How can the cooling of Venus near the surface be explained during the time that the specific heat capacity of $\rm CO_2$ quadruples from 60 - 40⁰ C? Imagine that a planet like Venus with an $\rm CO_2$ atmosphere of 92 bar near the surface could be cooled by shading it from the Sun.
The total heat content of the atmosphere with the initial 464⁰ C surface temperature would be the total $\rm CO_2$ mass multiplied by the specific heat capacity of $\rm CO_2$, and that multiplied by the the major part temperature of the $\rm CO_2$ mass in degree Kelvin.
With this peace software tool you can see that with a constant 92 bar pressure near the surface the specific isobar heat capacity of $\rm CO_2$ varies from about 1.1 - 2.5 kJ/kg.K from 464⁰ to 60⁰ C.
Then, by further cooling the planet only from 60 to 40⁰ C, the specific heat capacity near the surface would go up from 2.5 to  11.4 kJ/kg.K  !
So superficially it looks to me that from 60⁰ to 40⁰ C Venus' near surface atmosphere would more than quadruple its heat content !
How will  this strong capacity increase affect the total atmospheric heat content near the surface and thus its temperature?
Will the temperature have to drop to a quarter of its value to compensate for the quadruple heat capacity in order to keep a somewhat less heat content ?    (That content would normally diminish only by the temperature drop from 60⁰ to 40⁰ C by less than 7%)
 A: Your error is in thinking of the “total heat content” as the heat capacity multiplied by the temperature.  Heat capacity is a differential quantity.
That is, you seem to be thinking
$$
Q=mcT
$$
which is an incorrect extrapolation from
$$
\frac{\mathrm dQ}{\mathrm dT }=mc 
$$
or, as commonly presented in introductory textbooks,
$$
\Delta Q=mc\Delta T
$$
This finite-difference approximation is valid over temperature intervals where the heat capacity is approximately constant. If the heat capacity varies, you must cut the interval into smaller pieces so that the heat capacity is approximately constant. The limit of “add up the changes over many small intervals” is the integral,
$$
\Delta Q = \int_{T_1}^{T_2}\mathrm dQ
=  \int_{T_1}^{T_2}mc \,\mathrm dT
$$
If the heat capacity is discontinuous, you probably have a phase transition and have to deal with the latent heat. For total heat content, you must start this integral at absolute zero, which in your problem passes through several phase changes.
The only way that the heat content can decrease as temperature rises is if the heat capacity is negative. That can happen in some exotic systems, like spin systems or thermal systems whose primary interaction is gravitational. You are modeling carbon dioxide as a supercritical fluid above its critical pressure, which does not have these exotic properties.
In a comment, you write

it seems to me that when you want the heat content at a certain temperature there is no temperature range to be integrated, you suppose the atmospheric mass to be at a mean temperature. That has nothing to do with how that mass got heated in the past from T1 to T2

A counterexample, where the heat content is history-dependent, is molecular hydrogen, $\rm H_2$. The rotational states of the hydrogen molecule are split by a symmetry into even-$L$ and odd-$L$ bands, called “parahydrogen” and “orthohydrogen” respectively. Their coupling to each other depends very strongly on the hydrogen’s density, and is nearly zero for hydrogen vapor. At room temperature the two hydrogen species are indistinguishable in terms of density, heat capacity, mean $L$, etc.
However, the heat released in the ortho-to-para conversion is roughly the same as the latent heat of vaporization. So if you liquify room-temperature hydrogen, the increased coupling drives a rapid conversion of nearly all of the ortho to para, and the heat released boils the hydrogen away again.  If you condense recently-liquified (like, hours or days ago) hydrogen from room temperature, before the ortho up-conversion has had time to take place, then the hydrogen will stay liquid. The total heat of hydrogen vapor depends on its history, but its  heat capacity does not.
Most thermodynamics textbooks will clarify that “total heat” is not a well-defined quantity. The “total internal energy” is better-defined, but is not in general an observable. Thermodynamics is a game of differentials.  Temperature itself is a derivative. You could define a “heat capacity” which relates the total internal energy to the temperature, but it would not be a linear function of temperature, and it is not useful enough to be tabulated in reference texts.
A: Mostly to add some numbers to rob's answer.
The constant pressure heat capacity (C$_{p}$) represents how much energy per mole has to be put into (taken out of) the material to change the temperature 1K up (or down). So, to get at how much heat is removed to drop the temperature of a mole of material from T$_{high}$ to T$_{low}$ requires an integral over temperature accounting for the fact that C$_{p}$ is a function of temperature.
Now, for some numbers for CO$_{2}$ from Span and Wagner. Now, of course, they don't tabulate values in Table 35 for 92bar, but I will list some for both 80 and 100bar (8 and 10 MPa). Those table are also in K, not C, but I'll come close enough.
For the 8.00 MPa Isobar:
At 400K, $C_{p}$ is 1.2277 kJ/(kg K)
At 350K, $C_{p}$ is 1.5463 kJ/(kg K)
At 340K, $C_{p}$ is 1.7274 kJ/(kg K)
At 330K, $C_{p}$ is 2.0954 kJ/(kg K) (slightly below 60C)
At 320K, $C_{p}$ is 2.8750 kJ/(kg K)
At 310K, $C_{p}$ is 9.5864 kJ/(kg K) (slightly below 40C)
At 300K, $C_{p}$ is 3.9320 kJ/(kg K)
For the 10.00 MPa Isobar:
At 400K, $C_{p}$ is 1.3327 kJ/(kg K)
At 350K, $C_{p}$ is 1.9480 kJ/(kg K)
At 340K, $C_{p}$ is 2.4017 kJ/(kg K)
At 330K, $C_{p}$ is 3.5312 kJ/(kg K)
At 320K, $C_{p}$ is 7.6175 kJ/(kg K)
At 310K, $C_{p}$ is 4.4460 kJ/(kg K)
At 300K, $C_{p}$ is 2.9906 kJ/(kg K)

Remember, the critical point for CO$_{2}$ is 31.1C (304K) and 7.3773 MPa. The data clearly shows that as you near that point, the heat capacity of CO$_{2}$ can do some pretty weird things (on the 7.5 MPa isobar the $C_{p}$ is 67.567 kJ/(kg K) at 305K). For each degree of temperature drop, you need to pull out far more heat near the critical point than far from it.
