Conceptual question in thermodynamics about isothermal processes If a process is isothermal then Δ U is zero. So ΔQ is non zero. But isn't ΔQ=nCΔT. Implying that ΔQ is zero. Where am I wrong? (Considering ideal nature)
 A: The thing is that the relation $\Delta Q = nC\Delta T$ is true for some particular processes -- in an ideal gas, for instance, it is true when the system passes through a process at constant pressure (in which case $C = C_p$) or in constant volume (in which case $C = C_V$), but this is not true in general. In an adiabatic process, for example, you have $\Delta Q = 0$, but $\Delta T \neq0 $ -- and that does not contradict the relation above because neither pressure nor volume are constant in an adiabatic process in an ideal gas. The same holds for an isothermal process, which is the case you are interested in.
A: In freshman physics, we learned the $Q = nC\Delta T$, but in thermodynamics, we change the definition a little to reflect our understanding that, while Q depends on process path, C is supposed to be a physical property of the material experiencing the temperature change, independent of process path.  So we redefine C, not in terms of the heat exchanged, but rather in terms of the physical property changes in internal energy U and enthalpy H:
$$C_V=\left(\frac{\partial U}{\partial T}\right)_V\tag{1}$$
$$C_P=\left(\frac{\partial H}{\partial T}\right)_P\tag{2}$$
Eqn. 1 coincides with the old definition in cases where no work is done so that $\Delta V=0$.  Eqn. 2 coincides with the old definition in cases where the pressure is constant.  Otherwise, they don't; however, they are much more general and widely applicable than the old definition.
A: Your formula is only true when the heat is supplied at constant volume. In an isothermal process you can also supply heat  at constant pressure, in which a system does some work thereby using all the heat provided without raising the temperature at all 
