A black body at temperature `227^(@)C` radiates heat at the rate of 5 cal/`cm^(2)`, at a temperature of `27^(@)C`, the rate of heat radiated per unit area per unit time in cal/`cm^(2)`s, is

Options

A. 4
B. 1.5
C. 0.64
D. 0.25

Explanation

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.

### Step-by-Step Solution:

1. **Convert the temperatures from Celsius to Kelvin**:
- For the first temperature (T1 = 227°C):
\[ T_1 = 227 + 273 = 500 \, K \]
- For the second temperature (T2 = 27°C):
\[ T_2 = 27 + 273 = 300 \, K \]

2. **Use the Stefan-Boltzmann law**:
The law states:
\[ E \propto T^4 \]
This means that the rate of heat radiated per unit area (E) is proportional to the fourth power of the absolute temperature (T).

3. **Set up the ratio of the heat radiated at the two temperatures**:
Given that at T1 (500 K), the rate of heat radiated is 5 cal/cm²s, we can express this relationship as:
\[ \frac{E_1}{E_2} = \frac{T_1^4}{T_2^4} \]
Here, \(E_1 = 5 \, \text{cal/cm}^2\text{s}\) and we need to find \(E_2\).

4. **Substituting the values into the equation**:
\[ \frac{5}{E_2} = \frac{500^4}{300^4} \]

5. **Calculating the fourth powers**:
- Calculate \(500^4\) and \(300^4\):
\[ 500^4 = (5 \times 10^2)^4 = 5^4 \times 10^8 = 625 \times 10^8 \]
\[ 300^4 = (3 \times 10^2)^4 = 3^4 \times 10^8 = 81 \times 10^8 \]

6. **Setting up the ratio**:
\[ \frac{5}{E_2} = \frac{625 \times 10^8}{81 \times 10^8} \]
This simplifies to:
\[ \frac{5}{E_2} = \frac{625}{81} \]

7. **Cross-multiplying to solve for \(E_2\)**:
\[ 5 \cdot 81 = 625 \cdot E_2 \]
\[ 405 = 625 \cdot E_2 \]
\[ E_2 = \frac{405}{625} \]

8. **Calculating \(E_2\)**:
\[ E_2 = 0.648 \, \text{cal/cm}^2\text{s} \approx 0.64 \, \text{cal/cm}^2\text{s} \]

### Final Answer:
The rate of heat radiated per unit area per unit time at 27°C is approximately:
\[ \boxed{0.64 \, \text{cal/cm}^2\text{s}} \]