A spherical black body with radius 12 cm radiates 450 w power at 500 K. If the radius is halved and the temperature doubled, the power radiated in watts would be

Options

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

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 and the square of its radius. The formula for the power radiated is given by:

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

where:
- \( P \) is the power radiated,
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( A \) is the surface area of the sphere, and
- \( T \) is the absolute temperature.

### Step 1: Identify the initial conditions
Given:
- Initial radius \( r_1 = 12 \, \text{cm} = 0.12 \, \text{m} \)
- Initial power \( P_1 = 450 \, \text{W} \)
- Initial temperature \( T_1 = 500 \, \text{K} \)

### Step 2: Calculate the new conditions
The radius is halved:
- New radius \( r_2 = \frac{r_1}{2} = \frac{0.12}{2} = 0.06 \, \text{m} \)

The temperature is doubled:
- New temperature \( T_2 = 2 \cdot T_1 = 2 \cdot 500 = 1000 \, \text{K} \)

### Step 3: Use the Stefan-Boltzmann Law
Using the ratio of the powers based on the changes in radius and temperature:

\[ \frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^4 \cdot \left(\frac{r_2}{r_1}\right)^2 \]

### Step 4: Substitute the values
Substituting the known values into the equation:

\[ \frac{P_2}{450} = \left(\frac{1000}{500}\right)^4 \cdot \left(\frac{0.06}{0.12}\right)^2 \]

Calculating each part:
- Temperature ratio: \( \frac{1000}{500} = 2 \)
- Radius ratio: \( \frac{0.06}{0.12} = \frac{1}{2} \)

Now substituting these ratios:

\[ \frac{P_2}{450} = (2)^4 \cdot \left(\frac{1}{2}\right)^2 \]

Calculating powers:
- \( (2)^4 = 16 \)
- \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)

### Step 5: Combine the results
Now substituting back into the equation:

\[ \frac{P_2}{450} = 16 \cdot \frac{1}{4} = 4 \]

### Step 6: Solve for \( P_2 \)
Now, multiplying both sides by 450:

\[ P_2 = 4 \cdot 450 = 1800 \, \text{W} \]

### Final Answer
The power radiated when the radius is halved and the temperature is doubled is \( P_2 = 1800 \, \text{W} \).

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