Question

Problem 6. In space, thermal equilibrium is achieved when incoming radiation (e.g. from the Sun) is balanced against outgoing radiation (e.g. from the surface of Earth). The equilibrium achieved is a dynamic one, because there is still a net flow of heat in and out of the system, and the Sun and Earth never reach the same temperature (thankfully) because much of the radiation leaves the system.

(a) The Sun provides a heat to the surface of Earth with an intensity (power per unit area) of about 1000 W/m2 . Compute the total power received. (Hint: The correct area to use is the cross-sectional area of the Earth, because that is the size of the ‘shadow’ of solar radiation that is absorbed.) (In reality, about 1400 W/m2 reaches Earth and about 30% is reflected.)

(b) Suppose the Earth is a perfect black-body absorber and emitter of radiation, and has a uniform surface temperature. (This is not a great assumption.) Find the equilibrium temperature T of the surface in Kelvin and in Celsius, where Earth radiates exactly as much power as it receives from the Sun. Is it anywhere close to Earth’s average surface temperature?

(c) In fact, Earth’s atmosphere is not transparent to the outgoing radiation, which makes the emissivity of Earth imperfect. The result is a delicate balance that preserves a life-friendly temperature. What emissivity e is required to achieve the current 15◦C average surface temperature? What emissivity e would cause the temperature to rise by 2◦C? This is a vastly oversimplified model of Earth’s climate. More accurate models include multiple coupled layers with independent temperatures and emissivities; these models can fairly accurately predict the surface temperature as a function of greenhouse gas emissions (which determine the emissivity of the atmosphere). The net effect of adding carbon dioxide to the atmosphere is to reduce the amount of infrared emission at a given temperature, lowering e and raising the temperature.

Answer 1

Average energy on Earth = 1000 W/m^2

A =Cross sectional area of Earth = , where R = average Radius

A = x [6400 x 10^3 m]^2 = 1.28 x 10^14 m^2

Total energy incident = 1000 W/m^2 x 1.28 x 10^14 m^2 = 1.28 x 10^17 Watts

Part B

J =Average surface energy incident on Earth = 1000 J/s-m^2

J = T^4

= 5.67 x 10^(-8) W/m^2 .K^4

T = 4th root ( 1000 J /sec-m^2) / [ 5.67 x 10^(-8) W/m^2.K^4] = 364.4K

Actual average temperature = 279 Deg K , so order of magnitude is ok

Part C

(i)

Required average temp of Earth surface = 15+273 = 288 K

This will result average net incident energy at J' = 5.67 x 10^(-8) W/m^2.K^4 x (288K)^4 K^4 = 390.1 W/m^2

Emissivity = 390.1 WW/m^2 /1000 W/m^2 = 0.39

I

(ii)

t is proposed to raise temp by 2 DegC , so T = 290 Deg K

J'' = Energy required to radiate = 5.67 x 10^(-8) W/m^2K^4 x (290K) ^4 = 401 W/m^2

Emissivity = 401 W/m^2 / 1000 W/m^2 = 0.40