Radiant energy from the Sun reaches the Earth at a rate of 1.70 × 10¹⁷ W. When 30% of the radiant energy is reflected and 70% absorbed, the average temperature of the Earth's atmosphere is 253 K. If the atmosphere absorbed 100% of the solar radiant energy, then what would be the average temperature of the Earth's atmosphere?

Question

Quizplus · Accepted Answer

When 70% of the radiant energy is absorbed, the Earth's atmosphere absorbs (0.7) × (1.70 × 10^17 W)= 1.19 × 10^17 W. This energy is emitted by the Earth's atmosphere, which acts like a black body, according to the Stefan-Boltzmann law:
P = σA(T^4)
Where P is the power emitted, σ is the Stefan-Boltzmann constant, A is the surface area of the Earth, and T is the absolute temperature (in Kelvins) of the Earth's atmosphere. Solving for T, we get:
T = (P/(σA))^1/4
When the Earth's atmosphere absorbs 100% of the solar radiant energy, the power emitted by it is (1.70 × 10^17 W) - (0.3 × 1.70 × 10^17 W) = 1.19 × 10^17 W (since 30% is reflected). Therefore, we have:
T = ((1.19 × 10^17 W)/(σ × 4πR^2))^1/4
Where R is the radius of the Earth (taken as 6.4 × 10^6 m). Substituting the numerical values, we get:
T = (1.19 × 10^17/(5.67 × 10^-8 × 4π × (6.4 × 10^6)^2))^1/4 = 277 K
Therefore, the average temperature of the Earth's atmosphere would be about 277 K if it absorbed 100% of the solar radiant energy. The correct answer is (D).

Radiant Energy from the Sun Reaches the Earth at a Rate