Answer:
a) λ = 5,084 10⁻⁷ m
, b) P = 3.63 10²⁶ W
, c) P = 5.8 10²⁷ W and d) λ = 2.54 10⁻⁷ m
Explanation:
a) The maximum emission of the sun can be calculated using the Win equation
λ T = 2,898 10⁻³ m.K
λ = 2,898 10⁻³ / T
λ = 2,898 10⁻³ / 5700
λ = 5,084 10⁻⁷ m
λ = 5,084 10⁻⁷ m (1 10⁹ nm / 1m) =
λ = 5,084 10² nm = 508.4 nm
photon in the visible range
b) The emission of the Sun, is described by the Stefan equation
P = σ A e T⁴
Where σ is the Stefan-Boltzmann constant that vslue is 5,670 10-8 W/m²K⁴, A area of the Sun, and e the emissivity that for a perfect black body is 1
In order to use this equation, we must calculate the area of the sun, we consider it a perfect sphere
r = 695,000 km (1000m / 1 km) = 6.95 10⁸ m
Area of a sphere
A = 4π R²
A = 4π (6.95 10⁸8)²
A = 6.07 10¹⁸ m²
P = 5,670 10⁻⁸ 6.07 10¹⁸ 1 5700⁴
P = 3.63 10²⁶ W
c) The new temperature is double the previous one
T = 2 To
Let's substitute in the formula and calculate
P = σ A e (2To)⁴
P = σ A e T⁴ 2⁴
Po = σ A e T4 = 3.63 10 26 W
P = 16 Po
P= 16 (3.63 10²⁶)
P = 5.8 10²⁷ W
d) Let's calculate the explicit value of the temperature and use the Win equation
T = 2 5700
T = 11400K
λ = 2,898 10⁻³ / 11400
λ = 2.54 10⁻⁷ m
λ = 2.54 10²nm = 254 nm
photon in the UV range
