---
title: "Capillary porosity of hydrated cement (28 days)"
subject: "Concrete Technology and Masonry Structure"
chapter: 4
key: "semester_5/concrete_technology_and_masonry_structure:capillary-porosity-28-days"
source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/capillary-porosity-28-days
site: BCT Engineering
---

# Capillary porosity of hydrated cement (28 days)

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

- Asked in: 2067 Magh (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-28-days
- 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 condition (ASTM standard), calculate the percentage of capillary porosity in the hydrated cement after 28 days. Take W/C = 0.5.

## Short answer

Volume of water = 0.5 × 3.15 = 1.575 cm³ per cm³ cement; total paste = 2.575 cm³. Hydrated products = 2 cm³, so capillary pores = 0.575 cm³, giving porosity of 22.3%.

## 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 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).

## Check yourself

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

1. With w/c 0.5 and 1 cm^3 cement (SG 3.15) the water volume is:
   A. 1.575 cm^3
   B. 0.5 cm^3
   C. 3.15 cm^3
   D. 0.575 cm^3
   Answer: A. 0.5 x 3.15 = 1.575. (Key point: Water volume)

2. Hydrated products volume assuming full hydration is:
   A. 1.0 cm^3
   B. 2.00 cm^3
   C. 2.575 cm^3
   D. 0.575 cm^3
   Answer: B. 2 x 1 = 2.00. (Key point: Hydrated products)

3. Capillary pore volume is:
   A. 2.00 cm^3
   B. 1.575 cm^3
   C. 0.575 cm^3
   D. 0 cm^3
   Answer: C. 2.575 - 2.00 = 0.575. (Key point: Capillary pores)

4. Capillary porosity as % of paste volume is:
   A. 57.5%
   B. 28.7%
   C. 11.5%
   D. 22.3%
   Answer: D. 0.575/2.575 x 100 = 22.3%. (Key point: Porosity result)

## Close variants set in other exams

- [Capillary porosity with 75% hydration](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Acapillary-porosity-75-hydration) — Concrete Technology and Masonry Structure, chapter 4, asked 2×