When gypsum is heated to 393K, it forms:

We begin by recalling the composition of the mineral gypsum. Gypsum is chemically written as $$\text{CaSO}_{4}\,.\,2\text{H}_{2}\text{O}$$, that is, calcium sulphate containing two molecules of water of crystallisation.

Now, when gypsum is gently heated, a part of this water is driven off. The temperature at which such partial dehydration is carried out is about $$393\ \text{K}$$ (approximately $$120^{\circ}\text{C}$$). This temperature is deliberately kept below the point at which the entire water is lost, because our goal is to remove only a specific fraction of the water.

The chemical change can be written step by step. Starting with the full formula, we have

$$\text{CaSO}_{4}\,.\,2\text{H}_{2}\text{O} \xrightarrow[393\ \text{K}]{} \text{CaSO}_{4}\,.\,0.5\text{H}_{2}\text{O} + 1.5\text{H}_{2}\text{O}\uparrow$$

In words, one and a half molecules of water escape as vapour, and only half a molecule of water per formula unit remains within the crystal lattice. The resulting product $$\text{CaSO}_{4}\,.\,0.5\text{H}_{2}\text{O}$$ is known in everyday language as Plaster of Paris.

Thus, heating gypsum to $$393\ \text{K}$$ does not give anhydrous $$\text{CaSO}_{4}$$ (that would require complete dehydration at a higher temperature), nor does it leave the original dihydrate or form “dead-burnt” plaster. Instead, it produces the hemihydrate $$\text{CaSO}_{4}\,.\,0.5\text{H}_{2}\text{O}$$.

Hence, the correct answer is Option C.