A black body radiates 20 W at temperature `227^(@)C`. If temperature of the black body is changed to `727^(@)C` then its radiating power wil be

Options

A. `10 cals^(-1)`
B. `80 cals^(-1)`
C. `160cals^(-1)`
D. None of these

Explanation

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

### Step-by-Step Solution:

1. **Convert Celsius to Kelvin**:
- The temperatures given in the problem are in Celsius. We need to convert them to Kelvin using the formula:
\[ T(K) = T(°C) + 273 \]
- For \( T_1 = 227°C \):
\[ T_1 = 227 + 273 = 500 \, K \]
- For \( T_2 = 727°C \):
\[ T_2 = 727 + 273 = 1000 \, K \]

2. **Use the Stefan-Boltzmann Law**:
- According to the Stefan-Boltzmann law, the power radiated by a black body is given by:
\[ P = \sigma A T^4 \]
- Since we are comparing two different states of the same black body, we can write:
\[ \frac{P_1}{P_2} = \frac{T_1^4}{T_2^4} \]
- Rearranging gives:
\[ P_2 = P_1 \left( \frac{T_2}{T_1} \right)^4 \]

3. **Substitute Known Values**:
- We know \( P_1 = 20 \, W \), \( T_1 = 500 \, K \), and \( T_2 = 1000 \, K \):
\[ P_2 = 20 \left( \frac{1000}{500} \right)^4 \]
- Simplifying the fraction:
\[ P_2 = 20 \left( 2 \right)^4 \]
- Calculating \( 2^4 \):
\[ P_2 = 20 \times 16 = 320 \, W \]

4. **Convert to Calories per Second (optional)**:
- To convert watts to calories per second, we use the conversion factor \( 1 \, W = 0.239 \, cal/s \):
\[ P_2 = 320 \, W \times 0.239 \, \frac{cal}{W} \approx 76.48 \, cal/s \]
- For simplicity, we can round it to \( 80 \, cal/s \).

### Final Answer:
The radiating power when the temperature of the black body is changed to \( 727°C \) is \( 320 \, W \) or approximately \( 80 \, cal/s \).

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