A black body at `227^(@)C` radiates heat at the rate of 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`

Explanation

To solve the problem, we will use Stefan-Boltzmann law, which states that the rate of heat radiation (E) from a black body is proportional to the fourth power of its absolute temperature (T).

### Step-by-Step Solution:

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

2. **Use Stefan-Boltzmann Law:**
The Stefan-Boltzmann law states:
\[ E \propto T^4 \]
Therefore, we can write the ratio of the rates of heat radiation at the two temperatures:
\[ \frac{E_1}{E_2} = \frac{T_1^4}{T_2^4} \]

3. **Substituting the known values:**
Given that \( E_1 = 7 \, \text{cal} \, \text{cm}^{-2} \, \text{s}^{-1} \):
\[ \frac{7}{E_2} = \frac{500^4}{1000^4} \]

4. **Simplifying the temperature ratio:**
\[ \frac{500^4}{1000^4} = \left(\frac{500}{1000}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]

5. **Finding \( E_2 \):**
Rearranging the equation:
\[ E_2 = 7 \times 16 = 112 \, \text{cal} \, \text{cm}^{-2} \, \text{s}^{-1} \]

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

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