If each square cm of the sun's surface radiates energy at the rate of `6.42xx10^(3) Js^(-1) cm^(-2)` , calculate the temperature of the sun's surface in degree celsius, assuming Stefan's law applies to the radiation. (Stefan's constant = `5.67xx10^(-8)Wm^(-2)K^(-4)`)

Explanation

To calculate the temperature of the sun's surface using Stefan's law, we can follow these steps:

### Step 1: Convert the energy radiated per square centimeter to per square meter.
Given that each square centimeter of the sun's surface radiates energy at the rate of \(6.42 \times 10^3 \, \text{J/s/cm}^2\), we need to convert this to per square meter.

\[ \text{Energy per square meter} = 6.42 \times 10^3 \, \text{J/s/cm}^2 \times 10^4 \, \text{cm}^2/\text{m}^2 = 6.42 \times 10^7 \, \text{J/s/m}^2 \]

### Step 2: Use Stefan-Boltzmann Law to find the temperature.
According to the Stefan-Boltzmann Law, the power radiated per unit area of a black body is given by:

\[ P = \sigma T^4 \]

Where:
- \(P\) is the power per unit area (in W/m²),
- \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4\)),
- \(T\) is the absolute temperature in Kelvin.

We can rearrange this formula to solve for \(T\):

\[ T^4 = \frac{P}{\sigma} \]

Substituting the values we have:

\[ T^4 = \frac{6.42 \times 10^7 \, \text{J/s/m}^2}{5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4} \]

### Step 3: Calculate \(T^4\).
Calculating the right-hand side:

\[ T^4 = \frac{6.42 \times 10^7}{5.67 \times 10^{-8}} \approx 1.13 \times 10^{15} \, \text{K}^4 \]

### Step 4: Take the fourth root to find \(T\).
Now, we take the fourth root to find \(T\):

\[ T = \left(1.13 \times 10^{15}\right)^{1/4} \]

Calculating this gives:

\[ T \approx 5800 \, \text{K} \]

### Step 5: Convert Kelvin to Celsius.
To convert from Kelvin to Celsius, we use the formula:

\[ T_C = T_K - 273.15 \]

Thus,

\[ T_C = 5800 - 273.15 \approx 5726.85 \, \text{°C} \]

### Final Answer:
The temperature of the sun's surface is approximately \(5726.85 \, \text{°C}\).

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