Volume of Irregular Solids by Displacement

Introduction

A pebble, a key or a lump of rock has no simple shape, so its volume cannot be calculated with a ruler. Instead, chemists use a beautifully simple idea: when an object sinks completely into water, it pushes aside — displaces — a volume of water exactly equal to its own volume. Measuring the displaced water measures the object.

Core explanation

The principle. Two things cannot occupy the same space at once. When a solid is fully submerged, water must move out of the way to make room for it. The volume of water moved equals the volume of the solid.

Method 1 — measuring cylinder. For objects small enough to fit inside a measuring cylinder:

1. Half-fill the cylinder with water and record the volume (V₁). 2. Tilt the cylinder and slide the object in gently so it does not splash or crack the glass. 3. Make sure it is completely submerged and record the new volume (V₂). 4. Volume of object = V₂ − V₁.

Method 2 — eureka (displacement) can. For larger objects, fill the can with water until it overflows from the spout, and wait until dripping stops. Place a measuring cylinder under the spout, then lower the object into the can. The water that flows out is collected and its volume read directly — this is the object's volume.

Conditions for the method to work. - The object must sink and be fully submerged. For objects that float, such as cork, push them under with a thin wire, or use a "sinker" and subtract the sinker's own volume. - The object must not dissolve in or absorb water. For sugar or salt crystals, use a liquid they do not dissolve in, such as a light oil. - Air bubbles clinging to the object add to the reading, so tap the cylinder gently to release them.

Why the displaced water, not the object's weight? The rise in water level depends only on the space the object takes up, not on its mass. A 10 cm³ stone and a 10 cm³ lump of gold displace the same 10 cm³ of water, even though the gold is far heavier.

Step-by-step reasoning

A stone is lowered into a measuring cylinder:

1. Initial reading V₁ = 42 cm³. 2. Final reading V₂ = 55 cm³. 3. Volume of stone = V₂ − V₁ = 55 − 42 = 13 cm³. 4. Check: the stone was fully under water and no water splashed out.

Visual explanation

Picture two measuring cylinders side by side. The left one shows water at 42 cm³. The right one shows the same water with a stone at the bottom, and the level at 55 cm³. A bracket labelled "13 cm³ = volume of stone" spans the rise. A second diagram shows a eureka can with water streaming from its spout into a measuring cylinder as a rock is lowered in.

Real-world analogy

When you sit down in a full bath, water sloshes over the side. The volume that spills out equals the volume of the part of your body that went under the water. A eureka can is simply a tidy way of catching and measuring that overflow.

Real-world example

Geologists and gemmologists use displacement to find the volume of rough minerals and gemstones, which helps them calculate density and identify the material. A jeweller can check whether a "gold" ring is real by combining its mass with its displacement volume, without damaging it.

Why?

Why must the eureka can be filled until it overflows before the object is added? If the water is below the spout, the first part of the displaced water only raises the level to the spout and is not collected, so the measured volume would be too small. Starting at the overflow level means every bit of displaced water leaves through the spout.

Common misconception

Some students think a heavier object will always make the water rise more. The rise depends only on volume. A small, heavy steel ball bearing may raise the water less than a large, light plastic block that has been pushed under.

Worked example

Question: A eureka can is filled to the spout. When a key is lowered in, 4.5 cm³ of water is collected. A second key of the same metal collects 9.0 cm³. Compare the keys.

Reasoning: The collected volume equals each key's volume. The second key displaces twice as much water, so it has twice the volume.

Answer: Key 1 has a volume of 4.5 cm³ and key 2 has 9.0 cm³; key 2 is twice as large (and, being the same metal, about twice the mass).

Quick check

1. Water in a cylinder rises from 30 cm³ to 38 cm³ when a pebble is added. What is the pebble's volume? Answer: 38 − 30 = 8 cm³.

Exam focus

Describe the method in numbered steps, including recording the initial and final volumes and subtracting. Mention that the object must be fully submerged and must not dissolve or absorb water. Evaluation questions often ask about sources of error: splashing, air bubbles and drops left on the spout of a eureka can.

Advanced insight

The same principle underlies Archimedes' principle: a submerged object experiences an upward force (upthrust) equal to the weight of the fluid it displaces. That is why objects seem lighter in water and why ships float — ideas you will connect with density on page 45.

Summary

The volume of an irregular solid can be measured by displacement. In a measuring cylinder, volume = final reading − initial reading; with a eureka can, the collected overflow equals the volume. The object must sink fully and must not dissolve or absorb the liquid, and splashes and bubbles must be avoided.

Practice questions

1. Describe how to find the volume of a small, irregular rock using a measuring cylinder. Answer: Record the water level, gently slide the rock in until fully submerged, record the new level and subtract the first reading from the second. 2. Why can water displacement not be used to find the volume of a sugar lump? Answer: Sugar dissolves in water; a liquid it does not dissolve in, such as oil, must be used instead. 3. A cork floats. How could its volume be measured by displacement? Answer: Push it fully under the water with a thin wire (or attach a sinker of known volume and subtract that volume). 4. The water level rises from 60.0 cm³ to 72.5 cm³ when an object is added. Find its volume. Answer: 72.5 − 60.0 = 12.5 cm³.