---
title: "Capillary porosity with 75% hydration"
subject: "Concrete Technology and Masonry Structure"
chapter: 4
key: "semester_5/concrete_technology_and_masonry_structure:capillary-porosity-75-hydration"
source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/capillary-porosity-75-hydration
site: BCT Engineering
---

# Capillary porosity with 75% hydration

Concrete Technology and Masonry Structure · Chapter 4: Properties of hardened concrete

- Asked in: 2066 Bhadra (old course), 5 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/capillary-porosity-75-hydration
- Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-4

## Question

Assuming that 1 cm$^3$ of cement produces 2 cm$^3$ of hydrated products under the standard curing conditions (ASTM standard), calculate the percentage of capillary porosity in the hydrated cement paste after 28 days. Take w/c ratio as 0.5 and assume 75% hydration in 28 days.

## Short answer

Water volume 1.575 cm³, paste volume 2.575 cm³. With 75% hydration, products 1.5 cm³, unhydrated cement 0.25 cm³; capillary pores 0.825 cm³, porosity 32.0%.

## Answer

**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 $\alpha = 0.75$.

$$
\begin{aligned}
\text{Volume of water} &= 0.5\times3.15 = 1.575\ \text{cm}^3 \\
\text{Total paste volume} &= 1 + 1.575 = 2.575\ \text{cm}^3 \\
\text{Hydrated products} &= 2\times0.75 = 1.50\ \text{cm}^3 \\
\text{Unhydrated cement} &= 1 - 0.75 = 0.25\ \text{cm}^3 \\
\text{Capillary pores} &= 2.575 - 1.50 - 0.25 = 0.825\ \text{cm}^3 \\
\text{Capillary porosity} &= \frac{0.825}{2.575}\times100 = 32.0\%
\end{aligned}
$$

**Answer: capillary porosity = 32.0% of the paste volume.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For 1 cm^3 of cement with w/c = 0.5 and specific gravity 3.15, what is the volume of water?
   A. 0.5 cm^3
   B. 1.575 cm^3
   C. 0.25 cm^3
   D. 2.575 cm^3
   Answer: B. Water volume = w/c x specific gravity x cement volume = 0.5 x 3.15 = 1.575 cm^3. (Key point: Water volume)

2. At 75% hydration, how much hydrated product forms from 1 cm^3 of cement?
   A. 0.75 cm^3
   B. 2.00 cm^3
   C. 1.50 cm^3
   D. 0.25 cm^3
   Answer: C. Hydrated cement occupies 2 cm^3 per cm^3 of cement, so 2 x 0.75 = 1.50 cm^3. (Key point: Hydrated product volume)

3. Capillary pores in this method are taken as:
   A. Total paste volume minus hydrated products minus unhydrated cement
   B. Total paste volume minus the volume of water
   C. Only the gel pores inside the hydrated products
   D. Volume of water minus volume of cement
   Answer: A. Capillary pores are the space not occupied by hydrated products or unhydrated cement. (Key point: Capillary pore definition)

4. Capillary porosity at 75% hydration, w/c 0.5 is about:
   A. 25%
   B. 42%
   C. 8%
   D. 32%
   Answer: D. Capillary pores 0.825 / total paste 2.575 x 100 = 32.0%. (Key point: Final porosity)

## Close variants set in other exams

- [Capillary porosity of hydrated cement (28 days)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Acapillary-porosity-28-days) — Concrete Technology and Masonry Structure, chapter 4, asked 2×

## Related in Concrete Technology and Masonry Structure

- [Porosity at 50%, 75%, 90% hydration](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Aporosity-at-hydration-stages) — Concrete Technology and Masonry Structure, chapter 4