A black body, at a temperature of `227^@C`, radiates heat at a rate of 20 cal `m^(-2)s^(-1)`. When its temperature is raised to `727^@C`, the heat radiated by it in cal `M^(-2)s^(-1)` will be closest to

Options

A. 40
B. 160
C. 320
D. 640

Explanation

To solve the problem, we will use the Stefan-Boltzmann law, which states that the heat radiated by a black body is proportional to the fourth power of its absolute temperature (in Kelvin).

### Step-by-Step Solution:

1. **Convert the temperatures from Celsius to Kelvin**:
- The initial temperature \( T_1 = 227^\circ C = 227 + 273 = 500 \, K \)
- The final temperature \( T_2 = 727^\circ C = 727 + 273 = 1000 \, K \)

2. **Use the Stefan-Boltzmann law**:
- The heat radiated \( R \) is proportional to \( T^4 \):
\[ R \propto T^4 \]
- Therefore, we can write the ratio of the heat radiated at the two temperatures as:
\[ \frac{R_2}{R_1} = \left(\frac{T_2}{T_1}\right)^4 \]

3. **Substitute the known values**:
- Given \( R_1 = 20 \, \text{cal} \, m^{-2} s^{-1} \)
- Substitute \( T_1 \) and \( T_2 \):
\[ \frac{R_2}{20} = \left(\frac{1000}{500}\right)^4 \]

4. **Calculate the ratio**:
- Simplify \( \frac{1000}{500} = 2 \):
\[ \frac{R_2}{20} = 2^4 = 16 \]

5. **Solve for \( R_2 \)**:
- Multiply both sides by 20:
\[ R_2 = 20 \times 16 = 320 \, \text{cal} \, m^{-2} s^{-1} \]

6. **Conclusion**:
- The heat radiated at \( 727^\circ C \) is closest to \( 320 \, \text{cal} \, m^{-2} s^{-1} \).

### Final Answer:
The heat radiated by the black body when its temperature is raised to \( 727^\circ C \) will be closest to **320 cal m\(^{-2}\) s\(^{-1}\)**.

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