A black body emit heat at the rate of `20 W`, when its tempertaure is `227^(@)C` Another black body emits heat at the rate of `15 W`, when its temperature is `227^(@)C`. Compare the area of the surface of the two bodies, if the surrounding is at `NTP`

Options

A. `16:1`
B. `4:3`
C. `12:1`
D. `1:12`

Answer

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

### Step-by-Step Solution:

1. **Convert temperatures to Kelvin:**
- The temperature of both black bodies is given as \( 227^\circ C \).
- To convert to Kelvin, use the formula:
\[ T(K) = T(°C) + 273.15 \]
- Therefore,
\[ T_1 = T_2 = 227 + 273.15 = 500.15 \, K \approx 500 \, K \]
- The surrounding temperature at NTP (Normal Temperature and Pressure) is approximately \( 20^\circ C \) or \( 293 \, K \).

2. **Apply Stefan-Boltzmann Law:**
- The power emitted by a black body can be expressed as:
\[ E = \sigma \cdot \epsilon \cdot A \cdot (T^4 - T_0^4) \]
- Where:
- \( E \) = power emitted (in watts)
- \( \sigma \) = Stefan-Boltzmann constant
- \( \epsilon \) = emissivity (for a black body, \( \epsilon = 1 \))
- \( A \) = area of the surface
- \( T \) = absolute temperature of the body
- \( T_0 \) = absolute temperature of the surroundings

3. **Set up equations for both bodies:**
- For the first black body:
\[ E_1 = 20 \, W = \sigma \cdot A_1 \cdot (500^4 - 293^4) \]
- For the second black body:
\[ E_2 = 15 \, W = \sigma \cdot A_2 \cdot (500^4 - 293^4) \]

4. **Divide the equations:**
- Dividing the two equations gives:
\[ \frac{E_1}{E_2} = \frac{A_1 \cdot (500^4 - 293^4)}{A_2 \cdot (500^4 - 293^4)} \]
- This simplifies to:
\[ \frac{20}{15} = \frac{A_1}{A_2} \]
- Thus:
\[ \frac{A_1}{A_2} = \frac{4}{3} \]

5. **Conclusion:**
- The ratio of the areas of the surfaces of the two bodies is:
\[ A_1 : A_2 = 4 : 3 \]