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

The problem given involves finding the temperature of the Sun's surface using the Stefan-Boltzmann law. The Stefan-Boltzmann law states that the power radiated per unit area of a black body in terms of its temperature is given by:

\( P = \sigma \cdot A \cdot T^4 \cdot \varepsilon \)

where:

• \( P \) is the total power radiated,
• \( \sigma \) is the Stefan-Boltzmann constant, approximately \( 5.67 \times 10^{-8} \) W/m\(^2\cdot\)K\(^4\),
• \( A \) is the surface area,
• \( T \) is the temperature in Kelvin,
• \( \varepsilon \) is the emissivity of the surface. For a perfect black body, \( \varepsilon = 1 \).

Given:

• \( P = 3.9 \times 10^{26} \) W
• \( A = 6.1 \times 10^{18} \) m\(^2\) 
• \( \varepsilon = 1 \)

Using the formula:

\( P = \sigma \cdot A \cdot T^4 \cdot \varepsilon \)

Substituting in the known values:

\( 3.9 \times 10^{26} = 5.67 \times 10^{-8} \cdot 6.1 \times 10^{18} \cdot T^4 \cdot 1 \)

Solving for \( T \), we rearrange the equation:

\( T^4 = \frac{3.9 \times 10^{26}}{5.67 \times 10^{-8} \cdot 6.1 \times 10^{18}} \)

\( T = \left(\frac{3.9 \times 10^{26}}{5.67 \times 10^{-8} \cdot 6.1 \times 10^{18}}\right)^{1/4} \)

Calculating the right-hand side:

\( T = \left(\frac{3.9 \times 10^{26}}{3.46 \times 10^{11}}\right)^{1/4} \)

\( T = \left(1.127 \times 10^{15}\right)^{1/4} \)

Taking the fourth root:

\( T \approx 5800 \) K

The correct temperature of the Sun's surface is therefore 5800 K, matching the option: 5800 K.