan-Boltzmann Law**:
- The rate of radiation is proportional to the fourth power of the absolute temperature:
\[ E \propto T^4 \]
- Therefore, we can write the relationship for the two temperatures:
\[ \frac{E_1}{E_2} = \frac{T_1^4}{T_2^4} \]

3. **Substituting Known Values**:
- We know \( E_1 = E \), \( T_1 = 273 \, K \), and \( T_2 = 546 \, K \):
\[ \frac{E}{E_2} = \frac{(273)^4}{(546)^4} \]

4. **Rearranging for \( E_2 \)**:
- Rearranging the equation gives:
\[ E_2 = E \cdot \left(\frac{(546)^4}{(273)^4}\right) \]

5. **Simplifying the Ratio**:
- Notice that \( \frac{546}{273} = 2 \):
\[ E_2 = E \cdot \left(2^4\right) \]
- Calculating \( 2^4 \):
\[ 2^4 = 16 \]
- Therefore:
\[ E_2 = 16E \]

### Final Answer:
The rate of radiation of the black body at 273°C will be \( 16E \).