How do surface temperatures drop in response to optical depth?

I'm trying to come up with a super simple estimate for how atmospheric surface temperatures would be affected by a substance overhead with some optical depth $$\tau$$. I'll assume that the sun is always directly overhead, and that the parallel-plane approximation is safe (such that all extinctions are computed as integrals across the vertical $$z$$ only), and that the substance is the only thing that attenuates incoming radiation (ignoring the attenuation by the atmosphere).

In this case, the total radiative extinction is given by the Beer-Lambert law:

$$I = I_0e^{-\tau} \ \ \frac{\text{W}}{\text{m}^2}$$

where the incident irrandiance (flux per area) is $$I_0$$. I then define a flux density deficit after attenuation as

$$\Delta I = I - I_0 = I_0(e^{-\tau} - 1)$$

Now, I want to interpret this deficit as a change in temperature of the atmosphere near the surface. I do this by assuming that all of the lost energy $$\Delta I$$ would have gone to heat the surface, and that that heat would have been transferred to the atmosphere by conduction or longwave radiation. Thus, I define a heating rate $$q$$ for a patch of surface with area $$\sigma$$:

$$q = \sigma \Delta I \ \ \frac{\text{J}}{\text{s}}$$

if the mass of the atmosphere just above this patch of the surface (to some arbitrary height) is $$m$$, then the associated temperature tendency is

$$dT/dt = \frac{1}{c_p}\frac{q}{m} \ \ \frac{\text{K}}{\text{s}}$$

Doing this for reasonable values of $$\sigma$$ and $$m$$ give extremely high (negative) cooling rates for $$\tau=0.2$$ and any $$I_0$$ even remotely representative of true solar insolation. Do I have some bad assumptions in here? As stated, I assume that all attenuated energy would have, by one process or another, gone to heat a surface-layer of the atmosphere. Perhaps the process by which the surface exchanges heat to the atmosphere is not nearly that efficient, or quick, or something. Are there some fundamentals that I'm missing?

Of all the energy absorbed by the surface of the Earth, about 90% of it should be transferred to the atmosphere by one way or another. But I'm guessing that the biggest issue is that I assume the heat transfer is instantaneous (going from the third to last line above).