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.

Answer 2

This question uses the Stefan-Boltzmann law, which links how much power a hot surface radiates to its temperature. Instead of solving the equation from scratch, each option can be tested by plugging it back in and seeing which one reproduces the given power.

1. 2600 K: Raising 2600 to the fourth power gives roughly \( 4.6 \times 10^{13} \). Multiplying by \( \sigma A = 5.67 \times 10^{-8} \times 6.1 \times 10^{18} \approx 3.46 \times 10^{11} \) gives a power far smaller than the \( 3.9 \times 10^{26} \) W stated in the question, so this temperature is too low.
2. 3600 K: Its fourth power is around \( 1.68 \times 10^{14} \). Multiplying by the same \( \sigma A \) still lands well below \( 3.9 \times 10^{26} \) W, so this is still too low.
3. 4500 K: Its fourth power is close to \( 4.1 \times 10^{14} \), giving a radiated power still an order of magnitude short of the required value, so this temperature also falls short.
4. 5800 K: Its fourth power is close to \( 1.13 \times 10^{15} \). Multiplying by \( \sigma A \approx 3.46 \times 10^{11} \) gives about \( 3.9 \times 10^{26} \) W, matching the power stated in the question almost exactly.

Only 5800 K reproduces the Sun's actual radiated power when plugged into \( P = \sigma A T^4 \), given the surface area and emissivity stated in the question.

So the correct answer is 5800 K.