Capillary porosity of hydrated cement (28 days)

Assumptions: specific gravity of cement 3.15; 1 cm$^3$ of cement gives 2 cm$^3$ of hydrated products (gel, with gel pores); capillary pores are the space not occupied by hydrated products or unhydrated cement. Take 1 cm$^3$ of cement as the basis. Degree of hydration at 28 days under standard curing is taken as complete ($\alpha = 1$), since no value is given.

$$\begin{aligned} \text{Volume of water} &= 0.5\times3.15 = 1.575\ \text{cm}^3 \\ \text{Total paste volume} &= 1 + 1.575 = 2.575\ \text{cm}^3 \\ \text{Hydrated products} &= 2\times1.0 = 2.00\ \text{cm}^3 \\ \text{Unhydrated cement} &= 1 - 1.0 = 0.00\ \text{cm}^3 \\ \text{Capillary pores} &= 2.575 - 2.00 - 0.00 = 0.575\ \text{cm}^3 \\ \text{Capillary porosity} &= \frac{0.575}{2.575}\times100 = 22.3\% \end{aligned}$$

Answer: capillary porosity = 22.3% of the paste volume (if the hydration at 28 days is less than complete, the porosity is higher; e.g. 32.0% at 75% hydration).