Respuesta :
Answer:
(a) 1317.44 W/m²
(b) 1.74×10¹⁵ litres of heating oil
(c) -20.63°C
(d) Energy storage in the Earth and the air
Explanation:
The parameters given are;
Luminosity of the Sun = 3.828 × 10²⁶ Watt
Distance of the Earth from the Sun, d = 152.06 × 10⁶ km
The radius of the Sun = 696 × 10³ km
The radius of the Earth, [tex]r_E[/tex] = 6730 km
The surface area of the Sun = 12000 × Surface area of the Earth
The surface area of the Sun = 6.09 × 10¹² km²
Cross sectional area of the Earth = 1.27 × 10⁴ m² = 0.0127 km²
By the inverse square law, we have;
[tex]R = \dfrac{Luminosity \, of \, the \, Sun}{4 \pi d^2}[/tex]
Where:
R = Solar radiation reaching the Earth
Therefore;
[tex]R = \dfrac{3.828 \times 10^{26}}{4 \times \pi \times (1.5206 \times 10^{11})^2} = 1317.44 \ W/m^2[/tex]
Hence, the energy, E, reaching the Earth in a day is given as follows;
E = R × 4×π×[tex]r_E[/tex]²×60×60×24 = 1317.44 × 4 × π × 6730000² × 60×60×24
E = 6.479×10²² Joules
(b) The number of litres of heating oil is therefore;
6.479×10²² J ÷ (37.3×10⁶ J/litre) = 1.74×10¹⁵ litres of heating oil
(c) 30% of the Energy is reflected, therefore;
0.7 × 6.479×10²² Joules = 4.54×10²² Joules reaches the Earths surface
From Stefan-Boltzmann law, we have;
[tex]T = \left (\dfrac{\left (1 - \alpha \right ) \times R }{4\times \sigma } \right )^{\dfrac{1}{4}}[/tex]
Where:
α = 0.3
σ = Stefan-Boltzmann constant = 5.6704×10⁻⁸ W/(m²·K⁴)
Therefore;
[tex]T = \left (\dfrac{\left (1 - 0.3 \right ) \times 1317.44 }{4\times 5.6704 \times 10^{-8} } \right )^{\dfrac{1}{4}} = 252.52 \ K = -20.63 ^{\circ} C[/tex]
(d) The heat storage in the Earth and the air
