Calculate the temperature at which a perfect black body radiates at the rate of `1 W cm^(-2)` , value of Stefan's constant,
`sigma = 5.67 xx 10^(-5) W m^(-2) K^(-8)`

Options

A. `576 K`
B. `648 K`
C. `695 K`
D. `766 K`

Answer

To solve the problem of calculating the temperature at which a perfect black body radiates at the rate of \(1 \, \text{W cm}^{-2}\), we will use Stefan-Boltzmann's law, which states that the power radiated per unit area of a black body is given by:

\[ E = \sigma T^4 \]

where:
- \(E\) is the power radiated per unit area (in \( \text{W m}^{-2} \)),
- \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-5} \, \text{W m}^{-2} \text{K}^{-4}\)),
- \(T\) is the absolute temperature in Kelvin.

### Step 1: Convert the power from \( \text{W cm}^{-2} \) to \( \text{W m}^{-2} \)

Given:
\[ E = 1 \, \text{W cm}^{-2} \]

To convert this to \( \text{W m}^{-2} \):
\[ 1 \, \text{W cm}^{-2} = 1 \times 10^4 \, \text{W m}^{-2} \]

### Step 2: Substitute the values into the Stefan-Boltzmann equation

Now we can substitute the values into the equation:
\[ 1 \times 10^4 = (5.67 \times 10^{-5}) T^4 \]

### Step 3: Rearranging the equation to solve for \(T^4\)

Rearranging the equation gives:
\[ T^4 = \frac{1 \times 10^4}{5.67 \times 10^{-5}} \]

### Step 4: Calculate the right-hand side

Calculating the right-hand side:
\[ T^4 = \frac{10^4}{5.67 \times 10^{-5}} = 10^4 \times \frac{1}{5.67 \times 10^{-5}} = 10^4 \times 1.764 \times 10^4 \approx 1.764 \times 10^8 \]

### Step 5: Taking the fourth root to find \(T\)

Now, we take the fourth root of both sides to find \(T\):
\[ T = (1.764 \times 10^8)^{1/4} \]

Calculating this gives:
\[ T \approx 648 \, \text{K} \]

### Final Answer

Thus, the temperature at which a perfect black body radiates at the rate of \(1 \, \text{W cm}^{-2}\) is approximately:
\[ T \approx 648 \, \text{K} \]