N5 Heat & Properties of Matter

Question 1

What does the particle model (or kinetic theory) tell us about the particles of a substance?

Answer 1

Normally the particles of a substance are in constant random motion.

Question 2

Define temperature.

Answer 2

The temperature of a substance is a measure of the mean kinetic energy of its particles

Question 3

What can happen to the temperature of a substance when heat is gained or lost?

Answer 3

The temperature of a substance can increase when heat is gained

or decrease when heat is lost.

Question 4

Define specific heat capacity.

Answer 4

Specific heat capacity is the amount of heat energy required to change the temperature of 1 kg of a material by 1 °C.

Question 5

E_h = cmΔT

(Define symbols and units)

Answer 5

E_h - Heat energy (J)

c - specific heat capacity (J kg^-1 °C^-1 - see data sheet)

m - mass (kg)

ΔT - Change in temperature (°C)

Question 6

Example

Calculate the amount of heat energy required to boil a kettle of water (2 kg) from a room temperature of 20 °C.

Answer 6

E<sub>h</sub> = ?
c = 4180 J kg<sup>-1</sup> °C<sup>-1</sup> (from data sheet)
m = 2 kg
ΔT = 100 - 20 = 80 °C

*E<sub>h</sub> = cmΔT
E<sub>h</sub> = 4180 x 2 x 80
E<sub>h</sub> = 669 000 J (to 3sf)*

Question 7

What happens to a substance when heat is gained or lost at its melting or boiling point?

Answer 7

The substance will change state.

Question 8

Describe what happens to the particles of a substance when it changes state.

Answer 8

The bonds between particles are either loosened or strengthened (meaning the particles gain or lose potential energy).

Question 9

What happens to the temperature during a change of state?

Answer 9

There is no change in temperature during a change of state.

Question 10

E_h = ml

(Define symbols and units)

Answer 10

E_h - Heat energy (J)

m - mass (kg)

l - specific latent heat (J kg^-1 - see data sheet)

Question 11

Example

How much water will evaporate when 500 kJ of heat is applied?

Answer 11

E_h = 500 kJ = 500 x 10³ J
m - ?
l = 22·6 x 10⁵ J kg^-1 (from data sheet)

E_h = ml
500 x 10³ = m x 22·6 x 10⁵
m x 22·6 x 10⁵ = 500 x 10³
m = 500 x 10³/22·6 x 10⁵
m = 0·221 kg

Question 12

The heating curve (graph of temperature versus time) for a substance is shown.

What is happening in the section labelled A?

Answer 12

(The substance is a solid to begin with because there are only three states of matter with solid being the coolest).

The substance is increasing temperature as a solid.

Question 13

The heating curve (graph of temperature versus time) for a substance is shown.

What is happening in the section labelled B?

Answer 13

(The substance is a solid to begin with because there are only three states of matter with solid being the coolest).

The substance is changing state from solid to liquid.

(or the substance is melting)

Question 14

The heating curve (graph of temperature versus time) for a substance is shown.

What is happening in the section labelled C?

Answer 14

(The substance is a solid to begin with because there are only three states of matter with solid being the coolest).

The substance is increasing tempearture as a liquid.

Question 15

The heating curve (graph of temperature versus time) for a substance is shown.

What is happening in the section labelled D?

Answer 15

(The substance is a solid to begin with because there are only three states of matter with solid being the coolest).

The substance is changing state from liquid to gas.

(or the substance is boiling)

Question 16

The heating curve (graph of temperature versus time) for a substance is shown.

What is happening in the section labelled E?

Answer 16

(The substance is a solid to begin with because there are only three states of matter with solid being the coolest).

The substance is increasing temperature as a gas.

Question 17

The heating curve (graph of temperature versus time) for a substance is shown.

What is the melting point of the substance?

Answer 17

(The substance is a solid to begin with because there are only three states of matter with solid being the coolest. Thus it melts during section B.)

The melting point is 60 °C.

Question 18

The heating curve (graph of temperature versus time) for a substance is shown.

What is the boiling point of the substance?

Answer 18

(The substance is a solid to begin with because there are only three states of matter with solid being the coolest. Thus it boils during section D.)

The boiling point is 130 °C.

Question 19

What is the definition of the term pressure?

Answer 19

Pressure is defined as the force per unit area.

Question 20

p = F/A

(Define symbols and units)

Answer 20

p - pressure (Pa)

F - Force (N)

A - Area (m²)

Question 21

Example

Calculate the pressure exerted by a 20 kg box with a base measuring 20 cm by 30 cm.

Answer 21

p = ?
F = W = mg = 20 x 9.8 = 196 N
A = 20cm x 30xm = 0.2m x 0.3m = 0.06 m<sup>2</sup>

*p = F/A
p = 196/0.06
p = 3270 Pa (to 3 sig fig)*

Question 22

Explain how a snowshoe works

(in terms of pressure, force and area).

Answer 22

A snowshow spreads a force over a larger area - this causes the pressure on the snow to decrease.

Question 23

Account for the pressure of a gas, in terms of the particle model or kinetic theory.

Answer 23

• The particles of a gas are in constant random motion
• and will collide with their surroundings.
• Each collision contributes a small amount of force.
• The pressure = Total Force/Area.

Question 24

State the relationship between pressure and volume for a fixed mass of gas at constant temperature.

Answer 24

The pressure of a fixed mass of gas at constant temperature is inversely proportional to its volume.

Question 25

p₁V₁ = p₂V₂

(Define symbols and units)

Answer 25

p - pressure (Pa)

V - Volume (m³)

Other units can be used if consistent on both sides.

Question 26

Example

A gas container has a volume of 20 cm³ at atmospheric pressure (1.01 x 10⁵ Pa). Calculate the pressure inside when it expands to a volume of 1000 cm³.

Answer 26

p<sub>1</sub> = 1.01 x 10<sup>5</sup> Pa
V<sub>1</sub> = 20 cm<sup>3</sup>
p<sub>2</sub> = ?
V<sub>2</sub> = 1000 cm<sup>3</sup>

p₁V₁ = p₂V₂
​1.01 x 10⁵ x 20 = p₂ x 1000
p₂ x 1000 = 1.01 x 10⁵ x 20
p₂ = 1.01 x 10⁵ x 20 / 1000
p₂ = 2.02 x 10³ Pa

Question 27

State the relationship between pressure and temperature for a fixed mass of gas at constant volume.

Answer 27

The pressure of a fixed mass of gas at constant volume is directly proportional to its temperature measured in Kelvin.

Question 28

Explain what is meant by the absolute zero of temperature.

Answer 28

Absolute zero is the temperature (0 K) at which an ideal gas exerts zero pressure and occupies zero volume.

This is because it is the temperature at which the gas particles have no kinetic energy and stop moving.

Question 29

Describe how to convert a temperature in °C into K.

Answer 29

(Reminder: 0 K = -273 °C)

To convert a temperature from °C into K, add 273.

Question 30

Describe how to convert a temperature in K into °C.

Answer 30

(Reminder: 0 K = -273 °C)

To convert a temperature from K into °C, subtract 273.

Question 31

p₁/T₁ = p₂/T₂

(Define symbols and units)

Answer 31

p - pressure (Pa)

T - temperature (must be K)

Other units for p can be used if consistent on both sides.

Question 32

An aerosol can at room temperature (20 °C) contains gas at a pressure of 1·5 x 10⁵ Pa. Calculate the temperature when the can is sprayed and the gas emitted is at atmospheric pressure (1·01 x 10⁵ Pa).

Answer 32

p<sub>1</sub> = 1·5 x 10<sup>5</sup> Pa
T<sub>1</sub> = 20 °C = 20+273 = 293 K
p<sub>2</sub> = 1·01 x 10<sup>5</sup> Pa
T<sub>2</sub> = ?

p₁/T₁ = p₂/T₂
1·5 x 10⁵/293 = 1·01 x 10⁵/T₂
Cross Multiply
1·5 x 10⁵ x T₂ = 1·01 x 10⁵ x 293
T₂ = 1·01 x 10⁵ x 293 / 1·5 x 10⁵
T₂ = 197 K
(NB leave in K unless the question specifically asks for the answer in °C)

Question 33

State the relationship between volume and temperature for a fixed mass of gas at constant pressure.

Answer 33

The volume of a fixed mass of gas at constant pressure is directly proportional to its temperature measured in Kelvin.

Question 34

V₁/T₁ = V₂/T₂

(Define symbols and units)

Answer 34

V - volume (m³)

T - temperature (must be K)

Other units for V can be used if consistent on both sides.

Question 35

A gas is held in a flexible container so that its pressure is constant. It has an inital volume of 400 m³ and an inital temperature of 40 °C. What temperature would be required to double this volume?

Answer 35

V<sub>1</sub> = 400 m<sup>3</sup>
T<sub>1</sub> = 40 °C = 40 + 273 = 313 K
V<sub>2</sub> = 2 x 400 = 800 m<sup>3</sup>
T<sub>2</sub> = ?

V₁/T₁ = V₂/T₂
400/313 = 800/T₂
Cross multiply
400 x T₂ = 800 x 313
T₂ = 800 x 313 / 400
T₂ = 626 K
(NB leave in K unless the question specifically asks for the answer in °C)

Question 36

*pV/T = constant
p<sub>1</sub>V<sub>1</sub>/T<sub>1</sub> = p<sub>2</sub>V<sub>2</sub>/T<sub>2</sub>*

(Define symbols and units)

Answer 36

p - pressure (Pa)

V - volume (m³)

T - temperature (must be K)

Other units for p and V can be used if consistent on both sides.

Question 37

A 250 ml aerosol can at room temperature (20 °C) contains gas at a pressure of 5·0 x 10⁵ Pa. Calculate the temperature when the can is sprayed to empty and 1000 ml of gas is emitted at atmospheric pressure (1·01 x 10⁵ Pa).

Answer 37

p<sub>1</sub> = 5·0 x 10<sup>5</sup> Pa
V<sub>1</sub> = 250 ml
T<sub>1</sub> = 20 °C = 20 + 273 = 293 K
p<sub>2</sub> = 1·01 x 10<sup>5</sup> Pa
V<sub>2</sub> = 1000 ml
T<sub>2</sub> = ?

p₁V₁/T₁ = p₂V₂/T₂
5·0 x 10⁵ x 250/293 = 1·01 x 10⁵ x 1000/T₂
Cross Multiply
5·0 x 10⁵ x 250 x T₂ = 1·01 x 10⁵ x 1000 x 293
T₂ = 1·01 x 10⁵ x 1000 x 293 / 5·0 x 10⁵ x 250
T₂ = 237 K
(NB leave in K unless the question specifically asks for the answer in °C)

Question 38

Explain, in terms of particles, what happens to the pressure of a gas when its volume is increased?
(The mass and temperature of the gas are fixed.)

Answer 38

• When the volume of a gas is increased, the particles have more space to move around in
• The particles collide with the container walls less frequently
• The total force on the walls decreases and because pressure = Force/Area, the pressure decreases.

Question 39

Explain, in terms of particles, what happens to the pressure of a gas when its temperature is increased?
​(The mass and volume of the gas are fixed.)

Answer 39

• When the temperature of a gas is increased, the particles gain kinetic energy i.e. move faster
• The particles collide with the container walls more frequently and with greater individual force
• The total force on the walls increases and because pressure = Force/Area, the pressure increases.

Question 40

Explain, in terms of particles, what happens to the volume of a gas when its temperature is increased?
​(The mass and pressure of the gas are fixed.)

Answer 40

• When the temperature of a gas is increased, the particles gain kinetic energy i.e. move faster
• The particles collide with the container walls more frequently and with greater individual force
• The total force on the walls increases but the container is flexible so it expands i.e. the volume increases.