Let's start by computing heat capacities in a context that is a bit more general than polytropic processes; those defined by the constancy of some state variable $X$.
For concreteness, let's assume we are considering a thermodynamic system, like an ideal gas, whose state can be characterized by its temperature, pressure, and volume $(T,P,V)$ and for which the first law of thermodynamics reads
\begin{align}
dE = \delta Q - PdV
\end{align}
We further assume that there exists some equation of state which relates $T$, $V$, and $P$ so that the state of the system can in fact be specified by any two of these variables. Suppose that we want to determine the heat capacity of the system for a quasistatic process (curve in thermodynamic state space) for which some quantity $X=X(P,V)$ is kept constant. The trick is to first note that every state variable can be written (at least locally in sufficiently non-pathological cases), as a function of $T$ and $X$. Then the first law can be written as follows:
\begin{align}
\delta Q
&= dE + PdV \\
&= \left[\left(\frac{\partial E}{\partial T}\right)_{X}+P\left(\frac{\partial V}{\partial T}\right)_{X}\right]dT + \left[\left(\frac{\partial E}{\partial X}\right)_{T}+P\left(\frac{\partial V}{\partial X}\right)_{T}\right]dX
\end{align}
Now, we see that if we keep the quantity $X$ constant along the path, then $\delta Q$ is proportional to $dT$, and the proportionality function is (by definition) the heat capacity for a process at constant $X$;
\begin{align}
C_X = \left(\frac{\partial E}{\partial T}\right)_{X}+P\left(\frac{\partial V}{\partial T}\right)_{X}
\end{align}
Now, if you want an explicit expression for this heat capacity, then you simply need to determine the energy, pressure, and volume functions of $T$ and $X$ and then take the appropriate derivatives.
Consider, for example, a polytropic process like you originally described, and further, consider a monatomic ideal gas for which the energy and equation of state can be written as follows:
\begin{align}
E = \frac{3}{2} NkT, \qquad PV = NkT
\end{align}
For this process, we have
\begin{align}
X = PV^n
\end{align}
Using the equation of state and the definition of $X$, we obtain
\begin{align}
V = (NkT)^{1/(1-n)}X^{1/(n-1)}, \qquad P = (NkT)^{n/(n-1)}X^{1/(1-n)}
\end{align}
and now you can take the required derivatives in to obtain $C_X$ where $X$ is appropriate for an arbitrary polytropic process.
Moral of the story. If the system you care about can be written as a function of only two state variables, write all quantities in terms of $T$ and $X$, the variable you want to keep constant. Then, the first law takes the form $\delta Q = \mathrm{stuff}\,dT + \mathrm{stuff}\,dX$ and the $\mathrm{stuff}$ in front of $dT$ is, by definition, the desired heat capacity.
Work is force acting over a distance. That means there can only be work done on the gas if the piston moves, i.e. if the gas changes volume. In that case, $W = \int F \text{d} x = \int P\, \text{d}V$, where P is the pressure difference between the inside and outside of the calorimeter. Note, however, that in your example the gas is doing work on the piston, not the other way around, so the work done on the gas is negative, and the average KE per particle would decrease as a result of that part. Any transfer of energy into or out of the system not associated with a change of volume may be referred to as heat (positive if transferred in, negative if transferred out).
Your question suggests you may be confusing radiation with heat. In a good calorimeter on the time scale of the experiment there is negligible loss from the system due to radiation. In a system that is not so well isolated the radiation usually represents heat loss (i.e. negative heat in), but can also represent negative work if, for example, the radiation is strong enough to push out the piston. In general, though, heat may be lost from the system via conduction through the walls, or by other means if the system is not well-sealed (most importantly convection and evaporation). Depending how you do your accounting and what kind of reaction is going on, heat can also represent transfer of thermal energy out of or into your system as chemical energy (into = exothermic reaction, out of = endothermic).
Best Answer
In general it is the same idea as with ideal gases. This here is not what is formal answer, because specific heat is generally defined with entalpy and internal energy. This is rather the explanation, why there is a difference.
In order to change volume $V$ when the pressure is constant, some work $A$ has to be provided. In differential case (very small change): $dA=pdV$.
From conservation of energy we can than determine that: $$ dQ=dW+dA $$ $Q$ is internal energy of a system, and dW is energy added, and A is work done by the system.
So we can denote specific heat as $dQ/dT$:
$$ c_p=\frac{dQ}{dT}=\frac{dW+pdV}{dT} $$ and $$ c_V=\frac{dQ}{dT}=\frac{dW+pdV}{dT}=\frac{dW+p\cdot0}{dT}=\frac{dW}{dT} $$
You can see from here, that $c_p$ is greater than $c_v$. The relation between this two depends on equation of state and it can be quite ugly for liquids. But in general when we have isobaric conditions, some added energy is converted into work needed, to change the volume of system.
If you are not familiar with differentials $d$, they are just very small changes $\Delta$.