*E*. The Body is covered by three layers of gas. The Lower and Upper Gas are transparent to both short-wave and long-wave radiation. The Filter Gas is transparent to short-wave radiation but black to long-wave radiation. In later posts, we will see how a layer of transparent gas below a layer of black gas can be used to model the gradual absorption of long-wave radiation in the Earth's atmosphere. As before, we assume our system is in thermal equilibrium, which means that all temperatures are constant. No part of the system is heating up or cooling down.

_{S}By radiative symmetry, the Upper Gas radiates no heat. Nor can it transport heat into the vacuum by convection or conduction. The temperature of the Upper Gas is constant, so it cannot be gaining or losing heat. It must be at the same temperature as the upper surface of the Filter Gas, which is

*T*.

_{FU}Thermal equilibrium for the entire system dictates that

*Q*=

_{F}*E*=

_{FU}*E*. In our previous post we assumed

_{S}*E*= 350 W/m

_{S}^{2}for similarity with the Sun and the Earth, and we showed how Stefan's Law dictates that

*T*= 280 K.

_{FU}We also assumed

*Q*=

_{F}*K*(

_{F}*T*−

_{FL}*T*) and we chose

_{FU}*K*= 4.4 W/m

_{F}^{2}K. But this equation for

*Q*says nothing about the depth of the filter. Let us now relate the heat flowing through each gas layer to its depth with a single coefficient,

_{F}*k*, with units of W/mK.

*Q*=

_{F}*k*(

*T*−

_{FL}*T*)/

_{FU}*D*,

_{F}*Q*=

_{LG}*k*(

*T*−

_{B}*T*)/

_{FL}*D*

_{LG}For

*k*, let's choose 350 W/m

^{2}/ 80 K × 10 km = 44 kW/mK. We choose 80 K because that's roughly the drop across the Earth's troposphere, and we choose 10 km because that's roughly the altitude of the Earth's tropopause. With

*D*= 10 km and

_{F}*Q*= 350 W/m

_{F}^{2}we have

*T*−

_{FL}*T*= 80 K, so that

_{FU}*T*= 360 K.

_{FL}Let us consider thermal equilibrium at The Body surface. We must have

*E*+

_{S}*E*=

_{FL}*E*+

_{B}*Q*. We apply Stefan's Law to obtain the following.

_{LG}*E*+ σ(

_{S}*T*)

_{FL}^{4}= σ(

*T*)

_{B}^{4}+

*k*(

*T*−

_{B}*T*)/

_{FL}*D*

_{LG}Where σ = 5.7×10

^{−8}W/m

^{2}K

^{4}and

*k*= 44 kW/mK. Suppose we assume

*D*= 1 km. We have already calculated that

_{LG}*T*= 360 K, so we arrive at the following relation for

_{FL}*T*.

_{B}(

*T*)

_{B}^{4}.(5.7×10

^{−8}W/m

^{2}K

^{4}) +

*T*.(44 W/m

_{B}^{2}K) = 17 kW/m

^{2}

The solution to this equation is

*T*= 366.4 K = 94.5°C. With a vacuum in place of the Lower Gas, we found

_{B}*T*= 390 K = 117°C. Thus the addition of the Lower Gas reduced the temperature of The Body by 24°C. At this temperature,

_{B}*E*= 1.03 kW/m

_{B}^{2}and

*E*= 960 W/m

_{FL}^{2}, so that The Body is radiating 70 W/m

^{2}more heat to the Filter Gas than the Filter Gas is radiating back to The Body. But

*Q*= 280 W/m

_{LG}^{2}. Convection accounts for 80% of the net 350 W/m

^{2}passing from The Body to the Filter Gas. Imbalance in radiative transfer accounts for only 20%.

In future posts, we will see how our three-layer gas model can be used to approximate heat transfer through the Earth's atmosphere, and allow us to estimate the influence of carbon dioxide concentration upon the Earth's temperature.

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