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The sun’s intensity at the distance of the earth is 1370
Chapter 17, Problem 73P(choose chapter or problem)
Problem 73P
The sun’s intensity at the distance of the earth is 1370 W/m2. 30% of this energy is reflected by water and clouds; 70% is absorbed. What would be the earth’s average temperature (in °C) if the earth had no atmosphere? The emissivity of the surface is very close to 1. (The actual average temperature of the earth, about 15°C, is higher than your calculation because of the greenhouse effect.)
Questions & Answers
QUESTION:
Problem 73P
The sun’s intensity at the distance of the earth is 1370 W/m2. 30% of this energy is reflected by water and clouds; 70% is absorbed. What would be the earth’s average temperature (in °C) if the earth had no atmosphere? The emissivity of the surface is very close to 1. (The actual average temperature of the earth, about 15°C, is higher than your calculation because of the greenhouse effect.)
ANSWER:
Step 1 of 2
We have to find the Earth's average temperature if the Earth had no atmosphere.
At any given time only half the Earth faces the Sun, so the Earth receives \(1370 \mathrm{~W} / \mathrm{m}^{2}\) of energy from the Sun over a cross section of \(\pi R_{e}{ }_{e}{ }^{2}=\pi \times\left(6.37 \times 10^{6}\right)^{2}=1.275 \times 10^{14} \mathrm{~m}^{2}\) \(\left(R_{e}\right.\) is the radius of Earth \(\left.=6.37 \times 10^{6} \mathrm{~m}\right)\)
If 70 % of the incident power is absorbed by the Earth then the total power absorbed by the Earth is
\(\frac{Q}{\Delta t} =(0.70)\left(1370 \mathrm{~W} / \mathrm{m}^{2}\right)\left(1.275 \times 10^{14} \mathrm{~m}^{2}\right)\)
\(=1.222 \times 10^{7} \mathrm{~W}\)
This much power must also be radiated away from the Earth in equilibrium.