Question

From the measurement made on the Earth, it is known that the Sun has a surface area of \( 6.1 \times 10^{18} \) m\(^2\) and radiates energy at the rate of \( 3.9 \times 10^{26} \) W. Assuming that the emissivity of the Sun's surface is 1, the temperature of the Sun's surface is:

• A. 2600 K
• B. 3600 K
• C. 4500 K
• D. 5800 K

Hint:

Use the Stefan-Boltzmann Law to calculate the temperature of an object based on the power it radiates.

Answer 1

The Correct Option is D

Step 1: Find the power radiated per unit area (the flux).
\[ \frac{P}{A} = \frac{3.9 \times 10^{26}}{6.1 \times 10^{18}} \approx 6.39 \times 10^{7} \, \text{W/m}^2 \]

Step 2: Use the Stefan-Boltzmann law in terms of flux.
For a surface with emissivity 1, the flux equals \( \sigma T^4 \), so:
\[ T^4 = \frac{\text{flux}}{\sigma} = \frac{6.39 \times 10^{7}}{5.67 \times 10^{-8}} \approx 1.127 \times 10^{15} \]

Step 3: Take the fourth root in two stages.
First take the square root:
\[ \sqrt{1.127 \times 10^{15}} \approx 3.357 \times 10^{7} \]
Then take the square root again to get the fourth root overall:
\[ \sqrt{3.357 \times 10^{7}} \approx 5794 \]

Step 4: Final Answer.
\[ T \approx 5800 \, K \]
The Sun's surface temperature works out to about 5800 K.

Answer 2

Another way to check this question is to work with rounded powers of ten first, which makes it easy to see roughly where the answer should land before locking in the exact figure.

1. 2600 K: Rounding \( T^4 \) for 2600 K gives roughly \( 5 \times 10^{13} \), and multiplying by \( \sigma A \approx 3.5 \times 10^{11} \) gives about \( 1.7 \times 10^{25} \) W, more than ten times smaller than the required \( 3.9 \times 10^{26} \) W.
2. 3600 K: Rounding \( T^4 \) gives about \( 1.7 \times 10^{14} \), and multiplying by \( \sigma A \) gives roughly \( 6 \times 10^{25} \) W, still well short of the target power.
3. 4500 K: Rounding \( T^4 \) gives about \( 4.1 \times 10^{14} \), giving a power near \( 1.4 \times 10^{26} \) W, closer but still under the actual radiated power.
4. 5800 K: Rounding \( T^4 \) gives about \( 1.1 \times 10^{15} \), and multiplying by \( \sigma A \approx 3.5 \times 10^{11} \) lands right at \( 3.9 \times 10^{26} \) W, matching the given value.

Working through the powers of ten this way shows the radiated power climbs steeply with temperature, since it depends on the fourth power of \( T \). Only 5800 K brings the estimate in line with the Sun's actual output.

The correct answer is 5800 K.