A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be

Options

A. 225
B. 450
C. 1000
D. 1800

Answer

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 and the surface area of the body. The formula is given by:

\[ P = \sigma A T^4 \]

Where:
- \( P \) is the power radiated,
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( A \) is the surface area of the black body,
- \( T \) is the absolute temperature.

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- Initial radius \( R_1 = 12 \, \text{cm} = 0.12 \, \text{m} \)
- Initial power \( P_1 = 450 \, \text{W} \)
- Initial temperature \( T_1 = 500 \, \text{K} \)

2. **Calculate the initial surface area \( A_1 \):**
- The surface area of a sphere is given by \( A = 4\pi R^2 \).
- Thus, \( A_1 = 4\pi (R_1)^2 = 4\pi (0.12)^2 \).

3. **Calculate the new conditions:**
- New radius \( R_2 = \frac{R_1}{2} = \frac{0.12}{2} = 0.06 \, \text{m} \)
- New temperature \( T_2 = 2 \times T_1 = 2 \times 500 = 1000 \, \text{K} \)

4. **Calculate the new surface area \( A_2 \):**
- \( A_2 = 4\pi (R_2)^2 = 4\pi (0.06)^2 \).

5. **Use the Stefan-Boltzmann law to find the new power \( P_2 \):**
- The power radiated can be expressed as:
\[ P_1 = \sigma A_1 T_1^4 \]
\[ P_2 = \sigma A_2 T_2^4 \]

6. **Relate \( P_2 \) to \( P_1 \):**
- We can divide the two equations:
\[ \frac{P_2}{P_1} = \frac{A_2 T_2^4}{A_1 T_1^4} \]

7. **Substituting the areas:**
- Since \( A_1 = 4\pi (0.12)^2 \) and \( A_2 = 4\pi (0.06)^2 \), we can simplify:
\[ \frac{P_2}{P_1} = \frac{(0.06)^2 T_2^4}{(0.12)^2 T_1^4} \]
- This simplifies to:
\[ \frac{P_2}{P_1} = \frac{(0.06)^2}{(0.12)^2} \cdot \frac{(1000)^4}{(500)^4} \]

8. **Calculating the ratios:**
- \( \frac{(0.06)^2}{(0.12)^2} = \frac{1}{4} \)
- \( \frac{(1000)^4}{(500)^4} = \left(\frac{1000}{500}\right)^4 = 2^4 = 16 \)

9. **Combine the ratios:**
\[ \frac{P_2}{P_1} = \frac{1}{4} \cdot 16 = 4 \]

10. **Calculate \( P_2 \):**
\[ P_2 = 4 \cdot P_1 = 4 \cdot 450 = 1800 \, \text{W} \]

### Final Answer:
The power radiated when the radius is halved and the temperature is doubled is \( \boxed{1800 \, \text{W}} \).