When crystal of copper sulphate are strongly heated the Residue obtained is what?

The blue copper sulphate crystals found in most laboratories is the pentahydrate: CuSO4.5H2O. This undergoes a number of decomposition steps depending on the temperature:
It loses two water molecules upon heating at 63 °C
CuSO4.5H2O → CuSO4.3H2O + 2H2O
It loses a further two molecules of water on heating to 109 °C :
CuSO4.3H2O → CuSO4.H2O + 2H2O
It loses the final molecule of water on heating to 200 °C
CuSO4.H2O → CuSO4 + H2O
The anhydrous product is white
On further heating to 650 °C copper(II) sulfate decomposes into copper(II) oxide (CuO) and sulfur trioxide (SO3).
CuSO4 → CuO + SO3

To answer your question :
The final residue is copper oxide , CuO.

When crystal of copper(II) sulphate are strongly heated in laboratory, water of crystallization is lost and the residue is anhydrous copper(II) sulphate.
CuSO₄•5H₂O(s) → CuSO₄(s) + 5H₂O(g)


(When strongly heated at a much higher temperature that cannot be provided in laboratory, anhydrous copper(II) sulphate decomposes to copper(II) oxide and water vapour. Then, the residue is copper(II) oxide.)
(CuSO₄(s) → CuO(s) + H₂O(l))

Copper sulfate....

"Copper sulfate" usually is found as the pentahydrate, and gently heating the solid will drive off the water resulting in a while (or nearly white) residue of anhydrous copper(II) sulfate.
CuSO4•5H2O(s) --> CuSO4(s) + 5H2O(g)

But "strongly" heating anhydrous copper(II) sulfate will result in further decomposition.
2CuSO4(s) --> 2CuO(s) + 2SO2(g) + O2(g)

The only solid "residue" is solid copper(II) oxide, a black solid.

There is a colour change from blue to white.
Blue copper sulphate has the formula CuSO4.5HO
There are 5(five) molecules of water of hydration.

When you heat the colour changes to white (the residue) .
This is because you have driven off the water of hydration.
So the residue is 'CuSO4'. (Anhydrous copper sulphate).