Why is the change of heat non zero in a isothermal process? I was reading the definition of heat capacity and it says that
$$\Delta Q = C \ \Delta T$$ 
(introduction to statistical physics - K. Huang)
So my question is that, if we consider an isothermal process, becuase temperature remains the same, $\Delta Q$ would be zero. And $\Delta U= \Delta W$ by the first law, but thats wrong too because of what I have read on the internet. So, what am I doing wrong? Why do you have in a isothermal process $W=Q$ and not $U=W$?
 A: Suppose you take an ideal gas as your system. 
Then according to the equipartition theorem its total internal energy would be $\frac{1}{2}k_B T$ per degree of freedom per molecule. So if $f$ is the number of degrees of freedom then the total internal energy of your $N$ molecule ideal gas system would be $$\frac{f}{2}NK_BT$$ or $$\frac{f}{2}nRT$$
So as you can see the total internal energy only depends on the temperature. And moreover if $\Delta T$ is the change in temperature in a process then the corresponding change in total internal energy would be $$\Delta U = \frac{f}{2}nR\Delta T$$
As in isothermal process temperature remains constant, $\Delta T= 0 $.
So $$\Delta U =0$$
And the first of law of thermodynamics becomes 
$$W= -Q$$
(PS: Also, I want to point out a flaw in your question: 
Even when $T$ remains same, $\Delta Q$ need not be zero. Think of what happens in a phase change situation i.e latent heat.) 
A: In freshman physics, they did us a disservice by incorrectly teaching  us that heat capacity is defined by $Q=C\Delta T$ (or $Q=mC\Delta T$, where C is the heat capacity per unit mass) (or $Q=nC\Delta T$, where C is the heat capacity per mole).  This definition works fine as long as no work is done.  However, when work is done, this equation gives the wrong answer.  Moreover, in thermodynamics, we learn that Q represents a quantity that depends on path, while C is a physical property of the material that is independent of path.  So, in thermodynamics, they corrected their error by redefining heat capacity properly.  $$nC_v=\left(\frac{\partial U}{\partial T}\right)_V$$
For a process at constant volume, this remains consistent with the definition from freshman physics, and, moreover is a physical property of state (independent of path).  But for processes in which work is done, it gives the correct answer for all cases.
There is also another heat capacity property that is used in thermodynamics called the heat capacity at constant pressure $C_p$.  This is defined as $$nC_p=\left(\frac{\partial H}{\partial T}\right)_P$$where H is the enthalpy.  In this case, the relationship is consistent with the freshman definition specifically for situations in which work is done in a constant pressure process.  However, here too, the relationship is much more general than this.  
The real question is, "why didn't they teach this to us properly in the first case?"  One can only speculate on the answer, but it has been the source of confusion to thermodynamics students for centuries.
A: You are assuming the variable C to be constant. Actually ∆Q=∆(CT). Now as the thermodynamic system goes under a certain process its initial and terminal conditions must differ. And hence pressure and volume can change. Thus C is not a constant throughout the process and you can't claim that heat interaction with the surrounding is zero.
Moreover we define C=(∆Q/∆T). Hence C is undefined if ∆T is zero. You must have to specify the unchanged thermodynamic variable otherwise it doesn't make sense.
