A sphere has a surface area of `1.0 m^(2)` and a temperature of 400 K and the power radiated from it is 150 W. Assuming the sphere is black body radiator. The power in kilowatt radiated when the area expands to `2.0 m^(2)` and the temperature changes to 800 K

Options

A. 6.2
B. 9.6
C. 4.8
D. 16

Explanation

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

### Step 1: Write the formula for power radiated by a black body
The power \( P \) radiated by a black body is given by the formula:
\[ P = \epsilon \sigma A T^4 \]
For a black body, \( \epsilon = 1 \), so the formula simplifies to:
\[ P = \sigma A T^4 \]

### Step 2: Establish the relationship between initial and final power
We can express the ratio of the final power \( P_2 \) to the initial power \( P_1 \) as:
\[ \frac{P_2}{P_1} = \frac{A_2 T_2^4}{A_1 T_1^4} \]

### Step 3: Substitute the known values
From the problem, we know:
- Initial power \( P_1 = 150 \, \text{W} \)
- Initial area \( A_1 = 1.0 \, \text{m}^2 \)
- Final area \( A_2 = 2.0 \, \text{m}^2 \)
- Initial temperature \( T_1 = 400 \, \text{K} \)
- Final temperature \( T_2 = 800 \, \text{K} \)

Now substituting these values into the ratio:
\[ \frac{P_2}{150} = \frac{2.0 \times (800)^4}{1.0 \times (400)^4} \]

### Step 4: Simplify the equation
Calculating the temperature ratio:
\[ \frac{(800)^4}{(400)^4} = \left(\frac{800}{400}\right)^4 = 2^4 = 16 \]
Thus, we can rewrite the equation as:
\[ \frac{P_2}{150} = 2.0 \times 16 = 32 \]

### Step 5: Calculate the final power \( P_2 \)
Now we can find \( P_2 \):
\[ P_2 = 150 \times 32 = 4800 \, \text{W} \]

### Step 6: Convert power to kilowatts
To convert watts to kilowatts, we divide by 1000:
\[ P_2 = \frac{4800}{1000} = 4.8 \, \text{kW} \]

### Final Answer
The power radiated when the area expands to \( 2.0 \, \text{m}^2 \) and the temperature changes to \( 800 \, \text{K} \) is \( 4.8 \, \text{kW} \).

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