The Thermodynamic Approach
From an ongoing discussion here, it becomes worthwhile to demonstrate the correct solution to the plane parallel plate hotbox problem. A similar problem is called the “steel greenhouse” and was solved in my book “In the Cold Light of Day“, where with spherical geometry the climate alarmist radiative greenhouse effect mathematically formally disproves itself by ending with a solution where 1 = 2.
Dispensing with graphics, we simply consider a box which has a lid which is transparent to shortwave solar radiation, allowing solar radiant heating to enter the box. At the bottom of the box is a surface with some albedo or reflectivity α. What doesn’t get absorbed is reflected and just goes straight back out through the transparent lid.
The lid however is not transparent to the longwave radiation which the bottom surface will emit as IR thermal radiant energy. The question is: What is the equilibrium temperature of this system, if we consider that the box is insulated from physical energy loss and can only equilibrate via radiation?
All that we need to solve this problem is the First Law of Thermodynamics, i.e. the law of conservation of energy:
dU = Q + W = m Cp dT
This first law states that in order for an object to experience an increase in thermal energy, dU, then you need to supply heat (Q) and/or work (W), and then the resulting change in temperature for the object will be given by the terms ‘m Cp dT’ which are the object’s mass, thermal capacity, and change in temperature.
The equilibrium condition is defined as the state where the temperature is no longer changing, which is an obvious definition, and thus the end-state of the system is where dT = 0, which via the First Law requires that dU = 0, which thus also means that Q = 0 since we do not have any work occurring upon this system.
So it looks like we’ll need the equation for heat Q, which for radiant heat transfer requires another equation called the Stefan-Boltzmann Law. So, that’s three equations we’ll need in total: 1) the First Law of Thermodynamics, 2) the equation for radiant heat transfer for a plane parallel geometry, 3) the Stefan-Boltzmann Law.
The Stefan-Boltzmann Law simply gives the what the radiant thermal emission from a body will be given its temperature, and takes the form of a flux or energy flux density:
F = εσT4
where emissivity ε is the ability of the surface to emit energy, which is unity (1) when it can emit perfectly but otherwise some value between 0 and 1 for non-ideal emission. Here we will use ε = 1 just because it makes the equations simpler to read; it models the ideal scenario but doesn’t fundamentally change the underlying mechanics.
We can use this equation for example for the emission of the lid, given that this lid is opaque to the longwave emitted from the bottom and thus will attain some temperature from this radiation. The question simply becomes: if the interior of the box absorbs some energy, then what does the exterior of the lid have to emit in order to balance this absorption?
We define the solar input as S, and what is absorbed of this input is (1 – α)S because some energy is not absorbed. Thus, this absorbed input needs to equal the emitted output from the lid and its temperature, in order to satisfy conservation of energy:
(1 – α)S = σTL4
So solve for the temperature of the lid TL :
TL = [(1 – α)S / σ]1/4
A nice simple solution for the equilibrium temperature of the lid, given that the lid has to emit outward what is absorbed in the interior of the box.
What about the temperature of the bottom surface of the box, where the shortwave input initially warms the box? The equilibrium condition is again obviously where its temperature is no longer changing, and so we once again refer to the First Law but now applied to this bottom surface: dTB must equal zero, which means that Q must equal zero for the bottom surface of the box. So what is Q here?
Let’s consider the heat flow between the bottom and the lid of the box (QBL), which for plane parallel geometry takes the form:
QBL = σ(TB4 – TL4)
Let us consider the sign of QBL : since it is the bottom of the box which is initially heated, and first attains a temperature, then TL4 will always be less or at most equal to TB4. That is, QBL ≥ 0 since the lid can never emit more than what it receives from the bottom surface; it is greater than zero when temperatures are changing, and becomes equal to zero when temperature stops changing as per the mathematical definition of the First Law. Thus, since QBL ≥ 0, then that means that the ceiling never sends heat to the bottom of the box since it always only receives heat from the bottom of the box. And since heat is what is required to increase temperature as per the First law of Thermodynamics, then we never need to be concerned about the lid of the box raising the temperature of the bottom surface of the box. And now since we know that Q = 0 for equilibrium, then QBL = 0 gives us for the bottom of the box:
TB4 = TL4
TB = TL = [(1 – α)S / σ]1/4
We can confirm this solution using another approach, that of setting the heat flow between the input and the bottom surface (QIB) equal to zero, given that there is no heat flow from the lid to the bottom surface which we need to consider:
QIB = (1 – α)S – σTB4 = 0
which results in the same previous equation for TB. Very nice and consistent, Q = 0 for all objects, and input equals the output. There is no other way to solve this given the definition of the First Law of Thermodynamics.
The Climate Pseudoscience Radiative Greenhouse Approach
In climate science, which is a field of formal pseudoscience practiced at universities, the approach is taken to avoid reference to the equation and definition of heat flow, and all energy is instead treated as being able to increase temperature and cause heating, even if the energy comes from a cooler object. And so where we previously did not treat the emission from the lid as sending any heat to the bottom surface, now we treat all of the energy from the lid as being able to heat the bottom surface alongside the incoming shortwave radiation. While we note that the shortwave incoming radiation is high intensity and high frequency, and the emission from the lid will be longwave low intensity low frequency, in climate science this fact is put aside and all radiation is treated as having the same power to effect change and heat even if the frequencies are completely different.
And so because the climate science approach is to never use a heat flow equation, the approach here instead is to simply add the lid’s emission along with the shortwave input, and make this equal to the emission from the bottom surface:
σTB4 = (1 – α)S + σTL4
The climate pseudoscientists do at least still recognize that the lids output must equal the input to the box, and thus that (1 – α)S = σTL4 thereby giving
σTB4 = 2(1 – α)S
TB = [2(1 – α)S / σ]1/4
So they allow themselves to double the input in other words, and pretend that the lid is an additional source of heat and energy even though the lid only ever receives energy from the surface’s emission in the first place. There are of course other laws of thermodynamics, i.e. the Second Law, which explain that heat flow is not reversible, but nonetheless the climate pseudoscientists reverse the energy from the lid and put it back into the surface…
If climate pseudoscientists (i.e. climate scientists) would actually dare to refer to the First Law of Thermodynamics and the attendant equation for heat flow, as we will do here, then it becomes clear that this approach and its solution is inconsistent with the First Law of Thermodynamics. What did the First Law say about the condition of equilibrium and steady state temperatures?
dU = Q + W = m Cp dT
To have equilibrium and non-changing temperatures, then dT = 0, which requires Q = 0 (W is already 0) so that dU = 0. But if we dare apply the equation for heat flow between the bottom surface and the lid in this solution, we find:
QBL = σ(TB4 – TL4) = [2(1 – α)S – (1 – α)S] = (1 – α)S
Now since the input S is a constant greater than zero, then here QBL > 0, which means via the First Law that dT > 0, and hence, this is NOT an equilibrium solution even though it was intended to be solved as one! And so, the solution contradicts the logic of what it was supposed to solve for in the first place, thus exposing itself as a flawed and incorrect approach.
The approach is incorrect because it avoids the definition and equations for heat flow. And this is really the fundamental basis of the entire approach of climate science (pseudoscience) to studying the climate, given as they also avoid the fact that the Sun heats the Earth…notwithstanding that they also base all this on a flat Earth model!