A black body at a temperature of `227^(@)C` radiates heat energy at the rate of 5 cal/`cm^(2)`-sec. At a temperature of `727^(@)C`, the rate of heat radiated per unit area in cal/`cm^(2)`-sec will be

Options

A. 400
B. 80
C. 40
D. 15

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.

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin:**
- The temperatures given are in Celsius, so we need to convert them to Kelvin.
- \( T_1 = 227^\circ C + 273 = 500 \, K \)
- \( T_2 = 727^\circ C + 273 = 1000 \, K \)

2. **Use the Stefan-Boltzmann Law:**
- The rate of heat radiated per unit area (denoted as \( r \)) is given by:
\[ r = \sigma T^4 \]
- For the first temperature \( T_1 \):
\[ r_1 = \sigma T_1^4 \]
- For the second temperature \( T_2 \):
\[ r_2 = \sigma T_2^4 \]

3. **Set Up the Ratio of Heat Radiated:**
- We know that:
\[ \frac{r_2}{r_1} = \frac{T_2^4}{T_1^4} \]
- Rearranging gives:
\[ r_2 = r_1 \left( \frac{T_2}{T_1} \right)^4 \]

4. **Substitute Known Values:**
- We know \( r_1 = 5 \, \text{cal/cm}^2\text{s} \).
- Substitute \( T_1 = 500 \, K \) and \( T_2 = 1000 \, K \):
\[ r_2 = 5 \left( \frac{1000}{500} \right)^4 \]
- Simplifying the fraction:
\[ r_2 = 5 \left( 2 \right)^4 \]

5. **Calculate \( r_2 \):**
- Calculate \( 2^4 = 16 \):
\[ r_2 = 5 \times 16 = 80 \, \text{cal/cm}^2\text{s} \]

6. **Final Answer:**
- The rate of heat radiated per unit area at \( 727^\circ C \) is \( 80 \, \text{cal/cm}^2\text{s} \).

### Summary:
The final answer is \( 80 \, \text{cal/cm}^2\text{s} \).

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