Answer:
emissive power
E ≈ 3.87 x 10^26 W/m^2
Maximum wavelength
λmax ≈ 4.85 x 10^-7 m
total energy emitted by sun
Energy ≈ 2.36 x 10^45 Joules/sec
Step-by-step explanation:
To calculate the emissive power of the surface of the Sun, we will use Stefan-Boltzmann law, which states that the power radiated by a black body is proportional to the fourth power of its temperature.
E = σ * T^4
where E is the emissive power, σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/(m^2K^4)), and T is the temperature in Kelvin.
First, we need to convert the temperature of the Sun from Celsius to Kelvin:
5700°C + 273.15 = 5973.15 K
Now we can calculate the emissive power:
E = 5.67 x 10^-8 * (5973.15)^4
E ≈ 3.87 x 10^26 W/m^2
To calculate the wavelength for the maximum spectral intensity (λmax), we can use Wien's displacement law:
λmax = b / T
where b is Wien's displacement constant (approximately 2.898 x 10^-3 m·K).
λmax = 2.898 x 10^-3 / 5973.15
λmax ≈ 4.85 x 10^-7 m
Finally, to calculate the total energy emitted by the Sun per unit time, we need to consider its surface area. The formula is:
Energy = E * A
where E is the emissive power and A is the surface area.
The surface area of a sphere is given by:
A = 4πr^2
where r is the radius of the sphere.
Given the diameter of the Sun as 1.391 x 10^9 m, we can calculate the radius (r) as half of the diameter:
r = 1.391 x 10^9 / 2 = 6.955 x 10^8 m
Now we can calculate the surface area:
A = 4π(6.955 x 10^8)^2
A ≈ 6.087 x 10^18 m^2
Finally, we can calculate the total energy emitted by the Sun per unit time:
Energy = 3.87 x 10^26 * 6.087 x 10^18
Energy ≈ 2.36 x 10^45 Joules/sec