Now my question is ,if $Q_H$ is the heat supplied by the $Heat$
$Reservoir$ ,then the heat is used to do some work of say $dW=pdV$ and
if entire heat $Q_H$ gets used as Work, then how does it transfer $Q_c$
amount of heat to the cold reservoir?
I have the impression you think the heat $Q_C$ has to come from the heat $Q_H$. But that's not the case. The heat $Q_C$ comes from energy transferred to the system from the surroundings in the form of compression work done on the system by the surroundings. The compression work causes heat transfer by raising the system temperature infinitesimally greater than than the temperature of the cold reservoir.
Some of the drawbacks of Carnot Engine are 1)The Heat Transfer occurs only during isothermal process(compression and expansion)
Not only is it not a drawback, it is an advantage that all the heat transfers occur during the isothermal processes. The efficiency $\eta$ of the Carnot cycle, or any heat engine cycle for that matter, is the net work done divided by the gross heat added, or
$$\eta=\frac{W_{NET}}{Q_H}=\frac{Q_{H}-Q_{C}}{Q_H}=1-\frac{Q_C}{Q_H}$$
Figures 1 and 2 below are T-S diagrams for the Carnot heat engine cycle and an arbitrary reversible cycle both operating in the same temperature range.
In FIG 1 all the heat received in the Carnot cycle is from a single hot reservoir $T_H$ while all the heat rejected is to a single cold reservoir temperature $T_C$ of the reversible isothermal processes. The area enclosed in the cycle is the net work done.
In the arbitrary cycle of FIG 2, operating in the same range of temperatures as the Carnot cycle, but with multiple reservoirs instead of two, less total heat $Q_H$ is received and more heat $Q_C$ rejected, for a lower thermal efficiency.
...if it's at a different temperature(technically the temperature of
the working material is infinitesimally smaller)than the hot reservoir
,then some amount of heat get's used up to attain thermal equilibrium
(ie) it's irreversibly lost
But there is no finite temperature difference between the system and hot reservoir in the Carnot cycle, the difference being infinitesimal as you parenthetically acknowledged. That's what makes the cycle theoretically reversible instead of irreversible. That's the advantage of the Carnot (and any reversible) cycle.
Even if there were a finite temperature difference making the cycle irreversible, no heat from the hot reservoir would be irreversibly "lost" from the hot reservoir. What would be "lost" is the opportunity to do maximum net work due to the entropy generated by heat transfer over the finite temperature difference. To complete the cycle, that generated entropy would have to be transferred to the cold reservoir in the form of more heat $Q_C$ than for the reversible process. Since the net work equals $Q_{H}-Q_C$ less net work is done.
And it seems so odd to say that once the Isothermal processes are over
,it expands /contracts adiabatically ,are we physically removing the
sources here ?
The sources are not physically removed following the isothermal processes. The system is thermally insulated from the sources following the isothermal processes.
Is the only use of Adiabatic process in "Imaginary Carnot Cycle" to
produce a proper cycle?
They don't just complete the cycle, but complete it in such way to maximize thermal efficiency. Linking them by non isentropic (non constant S) processes lowers the efficiency, as shown for the arbitrary cycle of FIG 2.
Hope this helps.
