Considering the sun as a perfect sphere of radius `6.8xx10^(8)` m, calculate the energy radiated by it one minute. Take the temperature of sun as 5800 k and `sigma = 5.7xx10^(-8)` S.I. unit.

Explanation

To calculate the energy radiated by the Sun in one minute, we can follow these steps:

### Step 1: Identify the given values
- Radius of the Sun, \( r = 6.8 \times 10^8 \) m
- Temperature of the Sun, \( T = 5800 \) K
- Stefan-Boltzmann constant, \( \sigma = 5.7 \times 10^{-8} \) W/m²K⁴

### Step 2: Calculate the surface area of the Sun
The surface area \( A \) of a sphere is given by the formula:
\[ A = 4 \pi r^2 \]
Substituting the value of \( r \):
\[ A = 4 \pi (6.8 \times 10^8)^2 \]

### Step 3: Calculate the power radiated by the Sun
The power \( P \) radiated by a black body is given by the Stefan-Boltzmann law:
\[ P = \sigma A T^4 \]
Substituting the values of \( A \) and \( T \):
\[ P = 5.7 \times 10^{-8} \times A \times (5800)^4 \]

### Step 4: Calculate the energy radiated in one minute
Energy radiated in one minute \( Q \) can be calculated by multiplying the power by the time in seconds (1 minute = 60 seconds):
\[ Q = P \times 60 \]

### Step 5: Substitute and calculate
1. Calculate \( A \):
\[ A = 4 \pi (6.8 \times 10^8)^2 = 4 \times 3.14 \times (6.8^2 \times 10^{16}) \approx 4 \times 3.14 \times 46.24 \times 10^{16} \approx 3.95 \times 10^{17} \text{ m}^2 \]

2. Calculate \( P \):
\[ P = 5.7 \times 10^{-8} \times 3.95 \times 10^{17} \times (5800)^4 \]
First calculate \( (5800)^4 \):
\[ (5800)^4 = 1.086 \times 10^{15} \]
Now substitute:
\[ P \approx 5.7 \times 10^{-8} \times 3.95 \times 10^{17} \times 1.086 \times 10^{15} \approx 2.36 \times 10^{25} \text{ W} \]

3. Calculate \( Q \):
\[ Q = P \times 60 \approx 2.36 \times 10^{25} \times 60 \approx 1.416 \times 10^{27} \text{ J} \]

### Final Answer
The energy radiated by the Sun in one minute is approximately \( 1.416 \times 10^{27} \) Joules.

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