Parallel rays of light of intensity `I=912 WM^-2` are incident on a spherical black body kept in surroundings of temperature 300K. Take Stefan-Boltzmann constant `sigma=5.7xx10^-8`
`Wm^-2K^-4` and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to

Options

A. `330K`
B. `660K`
C. `990K`
D. `1550K`

Explanation

To find the final steady state temperature of a spherical black body when parallel rays of light are incident on it, we can follow these steps:

### Step 1: Understand the Energy Balance
In steady state, the power absorbed by the black body from the incident radiation equals the power emitted by the black body. The power absorbed can be calculated using the intensity of the incident light.

### Step 2: Calculate the Power Absorbed
The power absorbed \( P_{abs} \) by the black body can be calculated using the formula:
\[ P_{abs} = I \cdot A \]
where \( I \) is the intensity of the incident light (912 W/m²) and \( A \) is the surface area of the black body. For a sphere, the surface area \( A \) is given by:
\[ A = 4\pi r^2 \]
However, since we are looking for the temperature and the area will cancel out later, we can just denote the absorbed power as:
\[ P_{abs} = I \cdot A \]

### Step 3: Calculate the Power Emitted
The power emitted by the black body can be calculated using the Stefan-Boltzmann law:
\[ P_{emit} = \sigma \cdot A \cdot T^4 \]
where \( \sigma = 5.7 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \) is the Stefan-Boltzmann constant and \( T \) is the temperature of the black body.

### Step 4: Set the Absorbed Power Equal to the Emitted Power
At steady state, the absorbed power equals the emitted power:
\[ I \cdot A = \sigma \cdot A \cdot T^4 \]
We can cancel \( A \) from both sides (assuming \( A \neq 0 \)):
\[ I = \sigma T^4 \]

### Step 5: Solve for Temperature \( T \)
Rearranging the equation gives:
\[ T^4 = \frac{I}{\sigma} \]
Substituting the values:
\[ T^4 = \frac{912 \, \text{W/m}^2}{5.7 \times 10^{-8} \, \text{W/m}^2\text{K}^4} \]

### Step 6: Calculate \( T^4 \)
Calculating the right side:
\[ T^4 = \frac{912}{5.7 \times 10^{-8}} \approx 1.60 \times 10^{10} \]

### Step 7: Take the Fourth Root to Find \( T \)
Now, we take the fourth root to find \( T \):
\[ T = (1.60 \times 10^{10})^{1/4} \]
Calculating this gives:
\[ T \approx 330 \, \text{K} \]

### Final Answer
The final steady state temperature of the black body is approximately **330 K**.

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