A black body at `227^(@)C` radiates heat at the rate of `7 cal cm^(-2) s^(-1)`. At a temperature of `727^(@)C`, the rate of heat radiated in the same unit will be

Options

A. 60
B. 50
C. 112
D. 80

Answer

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

\[ R = \sigma T^4 \]

Where:
- \( R \) is the rate of heat radiation per unit area,
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( T \) is the absolute temperature in Kelvin.

### Step-by-Step Solution:

1. **Convert the given temperatures from Celsius to Kelvin:**
- For the first temperature \( T_1 = 227^\circ C \):
\[ T_1 = 227 + 273 = 500 \, K \]
- For the second temperature \( T_2 = 727^\circ C \):
\[ T_2 = 727 + 273 = 1000 \, K \]

2. **Use the Stefan-Boltzmann law to find the ratio of heat radiated at the two temperatures:**
- The ratio of the rates of heat radiation at two different temperatures is given by:
\[ \frac{R_2}{R_1} = \left(\frac{T_2}{T_1}\right)^4 \]

3. **Substituting the values of \( T_1 \) and \( T_2 \):**
\[ \frac{R_2}{R_1} = \left(\frac{1000}{500}\right)^4 = 2^4 = 16 \]

4. **Calculate \( R_2 \) using the known value of \( R_1 \):**
- Given \( R_1 = 7 \, \text{cal} \, \text{cm}^{-2} \, \text{s}^{-1} \):
\[ R_2 = 16 \times R_1 = 16 \times 7 = 112 \, \text{cal} \, \text{cm}^{-2} \, \text{s}^{-1} \]

5. **Final Result:**
- The rate of heat radiated at \( 727^\circ C \) is \( R_2 = 112 \, \text{cal} \, \text{cm}^{-2} \, \text{s}^{-1} \).

### Conclusion:
The answer to the question is \( 112 \, \text{cal} \, \text{cm}^{-2} \, \text{s}^{-1} \).

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