Starter

What is the name used for when a substance changes state from a gas into a liquid?

Give an example of a substance that does not have a fixed shape.

What is the boiling point of water?

When something melts, what do you think is happening to the particles?

Draw the particles arrangements for a solid, liquid and a gas.

Changes of State
The particle model is used to visualise how particles behave in a range of situations.

They are often shown as balls in a box, shown below:

This model is limited however, as it doesn’t show that the particles are moving (which they are), it doesn’t
bring in the idea that particles are moving in 3D (as it is a 2D picture), and there is no representation of forces
that act between the particles (which there are).

States of matter can change, usually depending on an increase or decrease of thermal energy of a substance
passing a threshold temperature.

These temperatures are either the Boiling Point, or the Melting Point.

Particles naturally vibrate, causing them to ‘jiggle’ or oscillate around a fixed point.

The greater the temperature, the faster the vibrations. Gas particles freely move around the space, and the
speed at which they do this also increases with temperature.

Solids have the strongest forces between the particles, whilst gases have little/no forces between them.
Because of this there’s nothing holding the particles together, allowing gas particles to randomly move freely.
Liquids have less, but still some forces between them, as the particles can move over each other, but not
separate from each other completely.

States change due to the forces between the particles either being overcome or forming.

Heating a substance and making it melt or evaporate, transfers thermal energy to the particles, causing them
to move faster and eventually, overcome the forces between them.

Cooling a substance and making it freeze or condense transfers thermal energy from the particles, making
them move slower and allowing the forces between them to be established.
Specific Latent Heat
When a substance is heated or cooled and changes state, two things are happening.
1. It is increasing/decreasing in temperature to either the boiling or melting point.
- This is due to the particles gaining/losing energy, and moving faster/slower.

2. Once the boiling or melting point is reached, the substance will begin to change state,
but the temperature won’t change.
o This is because temperature is a measure of how fast the particles vibrate.
- At hotter temperatures, particles vibrate faster. When they’re cooler, they vibrate
slower.
- Eventually due to the forces between the particles, they’ll reach a “top-speed” and
will be unable to vibrate faster than they currently are.
o As a result, temperature doesn’t change.
- The energy instead is used to break bonds (if heating), or given out when bonds are
formed (if cooling).

So there’s two amounts of energy involved in changing state:

- The energy needed to get to a melting/boiling point (by heating/cooling).
- The energy needed to overcome the forces between the particles.

This second part is what makes up specific Latent Heat.

Specific Latent Heat is defined as:

“The amount of energy needed to change the state of 1kg of a substance without changing its
temperature.”

See that bold part? – If the question is specifically asking about the definition for latent heat of
fusion or vaporisation, then you must specify WHAT states are involved. e.g. for specific
latent heat of vaporisation:

“The amount of energy needed to change the state of 1kg of a liquid substance into a gas
without changing its temperature.”

Fusion = Solid → Liquid

Vaporisation = Liquid → Gas

You must also specify 1kg.

Latent heat and state changes can be shown in graphs that look like this:

This is for a substance that is being heated.

From A to B, the substance is increasing in

temperature until the melting point.

From B to C, the substance is melting, but

temperature is not changing. We can see this
as the line is horizontal.

From C to D, the substance is increasing until

it reaches boiling point.
Specific Latent heat can be calculated using the equation:

Energy = Mass x Specific Latent Heat (of fusion or vaporisation)

(J) (kg) (J/kg)

This equation tells us how much energy is needed to change the state of something once it is at
melting or boiling point.

The energy needed to change 1kg of water into 1kg of steam may be more than this figure, as we’ll
also need to heat it to boiling point as well as factor in the specific latent heat amount.

Exam Questions

Q1.
Figure 1 shows solid ice on a car’s rear window.

Figure 1

© Captive cookies/iStock/Thinkstock

The glass window contains an electrical heating element.

(a) Use the particle model in Figure 2 to describe how the heating element causes the
arrangement of the ice particles to change as the ice melts.

Figure 2

You should include a description of how the particles are arranged in the solid ice
and in the water.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(6)

(b) A car manufacturer tests different heating elements by measuring how long it takes
ice to melt.

During the test some variables must be controlled.

Identify two control variables in the car manufacturer’s test.

Tick two boxes.

The colour of the car

The current in the heating

element

The mass of ice

The size of the car

The time taken for the ice to

melt
(2)
 (c) Some of the energy supplied by the heater causes the ice to melt without the
temperature of the ice increasing.

What is the name given to this energy supplied by the heater?

Tick one box.

Latent heat of freezing

Latent heat of fusion

Latent heat of vaporisation

(1)

(d) When the heater is supplied with 120 J of energy each second, the internal energy
of the ice increases by 45 J each second.

Use the following equation to calculate the efficiency of the heater.

Efficiency =

Give your answer to two decimal places.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Efficiency = ____________________
(2)
(Total 11 marks)

Q2.
Solid, liquid and gas are three different states of matter.

(a) Describe the difference between the solid and gas states, in terms of the
arrangement and movement of their particles.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
 ___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(4)

(b) What is meant by ‘specific latent heat of vaporisation’?

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(2)

(c) While a kettle boils, 0.018 kg of water changes to steam.

Calculate the amount of energy required for this change.

Specific latent heat of vaporisation of water = 2.3 × 106 J / kg.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Energy required = __________________ J

(2)

(d) The graph shows how temperature varies with time for a substance as it is heated.

The graph is not drawn to scale.

Explain what is happening to the substance in sections AB and BC of the graph.

Section AB _________________________________________________________

___________________________________________________________________

___________________________________________________________________
___________________________________________________________________

___________________________________________________________________

Section BC _________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(4)
(Total 12 marks)
Mark schemes

Q1.
(a) Level 3 (5–6 marks):
A clear, logical explanation containing accurate ideas presented in the correct order
with links between ideas.

Level 2 (3–4 marks):

Key ideas presented with some linked together to form a partial explanation.

Level 1 (1–2 marks):

Fragmented ideas, some may be relevant, insufficient links to form an explanation.

0 marks:
No relevant content.

Indicative content
• current in the wire causes heating
• increases temperature of the metal wires / ice

Solid
• arrangement of particles is regular
• particles vibrate about a fixed position

Melting
• internal energy of the ice increases, increasing the temperature to
melting
point
• so (as the temperature increases) particles vibrate faster
• eventually particles vibrate fast enough to break free from the (strong)
bonds
• therefore the arrangement of particles becomes irregular

Liquid
• arrangement of particles is irregular
• particles movement (translational) is random
6

(b) The current in the heating element

1

The mass of ice

1

(c) latent heat of fusion

1

45 / 120 = 0.375
1

0.38
allow 0.38 with no working shown for 2 marks
allow 0.375 with no working shown for 1 mark
1
[11]

Q2.
(a) solid
particles vibrate about fixed positions
1

closely packed
accept regular
1

gas
particles move randomly
accept particles move faster
accept freely for randomly
1

far apart
1

(b) amount of energy required to change the state of a substance from liquid to
gas (vapour)
1

unit mass / 1 kg
dependent on first marking point
1

(c) 41000 or 4.1 × 104 (J)

accept
41400 or 4.14 × 104
correct substitution of
0.018 × 2.3 × 106 gains 1 mark
2

(d) AB
changing state from solid to liquid / melting
1

at steady temperature
dependent on first AB mark
1

BC
temperature of liquid rises
1

until it reaches boiling point

dependent on first BC mark
1
[12]
Specific Heat Capacity
Different substances can heat up faster than others.

Why do you think we use metal for frying pans? It’s because they are very easily increased in
temperature.

To put it another way, you don’t have to put a lot of energy in in order for the substance to get
hot!

This property is called specific heat capacity. Think of it as “how hard is it to heat up or cool
down this substance?”.

Specific Heat Capacity can be defined as:

“The amount of energy required to change the temperature of 1kg of a substance by 1oC”

It is important to see that it says to “change” the temperature. This involves heating as well as cooling!

When a substance is heated, it gains energy.

When a substance is cooled, it loses energy.

Water has a specific heat capacity of 4200J/kgoC.

This means that for 1kg of water, raising the temperature from 20oC to 21oC requires 4200J of
energy to be absorbed by it.

Similarly, for water to cool from 21oC to 20oC, it must lose 4200J of energy.

To calculate how much energy is needed to raise/lower a mass of substance by a certain

temperature, we need to use the following equation:

Energy Required = Mass x Specific Heat Capacity x Temperature Change

Specific Heat Capacity is different for every substance, either you will be given this value to use the in
the equation in the information in the question, or it will be the answer you’re trying to calculate by
rearranging the above formula.

Also note that it is temperature CHANGE – how much has the temperature risen or decreased by, and
not the start/final temperature. You may need to do a small calculation to find out what this is.

The specific heat capacity calculated, particularly from experimental results, is likely to be
slightly inaccurate. This is because in order to find a substance’s specific heat capacity, we need
to rearrange the equation into:

𝐸𝑛𝑒𝑟𝑔𝑦 𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐻𝑒𝑎𝑡 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 =
𝑀𝑎𝑠𝑠 × 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑐ℎ𝑎𝑛𝑔𝑒

In doing this, we reach the required practical for this topic: Estimating the Specific Heat Capacity
of an unknown substance.
Required Practical - Specific Heat Capacity
To do this, look at the equation:

𝐸𝑛𝑒𝑟𝑔𝑦 𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐻𝑒𝑎𝑡 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 =
𝑀𝑎𝑠𝑠 × 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑐ℎ𝑎𝑛𝑔𝑒

We need to work out everything in that box there.

To do this:
- We work out the mass using a top-pan balance. This must be in kg.
- We work out the temperature change using a thermometer, and measuring the
initial temperature and the final temperature after heating.
- To find the energy needed, we use a joulemeter.
o This may not always be possible, as they are expensive.
o Instead of this, we will also need to fit in a voltmeter, ammeter and time the
length of time spent heating with a stopwatch.

This means that the practical setup could look like either of these two:

After finding the mass using a top-pan balance, we can then start the experiment.

We place an immersion heater inside the block we’re testing. This will heat up, and there’s an
additional hole to put a thermometer in.

We measure the initial temperature of the block now.

We switch on the power, and the number on the joulemeter will increase with every Joule of
energy used.

Stopping it after 3 minutes (180s), we read the final temperature of the block, as well as note
down the joulemeter reading.

We find the temperature change by doing final temperature – initial temperature.

Then we put all our values into the equation above.

If we’re using a voltmeter and ammeter set up:
- Carry out all above steps, including measuring the mass and initial temperature of the
block.
- Switch on circuit and start the stopwatch.
- Note down the voltage and current whilst the circuit is on.
- After 180s, switch off circuit and measure the final temperature.
- Calculate temperature change as before.
- With the voltmeter and ammeter readings, use the equation V x I = P to work out
power.
- Once we have the power, we multiply this with the time in seconds to get the energy
value for the equation.
o This is due to Energy = Power x time.

Specific heat capacities calculated in this way are likely fairly inaccurate, mainly due to
uncontrolled heat loss to the surroundings.

Energy is being put into the immersion heater, and recorded by our joulemeter, but not all of it
will be picked up by the thermometer, as heat energy can escape from the walls of the block.

To make this less of an issue, we can

insulate the block by wrapping it in foam.
This reduces the amount of heat energy
that escapes, keeping it inside the block
where it’s more likely to be picked up by
the thermometer.

We must also ensure that the immersion

Heat energy heater is FULLY inside the block, so that no
escaping this way will heat energy is lost through the top-side of
not reach the the heater itself (where the wires are).
thermometer.

Typically, specific heat capacity is calculated in labs under extremely controlled conditions, such
as super-insulated blocks.
Exam Questions

Q1.
A student made measurements to determine the specific heat capacity of vegetable oil.

Figure 1 shows the equipment used.

Figure 1

(a) Describe how the student could use the equipment shown in Figure 1 to determine
the specific heat capacity of vegetable oil.

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___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(6)
(b) Give one risk when using the equipment in Figure 1.

___________________________________________________________________

___________________________________________________________________
(1)

A different student did not have a joulemeter and calculated the energy transferred by the
electric heater.

Use the Physics Equations Sheet to answer parts (c) and (d).

(c) Write down the equation linking energy transferred (E), power (P) and time (t).

___________________________________________________________________
(1)

(d) The electric heater had a power output of 50 watts.

Calculate the time taken for the electric element to transfer 4750 joules of energy to
the vegetable oil.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Time taken = _______________ s

(3)
In a deep fryer, vegetable oil is heated by an electric heating element. Food is then
cooked in the hot vegetable oil.

The deep fryer contains an electrical component to monitor the temperature of the
vegetable oil.

Figure 2 shows how the resistance of this electrical component changes with
temperature.

Figure 2

(e) What electrical component is used to monitor the temperature of the vegetable oil?

___________________________________________________________________
(1)
(f) The electric heating element in the deep fryer automatically switches off when the
vegetable oil reaches a certain temperature.

Figure 3 shows how the temperature of the vegetable oil changed after the deep
fryer was switched on.

Figure 3

Determine the resistance of the electrical component when the electric heating
element automatically switched off.

Use Figure 2 and Figure 3.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Resistance = _______________ Ω
(2)

(g) Some chips were put in the deep fryer.

In the deep fryer, water in the chips underwent a physical change and became
steam.

Why is this a physical change?

Tick (✓) one box.

 All water can change to steam.

No chemicals are involved when water changes to steam.

The change from water to steam can be detected visually.

The water will recover its original properties if the steam is

cooled.
(1)
(Total 15 marks)

Q2.
A student investigated how the temperature of a metal block changed with time.

An electric heater was used to increase the temperature of the block.

The heater was placed in a hole drilled in the block as shown in Figure 1.

Figure 1

The student measured the temperature of the metal block every 60 seconds.
The table below shows the student’s results.

Time in s Temperature in °C
0 20.0
60 24.5
120 29.0
180 31.0
240 31.5
(a) Complete the graph of the data from the table above on the graph below.

• Choose a suitable scale for the x-axis.

• Label the x-axis.
• Plot the student’s results.
• Draw a line of best fit.

Figure 2

(4)

(b) The rate of change of temperature of the block is given by the gradient of the graph.

Determine the gradient of the graph over the first 60 seconds.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Gradient = _____________________
(2)
(c) The metal block had a mass of 1.50 kg

The specific heat capacity of the metal was 900 J/kg °C

Calculate the change in thermal energy of the metal during 240 seconds.

Use the Physics Equations Sheet.

Give your answer in kilojoules.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Change in thermal energy = __________________ kJ

(4)

(d) Another student repeated the investigation.

Give two variables this student would need to control to be able to compare their
results with the results in the table above.

1. _________________________________________________________________

2. _________________________________________________________________
(2)
(Total 12 marks)
Mark schemes

Q1.
(a) Level 3: The method would lead to the production of a valid outcome. All key
steps are identified and logically sequenced.
5−6

Level 2: The method would not necessarily lead to a valid outcome. Most
steps are identified, but the plan is not fully logically sequenced.
3−4

Level 1: The method would not lead to a valid outcome. Some relevant steps
are identified, but links are not made clear.
1−2

No relevant content
0

Indicative content

• measure mass of oil using the top pan balance

• measure start temperature of oil using the thermometer
• place beaker of oil on heater
• switch on heater to heat oil
• measure final temperature of oil using the thermometer
• measure energy transferred using joulemeter
• calculate increase in temperature (Δθ)
• use the equation E = mcΔθ to determine c

(b) burns / scalds

allow cuts from broken glass
ignore the heater / oil is hot
1

(c) power =

or

(d) 50 =

or

4750 = 50 × t
1

1
 t = 95 (s)
1

(e) thermistor
1

(f) 250 (Ω)

allow an answer in the range 240 (Ω) to 260 (Ω)
allow 1 mark for temperature = 160 (°C)
2

(g) the water will recover its original properties if the steam is cooled
1
[15]

Q2.
(a) x-axis labelled and suitable scale
1

points plotted correctly

allow 5 correctly plotted for 2 marks, 3−4
correctly plotted for 1 mark
allow ± ½ square
2

line of best fit

1

(b)
allow ecf from line of best fit in part (a)
1

0.075 (°C/s)
1
an answer of 0.075 (°C/s) scores 2 marks

(c) Δθ = 11.5 (°C)

a calculation using an incorrect temperature
scores max 3 marks
1

ΔE = 1.50 × 900 × 11.5

1

ΔE = 15 525 (J)
1

ΔE = 15.525 (kJ)
1
an answer of 15.525 (kJ) or 15.53 (kJ) or 15.5
(kJ) scores 4 marks
an answer of 15 525 (kJ) scores 3 marks

(d) any two from:

• mass of block*
• size / dimensions of block*
• material of block*
*allow same block for 1 mark

• current through heater

allow power of heater

• thickness of insulation*
• material of insulation*
*allow same insulation for 1 mark

• time interval

• starting temperature (of block / heater)

2
[12]
Internal Energy
Internal energy is defined as:
“The total energy stored inside a system by the particles that make up the system due to their motion and
positions relative to each other.”

The molecules within a substance possess two forms of energy:

- Kinetic energy (due to their random motion / vibration).
- Potential energy (due to their position relative to each other)

Together, these two form the total energy that makes up the internal energy of the system Molecules in a
substance have kinetic energy since they are in motion and potential energy from their position relative to
each other.

When a substance is heated, it’s internal energy store increases, due to the increase in kinetic energy of the
particles as they begin to move faster and faster.

State change, such as evaporation in which particles move away from the rest, cause cooling of the original
group of particles.
This is how sweating cools you down when you are warm.
1. Sweat/Water absorbs energy, so the particles begin moving faster.
2. Some particles gain enough energy and are moving so fast that they can escape from the surface of
the water into the atmosphere (they evaporate).
3. Due to this, there’s less energy in the remaining particles.
4. With lower energy, there’s a lower temperature.

Exam Question

Q1.
The diagram shows three cups A, B and C.

Energy is transferred from hot water in the cups to the surroundings.

(a) Some students investigated how the rate of cooling of water in a cup depends on
the surface area of the water in contact with the air.

They used cups A, B and C. They poured the same volume of hot water into each
cup and recorded the temperature of the water at regular time intervals.

The results are shown on the graph.

 Time in minutes

(i) What was the starting temperature of the water for each cup?

Starting temperature = ____________________ °C

(1)

(ii) Calculate the temperature fall of the water in cup B in the first 9 minutes.

______________________________________________________________

Temperature fall = ____________________ °C

(2)

(iii) Which cup, A, B or C, has the greatest rate of cooling?

Using the graph, give a reason for your answer.

______________________________________________________________

______________________________________________________________
(2)
 (iv) The investigation was repeated using the bowl shown in the diagram.

The same starting temperature and volume of water were used.

Draw on the graph in part (b) another line to show the expected result.
(1)

(v) After 4 hours, the temperature of the water in each of the cups and the bowl
was 20°C.

Suggest why the temperature does not fall below 20°C.

______________________________________________________________
(1)

(b) (i) The mass of water in each cup is 200 g.

Calculate the energy, in joules, transferred from the water in a cup when the
temperature of the water falls by 8°C.

Specific heat capacity of water = 4200 J / kg°C.

______________________________________________________________

______________________________________________________________

______________________________________________________________

Energy transferred = ____________________ J

(3)

(ii) Explain, in terms of particles, how evaporation causes the cooling of water.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________
(4)
(Total 14 marks)
Mark schemes

Q1.
(a) (i) 70
accept ± half a square
(69.8 to 70.2)
1

(ii) 15
accept 14.6 to 15.4 for 2 marks
allow for 1 mark 70 − 55
ecf from (b)(i) ± half a square
2

(iii) C
1

biggest drop in temperature during a given time

accept it has the steepest gradient this is a dependent
1

(iv) starting at 70 °C and below graph for C

must be a curve up to at least 8 minutes
1

(v) because 20 °C is room temperature

accept same temperature as surroundings
1

(b) (i) 6720

correct answer with or without working gains 3 marks
6 720 000 gains 2 marks
correct substitution of E = 0.2 × 4200 × 8 gains 2 marks
correct substitution of E = 200 × 4200 × 8 gains 1 mark
3

(ii) the fastest particles have enough energy

accept molecules for particles
1

to escape from the surface of the water

1

therefore the mean energy of the remaining particles decreases

accept speed for energy
1

the lower the mean energy of particles the lower the temperature (of the
water)
accept speed for energy
1
[14]
Summary Exam Questions

Q1.
A student investigated the effect of insulation on how quickly hot water cools.

Figure 1 shows the apparatus the student used. He added a different number of layers of
insulation around each beaker.

The student measured the temperature of the water every minute for 10 minutes as the
water cooled.

(a) State one variable the student should control during the investigation.

___________________________________________________________________
(1)

(b) Figure 2 shows a line graph of the results.

 What conclusion should the student make from Figure 2 about the effect of the
number of layers of insulation?

___________________________________________________________________

___________________________________________________________________
(1)

(c) (i) There is one anomalous result. Draw a ring around this result on Figure 2.
(1)

(ii) Suggest what the student might have done to cause this anomalous result.

______________________________________________________________

______________________________________________________________
(1)

(d) Draw a ring around the correct answer to complete the sentence.

Energy is transferred through the wall of the metal beaker by the process of

conduction. convection. infrared radiation.

(1)
(Total 5 marks)

Q2.
A piece of steel is heated until it has all melted.

(a) Calculate the change in thermal energy when the temperature of the piece of steel is
increased by 50 °C.

mass of steel = 4.0 kg

specific heat capacity of steel = 420 J/kg °C

Use the equation:

change in thermal energy = mass × specific heat capacity × temperature change

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Change in thermal energy = ____________________ J

(2)
(b) The internal energy of the steel increases as the steel is heated.

What is meant by ‘internal energy of the steel’?

Tick (✓) one box.

The change in energy of the steel particles as the steel

melts.

The energy added to the steel particles as they are

heated.

The total kinetic energy and potential energy of the

steel particles.
(1)

(c) Solid steel cannot be poured.

Which statement about the particles in a solid gives the reason why?

Tick (✓) one box.

The number of particles always stays the

same.

The particles are close together.

The particles are in fixed positions.

The particles have a fixed size.

(1)

(d) Complete the sentence.

Choose the answer from the box.

decreases stays the same increases

As the piece of solid steel melts, the mass of the steel ____________________.
(1)
(e) Which diagram shows how the arrangement of particles changes when a solid melts
and becomes a liquid?

Tick (✓) one box.

(1)

(f) The density of steel decreases as it melts.

How does the spacing of the particles change as the steel melts?

___________________________________________________________________

___________________________________________________________________
(1)

(g) Complete the sentence.

Choose the answer from the box.

chemical permanent physical

Melting is an example of a ____________________ change.

(1)
(h) Steel is a mixture of iron and a small amount of carbon.

The table below shows the mass of carbon in 1.0 kg of different types of steel.

Type of steel Mass of carbon in 1.0 kg of steel

Low carbon 2.0 g

Medium carbon 4.5 g

High carbon 7.0 g

A 4.0 kg piece of steel contains 18 grams of carbon.

Determine which type of steel the 4.0 kg piece is made from.

You should include a calculation in your answer.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Type of steel ____________________

(3)

(i) The 4.0 kg piece of solid steel was heated until it reached its melting point.

The additional energy required to melt the piece of steel was 280 000 J.

Calculate the specific latent heat of fusion of the steel.

Use the Physics Equations Sheet.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

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Specific latent heat of fusion of steel = ____________________ J/kg

(3)
(Total 14 marks)
Q3.
According to kinetic theory, all matter is made up of small particles. The particles are
constantly moving.

Diagram 1 shows how the particles may be arranged in a solid.

Diagram 1

(a) One kilogram of a gas has a much larger volume than one kilogram of a solid.

Use kinetic theory to explain why.

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(4)

(b) Diagram 2 shows the particles in a liquid. The liquid is evaporating.

Diagram 2

(i) How can you tell from Diagram 2 that the liquid is evaporating?

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(1)
(ii) The temperature of the liquid in the container decreases as the liquid
evaporates.

Use kinetic theory to explain why.

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(3)
(Total 8 marks)
Mark schemes

Q1.
(a) any one from:
• starting temperature
• volume of beaker
allow size / shape / material of beaker
• surface area (of beaker)
• volume of water
allow mass / amount of water
• type of insulation.
1

(b) the more layers of insulation the slower the temperature fell
accept the converse
answer must be a comparison
allow ‘heat’ for energy
accept when more layers are added less energy is lost

or
the more layers of insulation the smaller the temperature drop
accept the more layers (of insulaion) the slower the energy
loss
1

(c) (i) point at 7 minutes for 5 layers of insulation circled

1

(ii) any one from:

• misread the thermometer
• took the temperature at the wrong time
human / recording error needs to be qualified
• lifted the thermometer out of the water (when reading it)
• misplotted the point.
1

(d) conduction
1
[5]

Q2.
(a) ΔE = 4.0 × 420 × 50
1

ΔE = 84 000 (J)
1

(b) the total kinetic energy and potential energy of the steel particles
1

(c) the particles are in fixed positions

1

(d) stays the same

1
 (e)

(f) (the space between the particles) increases

allow the particles move further apart
1

(g) physical
1

(h)
1

mass per kg = 4.5 g

1

medium carbon
dependent on MP2

OR

mass in 4.0 kg of medium carbon steel = 4.5 × 4.0 (1)

allow mass in 4.0 kg of low carbon steel = 8 (g)
allow mass in 4.0 kg of high carbon steel = 28 (g)

mass in 4.0 kg of medium carbon steel = 18 g (1)

dependent on MP1

medium carbon (1)

dependent on MP2
1

(i) 280 000 = 4.0 × L

1

L = 70 000 (J/kg)
1
[14]

Q3.
(a) there are strong forces (of attraction) between the particles in a solid
accept molecules / atoms for particles throughout
accept bonds for forces
1
 (holding) the particles close together
particles in a solid are less spread out is insufficient
1

or

(holding) the particles in a fixed pattern / positions

but in a gas the forces between the particles are negligible

accept very small / zero for negligible
accept bonds for forces
1

so the particles spread out (to fill their container)

accept particles are not close together
gas particles are not in a fixed position is insufficient
1

(b) (i) particles are (shown) leaving (the liquid / container)

accept molecules / atoms for particles throughout
accept particles are escaping
particles are getting further apart is insufficient
1

(ii) accept molecules / atoms for particles throughout

accept speed / velocity for energy throughout

particles with most energy leave the (surface of the) liquid

accept fastest particles leave the liquid
1

so the mean / average energy of the remaining particles goes down

1

and the lower the average energy (of the particles) the lower the
temperature (of the liquid)
1
[8]