When $Q=mc\Delta T$? I've read the first answer in this question, by Mark Eichenlaub, where he is saying something like "heat is every change in energy of a system that cannot be attributed to work". My question is, then, if for some process where no heat is transferred (so all the energy change, including thermal energy, is produced by work), how can we express heat to be:
$$Q=mc\Delta T \ ?$$
Is it that this expression can only be used in a few cases, or is it that I have a misconception?
 A: No.  You are using the wrong definition of heat capacity.  In thermodynamics, we don't regard heat capacity as being directly related to the amount of heat Q.  It is directly related to the variations in internal energy U and enthalpy H with temperature:  $$\left(\frac{\partial U}{\partial T}\right)_V=mC_v$$and$$\left(\frac{\partial H}{\partial T}\right)_P=mC_P$$
A: 
Is it that [$Q=mc\Delta T$] can only be used in few cases, or is it that I have any misconception?

It can be used given the following condition: We are doing nothing but heating ($Q>0$) or cooling ($Q<0$) the system at equilibrium at the condition (e.g., constant volume, constant pressure) corresponding to the particular heat capacity $c$ being used.
Here's why. We can always write $dU=\delta Q+\delta W$ for a closed system; that is, we can increase the system energy infinitesimally either by heating the system or by performing work on it. (The specification of a closed system is part of the condition specified above, that we can do nothing but heat or cool the system; otherwise, we could change the system temperature even with $Q=0$ by jamming in more material.)
In addition, we can expand $dU$ in terms of the relevant state variables. For example, if we consider pressure–volume work only ($dU=\delta Q-P\,dV$), it's an introductory thermodynamics exercise to show that
$$dU=mc_V\,dT+(\alpha TK-P)\,dV,$$
where $n$ is the mass, $c_V$ is the constant-volume specific heat capacity, $\alpha$ is the constant-pressure thermal expansion coefficient, $T$ is temperature, $K$ is the constant-temperature bulk modulus, $P$ is pressure, and $V$ is volume.
(A related expression is
$$dH=mc_P\,dT+V(1-\alpha T)\,dP,$$
where $H$ is the enthalpy and $c_P$ is the constant-pressure specific heat capacity.)
Hopefully it's clear now from comparing the expressions $dU=\delta Q-P\,dV$ and $dU=mc_V\,dT+(\alpha TK-P)\,dV,$ that $Q=mc_V\Delta T$ can hold in general only if $dV$ is zero, i.e., if no work—only heating—is being performed, on a closed system.
(An interesting special case is the ideal gas, for which $\alpha TK-P$ and $1-\alpha T$ are always zero; thus, $\Delta U=mc_V\Delta T$ and $\Delta H=mc_P\Delta T$ always hold for an ideal gas.)
Relatedly, $Q=mc_P\Delta T$ can be used only if $dP$ is zero, and so on. For a much more complex system, we might define a heat capacity for heating at constant volume, constant magnetic field, and constant surface tension, for instance. That's the origin of the condition in bold above.
