Second law of thermodynamics and heat change existence It's a fact that $$  Q= nC \Delta T$$
Where $Q$ is the heat of the body and $C$ is the molar heat capacity while $T$ is the temperature. 
In an isothermal process we keep the temperature constant which means that internal energy is constant, but still according to the second law there is a net change is heat which is equal to the work done. According to our previous equation, there must be no heat absorbed, then how come the heat change exist according to the second law?
 A: 
According to our previous equation, there must be no heat absorbed

The equation $$Q=nC\Delta T$$ doesn't say that no heat is absorbed. Only that no heat absorbed causes a temperature change. Heat can be absorbed and work be done to cancel it at the same time. 
A: Let me jot down about ideal gas, first.
Thermal capacity at constant volume $\textrm C_V$ is defined as
$$\mathrm C_V ~=~\left(\frac{\partial U}{\partial T}\right)_V$$ where $U$ is the internal energy of the system.
For an ideal gas, $U=U(T);$ so, $$\mathrm C_V~\mathrm dT ~=~ \mathrm dU\,.\tag I$$
Substituting $\mathrm{(I)}$ in the First Law of Thermodynamics, 
$$\mathrm C_V~\mathrm dT +đw~=~  đQ,\tag{II} $$
$đw = p~\mathrm dV;$ this implies $$\mathrm C_V~\mathrm dT +p~\mathrm dV~=~  đQ.\tag{II.a} $$
Therefore, change in entropy $\mathrm dS_\textrm{ideal gas}$ for our system of ideal gas can be written as: $$\mathrm dS_\textrm{ideal gas} =\frac{đQ}T = \frac{\mathrm C_V}{T}~\mathrm dT + \underbrace{\frac{\mathcal R}V}_{pV~ =~\mathcal RT}~\mathrm dV . \tag{III}$$
When, $\mathrm dT = 0$ for isothermal transformation, then $$\mathrm dS_\textrm{ideal gas} = \frac{đQ}T = \frac{\mathcal R }{V}~\mathrm dV\tag{III.a}$$

Suppose, we choose $T, V$ as the independent variable to  define the state of a general system.
So, entropy change $\mathrm dS$ is given by $$\mathrm dS = \frac{đQ}T = \frac1T\left(\frac{\partial U}{\partial T}\right)_V ~\mathrm dT + \frac1T\left[\left(\frac{\partial U}{\partial V}\right)_T + p~\right]~\mathrm dV;\tag{IV}$$ when $\mathrm dT = 0,$ $$\mathrm dS =\frac{đQ}T =\frac1T\left[\left(\frac{\partial U}{\partial V}\right)_T + p~\right]~\mathrm dV.\tag{IV.a} $$
There is nothing contradictory here. $\mathrm dT= 0$ doesn't mean $đQ = 0$ as is evident above in $\mathrm{(II)}$.
A: To keep things simple, they fooled us in freshman physics.  They told us that $Q=nC\Delta T$.  But this equation is not correct when work is being done.  If they didn't want to confuse us, they should have introduced the internal energy U, and correctly defined the heat capacity (at least for ideal gases and incompressible solids and liquids) in terms of U by the equation $\Delta U=nC\Delta T$.  This equation still gives the correct result for Q when no work is being done.  But for cases in which work is being done, we obtain from the 1st law$$\Delta U=nC\Delta T=Q-W$$For an isothermal case, this reduces to Q = W.
