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 Stefan-Boltzmann Law relates the radiated power to the temperature of the emitter: \[ P = \sigma A T^4 \] where: - \( P \) is the power radiated, - \( \sigma \) is the Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} \, \text{W/m}^2 \text{K}^4 \)), - \( A \) is the surface area, - \( T \) is the temperature. Rearranging the equation to solve for \( T \): \[ T = \left( \frac{P}{\sigma A} \right)^{1/4} \] Substituting the given values: \[ T = \left( \frac{3.9 \times 10^{26}}{(5.67 \times 10^{-8})(6.1 \times 10^{18})} \right)^{1/4} \approx 5800 \, \text{K} \] Thus, the temperature of the Sun's surface is 5800 K.