8.1 Changes of state and the particle model — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check.

Warm-up: remember

Answer these from memory. They use ideas from Chapter 1 and Section 2.4.

Change $250\,\text{cm}^3$ into cubic metres.
A worker lifts a box of mass $20\,\text{kg}$ through a height of $1.5\,\text{m}$ . Take $g = 9.8\,\text{N/kg}$ . Calculate the energy gained by the gravitational potential store.

Answers

$1\,\text{cm}^3 = 10^{-6}\,\text{m}^3$ , so $250\,\text{cm}^3 = 2.5 \times 10^{-4}\,\text{m}^3$ .
$E = mgh = 20 \times 9.8 \times 1.5 = 294\,\text{J}$ .

If question 1 went wrong, remember that a cubic metre holds $10^6$ cubic centimetres. If question 2 went wrong, look again at Chapter 1: multiply the mass, the field strength and the height.

Practise with help

1. A block of metal has a mass of $0.60\,\text{kg}$ and a volume of $0.75 \times 10^{-3}\,\text{m}^3$ . Calculate its density.

Hint

You are given a mass and a volume, and you want the density. The equation $\rho = \dfrac{m}{V}$ links all three.

Next step

Divide the mass $0.60\,\text{kg}$ by the volume $0.75 \times 10^{-3}\,\text{m}^3$ .

Full working

\begin{aligned} \rho &= \frac{m}{V} \\ &= \frac{0.60}{0.75 \times 10^{-3}} \\ &= 800\,\text{kg/m}^3 \end{aligned}

2. A block of concrete has a density of $2400\,\text{kg/m}^3$ and a volume of $0.25\,\text{m}^3$ . Calculate its mass.

Hint

This time you want the mass. Rearrange $\rho = \dfrac{m}{V}$ first.

Next step

Write $m = \rho V$ , then multiply $2400\,\text{kg/m}^3$ by $0.25\,\text{m}^3$ .

Full working

\begin{aligned} m &= \rho V \\ &= 2400 \times 0.25 \\ &= 600\,\text{kg} \end{aligned}

3. A bottle holds $1.2\,\text{kg}$ of water. The density of water is $1000\,\text{kg/m}^3$ . Calculate the volume of the water.

Hint

You know the mass and the density, and you want the volume. Make $V$ the subject of $\rho = \dfrac{m}{V}$ .

Next step

Write $V = \dfrac{m}{\rho}$ , then divide $1.2\,\text{kg}$ by $1000\,\text{kg/m}^3$ .

Full working

\begin{aligned} V &= \frac{m}{\rho} \\ &= \frac{1.2}{1000} \\ &= 1.2 \times 10^{-3}\,\text{m}^3 \end{aligned}

4. A stone has a mass of $64.8\,\text{g}$ . It is lowered into a measuring cylinder. The water level rises from $50\,\text{cm}^3$ to $74\,\text{cm}^3$ . Calculate the density of the stone in $\text{g/cm}^3$ .

Hint

The rise in the water level is the volume of the stone. Read the two levels and subtract them.

Next step

Find the volume: $74 - 50 = 24\,\text{cm}^3$ . Then divide the mass by this volume.

Full working

\begin{aligned} V &= 74 - 50 = 24\,\text{cm}^3 \\ \rho &= \frac{m}{V} \\ &= \frac{64.8}{24} \\ &= 2.7\,\text{g/cm}^3 \end{aligned}

The stone has a density of $2.7\,\text{g/cm}^3$ , which is $2700\,\text{kg/m}^3$ .

5. Explain, using the particle model, why a gas has a much lower density than the same mass of a solid.

Hint

Compare how far apart the particles are in the two states.

Next step

Say where the particles are in each state, then link that to the volume the same mass fills.

Full working

In a gas the particles are far apart, with large empty spaces between them. In a solid the particles are packed closely together. The same mass of gas therefore fills a much larger volume than the solid, so the gas has a much lower density.

This answer is a model answer for you to compare with your own wording.

6. A student says that when ice melts, some of the ice is lost, so the mass of the water is less than the mass of the ice. Explain why the student is wrong.

Hint

Think about the particles. Are any of them lost during melting?

Next step

State what happens to the particles and what happens to the mass, then give the reason.

Full working

Melting is a change of state, not a chemical change. The same particles are present before and after melting. They move further apart and change their arrangement, but none are created or destroyed. The mass is therefore conserved, so the mass of the water equals the mass of the ice.

This answer is a model answer for you to compare with your own wording.

7. Say which measurements you would take, and which equation you would use, to find the density of a small irregular stone. Then calculate the density if the stone has a mass of $39\,\text{g}$ and raises the water level in a measuring cylinder from $20\,\text{cm}^3$ to $35\,\text{cm}^3$ .

Hint

An irregular stone has no simple length, width and height. Think about how to find its volume with water.

Next step

Measure the mass on a balance. Find the volume by displacement. Then use $\rho = \dfrac{m}{V}$ .

Full working

Measure the mass of the stone on a balance. Find the volume by lowering the stone into water in a measuring cylinder and reading the rise in the water level. Then use $\rho = \dfrac{m}{V}$ .

\begin{aligned} V &= 35 - 20 = 15\,\text{cm}^3 \\ \rho &= \frac{39}{15} \\ &= 2.6\,\text{g/cm}^3 \end{aligned}

The stone has a density of $2.6\,\text{g/cm}^3$ .