The Earth receives a certain amount of radiation from the Sun, and also generates a certain amount of heat from radioactivity in the core. It must radiate this amount of energy because otherwise it would heat up until the thermal emission matched the rate of energy input. It's this heat flux that you'd use when calculating how much your plate was heated by the Earth. The sums seem simple enough and I'm sure Google would be able to retrieve the values of heat received from the Sun and generated internally.
Given the heat flux from the Earth you can calculate a temperature using the Stefan Boltzmann law. This temperature wouldn't match the temperature of the Earth's surface, but then why should it? As you say, between the surface and space is a layer of insulating gas. The Stefan Boltzmann temperature would be some kind of average for the various parts of the Earth from which radiation is received. Increasing CO$_2$ or other greenhouse gases would change the temperature profiles between the surface and space, but not the overall amount of energy being emitted.
You can define an emissivity for the Earth, but I'm not sure how helpful this is. If you compare the temperature at the surface with the Stefan Boltzmann temperature you'd get an emissivity less than one, but then this isn't an especially useful comparison.
Wrong:
"since infrared waves have a shorter wavelength"
Infrared has longer wavelength than visible and visible longer wavelength than ultraviolet .
White is a term for visible light mixed wavelengths. In the plot you can see that almost half of the sun's radiated energy arrives as visible light. The white buildings reflect this visible light which otherwise,impinging on the surfaces would be absorbed and turned into infrared by the interactions, adding to the arriving infrared.
What is absorbed and what is reflected depends on the chemical bonds of the surfaces, whether the incoming radiation can excite molecular states of the materials. Infrared is in frequencies/wavelengths of the black body radiation of bodies in the temperature ranges comfortable for the human body, so they easily raise the vibrational and rotational levels of solids and liquids and the kinetic energy of gasses.
37C curve seen here practically all in infrared, and lower temperatures more so.
Thus white paint will not reflect infrared as efficiently as visible, a large part of infrared will be absorbed as also some part of visible will scatter at the surface and degrade to infrared. Infrared can be reflected by metal mirrors, from the collective fields in metals . If you put aluminum foil in front of a heater you are sheltered from most of the heat which is reflected, but some of it is absorbed as can be seen by touching the foil.
If I expose an object to EM radiation only from the infrared spectrum, will it only reflect back infrared?
Yes, but most of it will be absorbed ( except by mirror metal surfaces) because the materials have the receptors for these wavelengths. This is due to the fact that larger wavelengths have photons with less energy which cannot excite higher energy levels.The energy of the photons goes as h*c/lamda where h is plancks constant, lamda the wavelength and c the velocity of light.
Is this true for other types of EM radiation?
No.Visible and ultraviolet by scatterings degrade their energy down to infrared frequencies, depending on the material.
Is it possible to make an object that looks white and absorbs a lot of infrared radiation?
Usually most of the infrared will be absorbed except by mirror metal surfaces.
If an object reflects most of the EM radiation that it receives of a particular wavelength λ, will it also reflect most of the radiation it receives of wavelengths less than λ (and absorb most of the radiation of wavelengths larger than λ)? Is this why objects that reflect most visible light (and hence look white) also reflect most infrared radiation (since infrared waves have a shorter wavelength)?
There is no such rule. It depends on the material and its chemical bonds.
Best Answer
This kind of pictures is often misleading. When a paradox appears, common sense tells us that there is very likely something wrong with the assumptions. Here the bad assumption is to display the whole atmosphere as a single box what leads to an apparent paradox.
In reality the density varies with altitude and this has for consequence that the mean free path of photons varies with altitude too. Variations of all radiative parameters with altitude follow.
To see the difference with a picture where the atmosphere is represented by N layers, let us take an example for N=2. We suppose that a solid surface S is radiating F (W/m²) into an atmosphere constituted of 2 layers L1 and L2 and the system is in a steady state. We suppose farther that L1 absorbs all the radiation coming from S.
L1 must reemit all the absorbed radiation (steady state) and because of isotropy, F/2 is reemitted up and F/2 is reemitted down. A part P of the upgoing F/2 will be absorbed by the upper layer L2 and of course again reemitted P/2 up and out of the system and P/2 down. As this downgoing P/2 will be absorbed by L1 and again reemitted P/4 down and P/4 up, it is proven that the flux going from the atmosphere to the surface S is F/2 + P/4 + ... > F/2. so it is actually no paradox that the downgoing flux is more than a half of the flux emitted up by the surface. The values depend on the detailed radiation properties of the layers.
Of course N=2 is still not realistic enough and I have omitted the algebra that can be done by any interested reader but the purpose was to show that an N layer atmosphere doesn't behave like a 1 layer atmosphere. Among others I have omitted the source of energy that is necessary to make up for the difference between F that S emits and F/2 + ... etc that S absorbs from the atmosphere.
Regarding the reflection, there is none in this particular case. The infrared radiation emitted by the surface between 0°C and 30°C is mostly absorbed by the vibrational spectrum of H20 molecules (partly by CO2 too). So this radiation is indeed absorbed and exactly reemitted with the exception of a small window that goes directly to space. Obviously there is always some scattering (clouds) but the processes in the dense and warm lower atmosphere are dominated by absorption and emission in the IR spectrum.