Can any solid material with a low heat capacity exist that feels closer to human body temperature than another solid material with a higher heat capacity; where both materials were previously kept in either a mundane oven or freezer for a sustained period?
Let me rephrase to:
Is there any solid which disobeys the inverse proportionality of thermal conductivity and specific heat capacity?
Consider $1000kg$ of wood and $1000kg$ of aluminium, both at $320K$ (very warm). At the instant you place a finger on such large thermal masses, your perception of temperature comparison is dependent on heat conductivity of the materials, not their heat capacity (their masses are so large compared to your finger, their temperature is almost constant depsite losing heat to your finger). Using such large masses and (equal masses for that matter) is necessary since otherwise I can instantly answer yes to your question by giving you 100g of wood and 1g of gold (beaten to the same surface area of the wood) just taken from the freezer and you would perceive gold being closer to body temperature than the wood after a second. So lets define the question by specific heat capacity, and instantaneous perception of heat transfer.
To answer it though, there is in fact no metal which disobeys this relation due to the electron sea being the majority carrier of kinetic energy in the bulk metal. Their having large mean free paths and low masses allow them to attain very high velocities (which is a property of high temperature) and therefore are able to transfer energy quickly in the bulk material. In other words, if metals used anything heavier to transmit heat, like their nuclei, it would not only take much more heat to accelerate them to the same velocities the electrons could attain (resulting in higher heat capacity), but the rate at which that kinetic energy is transmitted across the material is accordingly slower (lower thermal conductivity). In fact the lattice of metal nuclei do in fact contribute to both properties via phonons not translational kinetic energy like in gases, but phonons are still greatly superseded by the effect from electrons. Therefore the inverse relation between thermal conductivity and heat capacity is valid for metals.
What you are looking for is a non conductor with both higher heat capacity and thermal conductivity than a conductor. For that I give you diamond (figuratively...I can't afford one), which has a specific heat capacity of $0.5 J/gK$, higher than that of any metal denser than vanadium (which is almost all of them), but has a thermal conductivity of $>900W/mK$, trumping silver's $421W/mK$ which is tops for all pure metals.
Indeed, $1kg$ of silver would feel much closer to body temperature than $1kg$ of diamond (that's alot of diamond!) despite diamond having a higher heat capacity.
Today I heard a thermodynamic argument for about this. Since there is little work done by the system in solid and liquid phase. The heat capacity must be (roughly) same for the solid and liquid phase.
This one does not convince me at all. Especially, because no work is done by the system in the case one considers the constant volume heat capacity $c_V$. I think the argument you heard was more likely that $c_V \approx c_p$ for liquids and solids (as their volume expansion coefficients are small, and thus $W = p \Delta V$ is small in the constant pressure case), while the values differ relevantly for gases (as they expand relevantly).
So I googled and found that one do expect that heat capacity of liquid be more (i.e. for given heat small temp change) than that of the solids and gases as the majority of contribution to heat capacity, in solids come from 3 vibrational degrees of freedom and 3 translational degrees of freedom while in liquids both are significant hence the energy gets distributed in 3 + 3 = 2 × 3 degrees of freedom. Hence we expect some liquid (at least).
While the I do not see how the math fits here, this argument is much more convincing. At high temperatures (depending on the system, this usually means above some $10\,\text{K}$), each vibrational degree of freedom takes energy $T$, each translational degree of freedom takes energy $T/2$, rotational degrees of freedom do also take energy $T/2$.
In a solid state system we have a number of phonon modes, depending on the number of atoms $n$ in the unit cell (specifically $3n$). Thus the high temperature limit of $c$ (per amount of substance) will be $3n$ (when counted per unit cell, as with ionic solids), in the case of water it will be $3$ (although water ice has more than one molecule per unit cell).
As an atom cannot rotate (without being electronically excited, which takes lots of energy), the number of rotational degrees of freedom depends on the form of the molecule (a two atom molecule having two rotations, a non-linear three atom molecule having three rotations).
This gives for gases: $c = 3/2$ for atomic gases, $c = 5/2$ for handles and $c = 3$ for three atom molecules (like water). (Vibrations of molecules usually have higher energies than our environment, thus the degrees of freedom are "frozen out").
In a liquid it is more complicated (and I am not sure about the correctness of my statements here), but we could probably argue we have three translational and vibrational degree of freedom (the longitudinal phonon) rotations are usually irrelevant in a liquid due to dense packing. Additionally, in liquid water energy can be dispensed by breaking hydrogen bonds (and exactly this effect causes the negative volume expansion coefficent of water near $0^\circ C$).
As you can see this seems to predict that $c_s = c_g$ for water but not that $c_l = 2c_s$. In other words it shows, that the case of the liquid is much more complicated. Overmore it shows that the factors are not random, but also, that the rules are not general for all materials (but depend on the structure).
Especially, you can not get away with simple counting of degrees of freedom (as the energy in the high temperature limit depends on the kind of degree of freedom).
Best Answer
For a large temperature change, specific heat is a function of temperature. Once that function has been characterized, one can obtain the total heat input as an enthalpy change,$$ \Delta H = \int_{T_1}^{T_2} c\,\mathrm{d}T. $$Note that it is common to curve fit specific heat as a function of temperature, such that specific heat is a polynomial function of temperature (which makes it very easy to integrate).