1
(a)
Lead has a specific heat capacity of 130 J kg−1 K−1.
Explain what is meant by this statement.
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(1)
(b)
Lead of mass 0.75 kg is heated from 21 °C to its melting point and continues to be heated
until it has all melted.
Calculate how much energy is supplied to the lead.
Give your answer to an appropriate number of significant figures.
melting point of lead = 327.5 °C
specific latent heat of fusion of lead = 23 000 J kg−1
energy supplied ................................................... J
(3)
(Total 4 marks)
2
The mass of a car and its passengers is 950 kg. When the brakes are applied the car
decelerates uniformly from a speed of 25 m s–1 to a speed of 15 m s–1 in 2.5 s.
(a)
Calculate the decelerating force developed by the brakes.
(2)
(b)
Calculate the work done in decelerating the car.
(3)
(c)
Calculate the rate of energy dissipation by the brakes.
(2)
Page 1 of 80
(d)
There are four brake discs, each of mass 1.2 kg. The material from which the discs are
made has a specific heat capacity of 510 J kg–1 K–1.
(i)
Assuming that all the energy dissipated during braking is converted into internal
energy of the brake discs equally, calculate the temperature rise of the discs.
(3)
(ii)
State and explain the effect on the temperature rise of one factor that has not been
taken into account in the assumption in part (i).
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(2)
(Total 12 marks)
3
The Carnot cycle is the most efficient theoretical cycle of changes for a fixed mass of gas in a
heat engine.
The graph below shows the pressure−volume (p−V) diagram for a gas undergoing a Carnot
cycle of changes ABCDA.
Page 2 of 80
(a)
(i)
Show that during the change AB the gas undergoes an isothermal change.
(3)
(ii)
Explain how the first law of thermodynamics applies to the gas in the change BC.
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(3)
(iii)
Determine the ratio
,
where TA is the temperature of the gas at A and TC is the temperature of the gas at
C.
ratio .....................................................
(3)
(b)
Show that the work done during the change AB is about 110 J.
(2)
Page 3 of 80
(c)
When running at a constant temperature, one practical engine goes through 2400 cycles
every minute. In one complete cycle of this engine, 114 J of energy has to be removed by a
coolant so that the engine runs at a constant temperature. The temperature of the coolant
rises by 18 °C as it passes through the engine.
Calculate the volume of the coolant that flows through the engine in one second.
specific heat capacity of coolant = 3.8 × 103 J kg–1 K–1
density of coolant = 1.1 × 103 kg m–3
volume flowing in one second ................................................ m3
(3)
(Total 14 marks)
4
A car of mass M travelling at speed V comes to rest using its brakes. Energy is dissipated in the
brake discs of total mass m and specific heat capacity c. The rise in temperature of the brake
discs can be estimated from
A
B
C
D
(Total 1 mark)
Page 4 of 80
5
A raindrop of mass m falls to the ground at its terminal speed v. The specific heat capacity of
water is c and the acceleration of free fall is g. Given that 25% of the energy is retained in the
raindrop when it strikes the ground, what is the rise in temperature of the raindrop?
A
B
C
D
(Total 1 mark)
6
A 1.0 kΩ resistor is thermally insulated and a potential difference of 6.0 V is applied to it for 2.0
minutes. The thermal capacity of the resistor is 9.0 J K–1. The rise in temperature, in K, is
A
1.3 × 10–3
B
8.0 × 10–3
C
0.48
D
0.80
(Total 1 mark)
7
A thermometer has a thermal capacity of 1.3 J K–1. The initial temperature of the thermometer is
20°C. When used to measure the temperature of 40 g of water, it measures 37°C.
(a)
Determine the energy absorbed by the thermometer when it is placed in the water.
(2)
(b)
Calculate the temperature change of the water as a result of introducing the thermometer.
specific heat capacity of water = 4.2 × 103 J kg–1 K–1
(2)
(Total 4 marks)
8
Nitrogen at 20°C and a pressure of 1.1 × 105 Pa is held in a glass gas syringe as shown in
Figure 1. The gas, of original volume 8.5 × 10–5 m3, is compressed to a volume of 5.8 × 10–5 m3
by placing a mass on to the plunger of the syringe. The change in pressure of the gas is
adiabatic. The new pressure of the gas is 1.9 × 105 Pa.
Page 5 of 80
Figure 1
(a)
(i)
Calculate the new temperature of the nitrogen. Give your answer in °C.
(3)
(ii)
Calculate the number of moles of nitrogen present in the syringe.
molar gas constant, R = 8.3 J mol–1 K–1
(3)
(iii)
The mass of the nitrogen in the syringe is 1.1 × 10–4 kg. Calculate the mean square
speed of the molecules when the gas has been compressed.
(2)
(iv)
Explain why the change in pressure of the nitrogen is adiabatic.
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(2)
Page 6 of 80
(v)
Explain, in terms of the behaviour of the nitrogen molecules, how the gas exerts a
greater pressure than it did before it was compressed.
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(3)
(b)
After the adiabatic compression, the nitrogen is allowed to cool at constant volume.
Figure 2 shows the variation of pressure with volume for the adiabatic compression and
the subsequent cooling. The dotted line represents the isothermal compression that would
have achieved the same final state.
Figure 2
(i)
Draw arrows on the graph to show the directions of the changes. Label your arrows
adiabatic compression and cooling as appropriate.
(1)
(ii)
State the significance of the shaded area of the graph.
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(1)
(Total 15 marks)
9
(a)
State the principle of conservation of linear momentum for two colliding bodies.
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(2)
Page 7 of 80
(b)
A bullet of mass 0.010 kg travelling at a speed of 200 m s–1 strikes a block of wood of mass
0.390 kg hanging at rest from a long string. The bullet enters the block and lodges in the
block. Calculate
(i)
the linear momentum of the bullet before it strikes the block,
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(ii)
the speed with which the block first moves from rest after the bullet strikes it.
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(4)
(c)
During the collision of the bullet and block, kinetic energy is converted into internal energy
which results in a temperature rise.
(i)
Show that the kinetic energy of the bullet before it strikes the block is 200 J.
...............................................................................................................
(ii)
Show that the kinetic energy of the combined block and bullet immediately after the
bullet has lodged in the block is 5.0 J.
...............................................................................................................
(iii)
The material from which the bullet is made has a specific heat capacity of
250 J kg–1 K–1. Assuming that all the lost kinetic energy becomes internal energy in
the bullet, calculate its temperature rise during the collision.
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(5)
Page 8 of 80
(d)
The bullet lodges at the centre of mass G of the block. Calculate the vertical height h
through which the block rises after the collision.
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(2)
(Total 13 marks)
10
(a)
Explain the meaning of the statement the specific heat capacity of ice is 2100 J kg –1K–1.
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(2)
(b)
An engineer is designing an ice-making machine. Water will enter the device at 18°C and
the ice cubes are to be cooled to –5°C before release.
(i)
Show that about 0.4 MJ of energy must be removed from 1.0 kg of water at 18°C to
change it into ice at –5°C.
Set out the stages in your answer clearly.
specific heat capacity of water
= 4.2 ×103 J kg–1 K–1
specific heat capacity of ice
= 2.1 ×103 J kg–1 K–1
specific latent heat of fusion of ice
= 3.3 ×105 J kg–1
(3)
Page 9 of 80
(ii)
The design brief requires that 1.5 kg of water is frozen in 300 s. Calculate the rate at
which energy must be removed by the machine.
(2)
(Total 7 marks)
11
In a power station, water is heated in a boiler to create steam. This steam is passed through a
turbine before being cooled by river water to condense it back into the liquid state. The cooled
water is then pumped back to the boiler for re-use.
(a)
The cooling water enters the condenser at 16°C and is returned to the river at 40°C. Every
second, 35 × 103 kg of water are removed from the river.
Calculate the power required to heat this water.
Specific heat capacity of water, c, = 4200 J kg−1 K−1
(2)
(b)
Assume that all the energy transferred by the river water came from steam changing to
water without a change in temperature.
Calculate the mass of steam passing through the turbine per second.
Specific latent heat of vaporisation of water = 2.4 × 106 J kg−1
(2)
(c)
The turbine is linked to a generator that produces 800 MW of electrical power. Calculate the
efficiency of conversion of the internal energy of the steam into electrical energy.
(2)
(Total 6 marks)
12
A meteorite of mass 1.2 × 104 kg enters the Earth's atmosphere at a speed of 2.5 km s–1.
(a)
Calculate the kinetic energy of the meteorite as it enters the atmosphere.
(2)
(b)
The meteorite is initially at a temperature of 25°C. It is made of iron of specific heat
capacity 110 J kg–1 K–1 and of melting point 1810°C. As the meteorite travels through the
Earth’s atmosphere friction causes its temperature to rise. Calculate the energy needed to
raise the temperature of the meteorite to its melting point. You should assume that the
specific heat capacity of iron remains constant.
(3)
Page 10 of 80
(c)
When it reaches the surface of the Earth the mass of the meteorite has fallen to
5.5 × 103 kg and its speed to 150 m s–1 so that its kinetic energy is only 6.2 × 107 J. On
striking the Earth this mass penetrates the Earth’s crust and is brought to rest in a distance
of 25 m. Calculate:
(i)
the average force acting on the meteorite as it penetrates the Earth’s crust;
(2)
(ii)
the time it takes for the meteorite to be brought to rest by the Earth’s crust;
(2)
(iii)
the rate at which the meteorite dissipates energy in this time.
(2)
(Total 11 marks)
13
(a)
Explain what is meant by
(i)
the specific heat capacity of water,
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(ii)
the specific latent heat of fusion of ice.
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(4)
Page 11 of 80
(b)
A sample of solid material, which has a mass of 0.15 kg, is supplied with energy at a
constant rate. The specific heat capacity of the material is 1200 J kg–1 K–1 when in the solid
state. During heating, its temperature is recorded at various times and the following graph
is plotted.
Assume there is no heat exchange with the surroundings.
(i)
Show that energy is supplied to the material at a rate of 24 W.
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(ii)
Calculate the specific latent heat of fusion of the material.
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(iii)
Calculate the specific heat capacity of the material when in the liquid state.
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(6)
(Total 10 marks)
Page 12 of 80
14
(a)
(i)
State what is meant by thermal equilibrium.
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(ii)
Explain thermal equilibrium by reference to the behaviour of the molecules when a
sample of hot gas is mixed with a sample of cooler gas and thermal equilibrium is
reached.
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(3)
(b)
A sealed container holds a mixture of nitrogen molecules and helium molecules at a
temperature of 290 K. The total pressure exerted by the gas on the container is 120 kPa.
(i)
molar mass of helium
=
4.00 × 10–3 kg mol–1
molar gas constant R
the Avogadro constant NA
=
=
8.31 J K–1 mol–1
6.02 × 1023 mol–1
Calculate the root mean square speed of the helium molecules.
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(ii)
Calculate the average kinetic energy of a nitrogen molecule.
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(iii)
If there are twice as many helium molecules as nitrogen molecules in the container,
calculate the pressure exerted on the container by the helium molecules.
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(6)
(Total 9 marks)
Page 13 of 80
15
(a)
(i)
Explain what is meant by the specific latent heat of vaporisation of a liquid.
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(ii)
Suggest why the specific latent heat of vaporisation of water is much greater than the
specific latent heat of fusion of water.
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(3)
(b)
A cup contains 0.25 kg of water at a temperature of 15 °C. The water is heated by passing
steam at 100 °C into it.
specific heat capacity of water
= 4200 J kg–1 K–1
specific latent heat of vaporisation of water
= 2.3 × 106 J kg–1
boiling point of water
= 100 °C
(i)
Use the above data to calculate the minimum mass of water that is in the cup when
the temperature of the water reaches its boiling point.
(ii)
Explain why there is likely to be a greater mass of water in the cup than you have
calculated in part (b)(i).
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(4)
(Total 7 marks)
Page 14 of 80
16
(a)
A 500 µF capacitor and a 1000 µF capacitor are connected in series. Calculate the total
capacitance of the combination.
(2)
(b)
The figure below shows a diagram of an arrangement used to investigate the energy stored
by a capacitor.
The bundle of constantan wire has a resistance of 8.5 Ω. The capacitor is initially charged
to a potential difference of 9.0 V by closing S1.
(i)
Calculate the charge stored by the 0.25 F capacitor.
(ii)
Calculate the energy stored by the capacitor.
(iii)
Switch S1 is now opened and S2 is closed so that the capacitor discharges through
the constantan wire. Calculate the time taken for the potential difference across the
capacitor to fall to 0.10 V.
(7)
Page 15 of 80
(c)
The volume of constantan wire in the bundle in the figure above is 2.2 × 10–7 m3.
density of constantan = 8900 kg m–3
specific heat capacity of constantan = 420 J kg–1 K–1
(i)
Assume that all the energy stored by the capacitor is used to raise the temperature of
the wire. Use your answer to part (b)(ii) to calculate the expected temperature rise
when the capacitor is discharged through the constantan wire.
(ii)
Give two reasons why, in practice, the final temperature will be lower than that
calculated in part (c)(i).
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(5)
(Total 14 marks)
17
A bicycle and its rider have a total mass of 95 kg. The bicycle is travelling along a horizontal road
at a constant speed of 8.0 m s–1.
(a)
Calculate the kinetic energy of the bicycle and rider.
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(2)
Page 16 of 80
(b)
The brakes are applied until the bicycle and rider come to rest. During braking, 60% of the
kinetic energy of the bicycle and rider is converted to thermal energy in the brake blocks.
The brake blocks have a total mass of 0.12 kg and the material from which they are made
has a specific heat capacity of 1200 J kg–1 K–1.
(i)
Calculate the maximum rise in temperature of the brake blocks.
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(ii)
State an assumption you have made in part (b)(i).
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(4)
(Total 6 marks)
18
(a)
A student immerses a 2.0 kW electric heater in an insulated beaker of water. The heater is
switched on and after 120 s the water reaches boiling point.
The table below gives data collected during the experiment.
initial mass of beaker
initial mass of beaker and water
initial temperature of water
final temperature of water
25 g
750 g
20 °C
100 °C
Calculate the specific heat capacity of water if the thermal capacity of the beaker is
negligible.
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(4)
(b)
The student in part (a) continues to heat the water so that it boils for 105 s. When the mass
of the beaker and water is measured again, it is found that it has decreased by 94 g.
(i)
Calculate a value for the specific latent heat of vaporisation of water.
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Page 17 of 80
(ii)
State two assumptions made in your calculation.
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(4)
(Total 8 marks)
19
Use the following data to answer the question below.
specific latent heat of fusion of lead = 23 kJ kg–1
molar mass of lead = 0.21 kg mol–1
(a)
Estimate the mass of a lead atom.
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(b)
Estimate the energy supplied to an atom of lead when solid lead melts.
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(c)
Calculate the speed of a lead atom with the same kinetic energy as the energy supplied in
part (b).
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(Total 5 marks)
Page 18 of 80
20
An electrical immersion heater supplies 8.5 kJ of energy every second. Water flows through the
heater at a rate of 0.12 kg s–1 as shown in the figure below.
(a)
Assuming all the energy is transferred to the water, calculate the rise in temperature of the
water as it flows through the heater.
specific heat capacity of water = 4200 J kg–1 K–1
answer = ....................................... K
(2)
(b)
The water suddenly stops flowing at the instant when its average temperature is 26 °C.
The mass of water trapped in the heater is 0.41 kg.
Calculate the time taken for the water to reach 100 °C if the immersion heater continues
supplying energy at the same rate.
answer = ....................................... s
(2)
(Total 4 marks)
21
The following data refer to a dishwasher.
Page 19 of 80
(a)
power of heating element
2.5 kW
time to heat water
360 s
mass of water used
3.0 kg
initial temperature of water
20°C
final temperature of water
60°C
Taking the specific heat capacity of water to be 4200 J kg–1 K–1, calculate
(i)
the energy provided by the heating element,
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(ii)
the energy required to heat the water.
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(4)
(b)
Give two reasons why your answers in part (a) differ from each other.
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(2)
(Total 6 marks)
22
An electrical heater is used to heat a 1.0 kg block of metal, which is well lagged. The table shows
how the temperature of the block increased with time.
temp / °C
time / s
20.1
23.0
26.9
30.0
33.1
36.9
0
60
120
180
240
300
Page 20 of 80
(a)
Plot a graph of temperature against time on the grid provided.
(3)
(b)
Determine the gradient of the graph.
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(2)
Page 21 of 80
(c)
The heater provides thermal energy at the rate of 48 W. Use your value for the gradient of
the graph to determine a value for the specific heat capacity of the metal in the block.
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(2)
(d)
The heater in part (c) is placed in some crushed ice that has been placed in a funnel as
shown.
The heater is switched on for 200 s and 32 g of ice are found to have melted during this
time. Use this information to calculate a value for the specific latent heat of fusion for water,
stating one assumption made.
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(3)
(Total 10 marks)
23
(a)
A 3.0 kW electric kettle heats 2.4 kg of water from 16°C to 100°C in 320 seconds.
(i)
Calculate the electrical energy supplied to the kettle.
...............................................................................................................
...............................................................................................................
(ii)
Calculate the heat energy supplied to the water.
specific heat capacity of water = 4200 J kg–1 K–1
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Page 22 of 80
(iii)
Give one reason why not all the electrical energy supplied to the kettle is transferred
to the water.
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(4)
(b)
The potential difference supplied to the kettle in part (a) is 230 V.
(i)
Calculate the resistance of the heating element of the kettle.
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(ii)
The heating element consists of an insulated conductor of length 0.25 m and
diameter 0.65 mm. Calculate the resistivity of the conductor.
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(5)
(Total 9 marks)
24
A tray containing 0.20 kg of water at 20 °C is placed in a freezer.
(a)
The temperature of the water drops to 0 °C in 10 minutes.
specific heat capacity of water = 4200 J kg–1 K–1
Calculate
(i)
the energy lost by the water as it cools to 0 °C,
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.............................................................................................................
(ii)
the average rate at which the water is losing energy, in J s–1.
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(3)
Page 23 of 80
(b)
(i)
Estimate the time taken for the water at 0 °C to turn completely into ice.
specific latent heat of fusion of water = 3.3 × 105 J kg–1
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(ii)
State any assumptions you make.
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(3)
(Total 6 marks)
25
(a)
An electric shower heats water from 15°C to 47°C when water flows through it at a rate of
0.045 kg s–1.
(i)
Calculate the energy supplied to the water each second by the heating element in the
shower.
specific heat capacity of water = 4200 J kg–1 K–1
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(ii)
Show that the power of the heating element is 6.0 kW. Assume there is no heat loss
to the surroundings.
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(3)
Page 24 of 80
(b)
(i)
The heating element in part (a) is connected to an alternating supply at 230 V rms.
Calculate the rms current passing through the heating element in normal operation.
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(ii)
The live wire and the neutral wire in the connecting cable are insulated copper wires
of diameter 2.4 mm. Calculate the resistance per metre length of copper wire of this
diameter.
resistivity of copper = 1.7 × 10−8 Ωm
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(iii)
Show that in normal operation, the potential drop per metre along the cable is
0.20 V m–1.
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(iv)
Electrical safety regulations require the potential drop along the cable to be less than
6.0 V.
Calculate the maximum safe distance along the cable from the distribution board to
the heating element.
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(9)
(Total 12 marks)
Page 25 of 80
26
In an experiment to measure the temperature of the flame of a Bunsen burner, a lump of copper
of mass 0.12 kg is heated in the flame for several minutes. The copper is then transferred quickly
to a beaker, of negligible heat capacity, containing 0.45 kg of water, and the temperature rise of
the water measured.
specific heat capacity of water = 4200 J kg–1 K–1
specific heat capacity of copper =
(a)
390 J kg–1 K–1
If the temperature of the water rises from 15 °C to 35 °C, calculate the thermal energy
gained by the water.
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(2)
(b)
(i)
State the thermal energy lost by the copper, assuming no heat is lost during its
transfer.
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(ii)
Calculate the fall in temperature of the copper.
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.............................................................................................................
(iii)
Hence calculate the temperature reached by the copper while in the flame.
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(4)
(Total 6 marks)
Page 26 of 80
27
(a)
Suggest two reasons why an α particle causes more ionisation than a β particle of the
same initial kinetic energy.
You may be awarded marks for the quality of written communication in your answer.
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(2)
(b)
A radioactive source has an activity of 3.2 × 109 Bq and emits α particles, each with kinetic
energy of 5.2 Me V. The source is enclosed in a small aluminium container of mass
2.0 × 10–4 kg which absorbs the radiation completely.
(i)
Calculate the energy, in J, absorbed from the source each second by the aluminium
container.
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.............................................................................................................
(ii)
Estimate the temperature rise of the aluminium container in 1 minute, assuming no
energy is lost from the aluminium.
specific heat capacity of aluminium = 900 J kg–1 K–1
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(5)
(Total 7 marks)
Page 27 of 80
28
A female runner of mass 60 kg generates thermal energy at a rate of 800 W.
(a)
Assuming that she loses no energy to the surroundings and that the average specific heat
capacity of her body is 3900 J kg–1K–1, calculate
(i)
the thermal energy generated in one minute,
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.............................................................................................................
(ii)
the temperature rise of her body in one minute.
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.............................................................................................................
.............................................................................................................
(3)
(b)
In practice it is desirable for a runner to maintain a constant temperature. This may be
achieved partly by the evaporation of sweat. The runner in part (a) loses energy at a rate of
500 Wby this process.
Calculate the mass of sweat evaporated in one minute.
specific latent heat of vaporisation of water = 2.3 × 106 J kg–1
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(c)
Explain why, when she stops running, her temperature is likely to fall.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 8 marks)
Page 28 of 80
29
An electric shower heats the water flowing through it from 10°C to 42°C when the volume flow
rate is 5.2 × 10−5 m3 s−1.
(a)
(i)
Calculate the mass of water flowing through the shower each second.
density of water = 1000 kg m−3
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
Calculate the power supplied to the shower, assuming all the electrical energy
supplied to it is gained by the water as thermal energy.
specific heat capacity of water = 4200 J kg−1 K−1.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(4)
(b)
A jet of water emerges horizontally at a speed of 2.5 m s−1 from a hole in the shower head.
The hole is 2.0 m above the floor of the shower. Calculate the horizontal distance travelled
by this jet. Assume air resistance is negligible.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(Total 7 marks)
Page 29 of 80
30
(a)
A 2.0 kW heater is used to heat a room from 5 °C to 20 °C. The mass of air in the room
is 30 kg. Under these conditions the specific heat capacity of air = 1000 J kg–1 K–1.
Calculate
(i)
the gain in thermal energy of the air,
.............................................................................................................
.............................................................................................................
(ii)
the minimum time required to heat the room.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(b)
State and explain one reason why the actual time taken to heat the room is longer than the
value calculated in part (a)(ii).
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 6 marks)
Page 30 of 80
31
In a nuclear reactor the mean energy produced by each uranium-235 nucleus that undergoes
induced fission is 3.0 × 10–11 J. In one pressurised water reactor, PWR, the fuel rods in the
reactor contain 2.0 × 104 kg of uranium-235 and 40% of the energy produced per second is
converted to 500 MW of electrical output power. It is assumed that all the energy produced in the
reactor core is removed by pressurised water in the coolant system. The pressure of the water is
approximately 150 times greater than normal atmospheric pressure. The water enters the reactor
at a temperature of 275 °C ad leaves at a temperature of 315 °C. Under the operational
conditions of the reactor the mean density of water in the coolant circuit is 730 kg m–3 and the
specific heat capacity of water is approximately 5000 J kg–1 K–1.
normal atmospheric pressure = 1.0 × 105 Pa
molar mass of uranium-235 = 0.235 kg
(a)
The equation below gives one induced fission reaction that takes place in a reactor.
(i)
State the name of the particle represented by X.
...............................................................................................................
(1)
(ii)
State the proton and nucleon numbers represented by p and n.
p ..........................................................
n ..........................................................
(2)
(b)
(i)
Calculate the number of fission reactions that occur in the reactor each second.
number of fission reactions per second ............................................
(2)
Page 31 of 80
(ii)
The reactor fuel rods contain 2.0 × 104 kg of uranium-235. Assume that all this
uranium-235 could be used.
Calculate the maximum time, in years, for which the reactor could operate.
time .........................................................years
(4)
(iii)
Suggest why it is not possible to use all the uranium-235 in the reactor fuel rods.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(c)
Calculate the force exerted by the pressurised water on each square centimetre of the wall
of the reactor.
force .........................................................N
(2)
(d)
Calculate, in m3 s–1, the flow rate of the water through the PWR reactor.
You will need to use data from the passage at the beginning of the question.
flow rate ......................................................... m3 s–1
(4)
Page 32 of 80
(e)
In a PWR the cooling water also acts as the moderator in the reactor and boron rods are
used to control the power output. Describe the physical processes that take place in the
moderator and control rods.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(4)
(Total 21 marks)
32
An electrical heater is placed in an insulated container holding 100 g of ice at a temperature of
–14 °C. The heater supplies energy at a rate of 98 joules per second.
(a)
After an interval of 30 s, all the ice has reached a temperature of 0 °C. Calculate the
specific heat capacity of ice.
answer = ............................J kg–1K–1
(2)
Page 33 of 80
(b)
Show that the final temperature of the water formed when the heater is left on for a further
500 s is about 40 °C.
specific heat capacity of water = 4200 J kg–1K–1
specific latent heat of fusion of water = 3.3 × 105 J kg–1
(3)
(c)
The whole procedure is repeated in an uninsulated container in a room at a temperature of
25 °C.
State and explain whether the final temperature of the water formed would be higher or
lower than that calculated in part (b).
.....................................................................................................................
.....................................................................................................................
.....................................................................................................................
.....................................................................................................................
(2)
(Total 7 marks)
33
(a)
Calculate the energy released when 1.5 kg of water at 18 °C cools to 0 °C and then
freezes to form ice, also at 0 °C.
specific heat capacity of water = 4200 J kg–1 K–1
specific latent heat of fusion of ice = 3.4 × 105 J kg–1
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(4)
Page 34 of 80
(b)
Explain why it is more effective to cool cans of drinks by placing them in a bucket full of
melting ice rather than in a bucket of water at an initial temperature of 0 °C.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 6 marks)
34
In a geothermal power station, water is pumped through pipes into an underground region of hot
rocks. The thermal energy of the rocks heats the water and turns it to steam at high pressure.
The steam then drives a turbine at the surface to produce electricity.
(a)
Water at 21°C is pumped into the hot rocks and steam at 100°C is produced at a rate of
190 kg s–1.
(i)
Show that the energy per second transferred from the hot rocks to the power station
in this process is at least 500 MW.
specific heat capacity of water = 4200 J kg–1 K–1
specific latent heat of steam
= 2.3 × 106 J kg–1
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
Page 35 of 80
(ii)
The hot rocks are estimated to have a volume of 4.0 × 106 m3. Estimate the fall of
temperature of these rocks in one day if thermal energy is removed from them at the
rate calculated in part (i) without any thermal energy gain from deeper underground.
specific heat capacity of the rocks = 850 J kg–1 K–1
density of the rocks
= 3200 kg m–3
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(7)
Page 36 of 80
(b)
Geothermal energy originates as energy released in the radioactive decay of the
uranium isotope
U deep inside the Earth. Each nucleus that decays releases 4.2 MeV.
Calculate the mass of
half-life of
U
molar mass of
U that would release energy at a rate of 500 MW.
= 4.5 × 109 years
U = 0.238 kg mol–1
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(5)
(Total 12 marks)
Page 37 of 80
35
Molten lead at its melting temperature of 327°C is poured into an iron mould where it solidifies.
The temperature of the iron mould rises from 27°C to 84°C, at which the mould is in thermal
equilibrium with the now solid lead.
mass of lead = 1.20 kg
specific latent heat of fusion of lead = 2.5 × 104 J kg–1
mass of iron mould = 3.00 kg
specific heat capacity of iron = 440 J kg–1K–1
(a)
Calculate the heat energy absorbed by the iron mould.
answer = ..................................... J
(2)
(b)
Calculate the heat energy given out by the lead while it is changing state.
answer = ...................................... J
(1)
(c)
Calculate the specific heat capacity of lead.
answer = ...................................... J kg–1 K–1
(3)
Page 38 of 80
(d)
State one reason why the answer to part (c) is only an approximation.
......................................................................................................................
......................................................................................................................
......................................................................................................................
(1)
(Total 7 marks)
36
A liquid flows continuously through a chamber that contains an electric heater. When the steady
state is reached, the liquid leaving the chamber is at a higher temperature than the liquid entering
the chamber. The difference in temperature is Δt.
Which of the following will increase Δt with no other change?
A
Increasing the volume flow rate of the liquid
B
Changing the liquid to one with a lower specific heat capacity
C
Using a heating element with a higher resistance
D
Changing the liquid to one that has a higher density
(Total 1 mark)
37
The temperature of a hot liquid in a container falls at a rate of 2 K per minute just before it begins
to solidify. The temperature then remains steady for 20 minutes by which time all the liquid has all
solidified.
What is the quantity
?
A
B
C
10 K–1
D
40 K–1
(Total 1 mark)
Page 39 of 80
38
(a)
Define the specific latent heat of vaporisation of water.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
Page 40 of 80
(b)
An insulated copper can of mass 20 g contains 50 g of water both at a temperature of
84 °C. A block of copper of mass 47 g at a temperature of 990 °C is lowered into the water
as shown in the figure below. As a result, the temperature of the can and its contents
reaches 100 °C and some of the water turns to steam.
specific heat capacity of copper = 390 J kgsup class="xsmall">–1 Ksup
class="xsmall">–1
specific heat capacity of water = 4200 J kgsup class="xsmall">–1 Ksup
class="xsmall">–1
specific latent heat of vaporisation of water = 2.3 × 106 J kgsup class="xsmall">–1
Before placement
(i)
After placement
Calculate how much thermal energy is transferred from the copper block as it cools to
100 °C.
Give your answer to an appropriate number of significant figures.
thermal energy transferred ........................................... J
(2)
Page 41 of 80
(ii)
Calculate how much of this thermal energy is available to make steam.
Assume no heat is lost to the surroundings.
available thermal energy ........................................... J
(2)
(iii)
Calculate the maximum mass of steam that may be produced.
mass ......................................... kg
(1)
(Total 7 marks)
39
A cola drink of mass 0.200 kg at a temperature of 3.0 °C is poured into a glass beaker. The
beaker has a mass of 0.250 kg and is initially at a temperature of 30.0 °C.
specific heat capacity of glass = 840 J kg–1K–1
specific heat capacity of cola = 4190 J kg–1K–1
(i)
Show that the final temperature, Tf, of the cola drink is about 8 °C when it reaches thermal
equilibrium with the beaker.
Assume no heat is gained from or lost to the surroundings.
(2)
Page 42 of 80
(ii)
The cola drink and beaker are cooled from Tf to a temperature of 3.0 °C by adding ice at a
temperature of 0 °C.
Calculate the mass of ice added.
Assume no heat is gained from or lost to the surroundings.
specific heat capacity of water = 4190 J kg–1 K–1
specific latent heat of fusion of ice = 3.34 × 105 J kg–1
mass .......................................... kg
(3)
(Total 5 marks)
Page 43 of 80
Mark schemes
1
(a)
(it takes) 130 J / this energy to raise (the temperature of) a mass of 1 kg (of lead) by 1 K / 1
°C (without changing its state) ✓
1 kg can be replaced with unit mass.
Marks for 130J or energy.
+1 kg or unit mass.
+1 K or 1 °C.
Condone the use of 1 °K
1
(b)
(using Q = mcΔT + ml)
= 0.75 × 130 × (327.5 ‒ 21) + 0.75 × 23000 ✓
(= 29884 + 17250)
= 47134 ✓
= 4.7 × 104 (J) ✓
For the first mark the two terms may appear separately i.e. they do
not have to be added.
Marks for substitution + answer + 2 sig figs (that can stand alone).
3
[4]
2
(a)
Ft = Δ(mv)
or F = ma and a = (v – u) / t
C1
3800 N
A1
(2)
(b)
work done = change in KE
C1
or
appropriate equation of motion for s
or
work done = Fs
Calculation of one KE correctly
or s calculated correctly (50 m)
C1
1.9 ×
105
J (condone N m) e.c.f. for F
A1
(3)
(c)
power = {force from (a)} × any velocity
C1
or
power = change in KE / time
76 kW (kJ s–1)
ecf from (b) or ecf from (a) for use of P = Fv (their F × 20)
A1
(2)
Page 44 of 80
(d)
(i)
(their (b)) = 4.8 (or 1.2) × 510 × Δθ (allow use of 1.2 instead of 4.8 for this mark)
C1
appreciation of 4 discs evident in the calculation
C1
77.6 (78) K (or °C) or their (b) / 2450
A1
(3)
(ii)
temperature rise will be lower
B1
there will be air resistance
some energy becomes internal energy of the air
B1
OR
other components of the braking system (including answers involving friction of
tyres with road)
these will use some of the energy to increase temperature
B1
OR
heat / energy transfer to the surroundings
since surroundings at lower temperature or
temperature or internal energy of surroundings rises
B1
(2)
[12]
3
(a)
(i)
Appreciates pV should be constant for isothermal change (by working or
statement) W =pΔV is TO
Allow only products seen where are approximately 150
for 1 mark
Penalise J as unit here
M1
Demonstrates pV = constant using 2 points (on the line) set
equal to each other or conclusion made or shows that for V
doubling that p halves (worth 2 marks)
need to see values for p and V
Products should equal 150 to 2 sf
Accept statement that products are slightly different so
not quite isothermal
A1
Demonstrates pV = constant using 3 points (on the line) with
conclusion
Need to see values for p and V
Products should equal 150 to 2 sf
Accept statement that products are slightly different so
not quite isothermal
A1
3
Page 45 of 80
(ii)
Adiabatic therefore no heat transfer or
Adiabatic therefore Q = 0
B1
Work is done by gas therefore W is negative or
Work is done by gas therefore energy is removed from the
system
B1
ΔU is negative therefore internal energy of gas decreases or
energy is removed from the system therefore internal energy
of gas decreases or work done by the gas so internal energy
decreases
Allow
−ΔU = −W or ΔU = −W
B1
3
(iii)
Uses pV
/ T = constant or uses pV=nRT or uses pV=NkT
e.g. makes T subject or substitutes into an equation with pA
and VAor pC and VC (condone use of n = 1) or their
Va read off range
= 2.5 to 2.6 (× 10−4)
pA = 600 × 103
VC read off range
= 8.5 to 8.6 (× 10−4)
pC = 140 × 103
C1
Correct substitution of coordinates (inside range) into
With consistent use of powers of 10
(pV)A range is 150 to 156 and (pV)C range is 119 to 120.4
C1
1.2(5) Allow range from 1.2 to 1.3
Accept decimal fraction : 1
A1
3
Page 46 of 80
(b)
Energy per large square = 10(J) or states that work done is equal to
area under curve (between A and B)
or energy per small square = 0.4(J)
or square counting seen on correct area
Must be clear that area represents energy either by subject of
formula or use of units on 10 or 0.4
Alternative:
W = area of a trapezium
(with working)
or W = Pmean × ΔV or
W = 450 × 103 × 2.5 × 10−4
or W = area of a rectangle + area of a triangle (with working)
B1
Number of large squares = 10.5 to 11.5 seen and (W) = number of
squares × area of one square (using numbers)
Range = 105 to 115 (J)
Or
Number of small squares = 263 to 287 seen and (W) = number of
squares × area of one square (using numbers)
Range = 105 to 115 (J)
States that actual work done would be lower because of
curvature of line
B1
2
(c)
(Total energy removed per s =) 4560 (J)
or number of cycles per s = 40
or (Mass per second =) 114 ÷ 68400 in rearranged form
or their energy ÷ (c ΔT ) or their energy ÷ 68400
C1
0.067 (kg) seen Allow 0.066 (kg) here
or allow V / t = 1.67 × 10−3 ÷ 1100
or (
)=
and correct substitution seen
Condone E = 114 (J) or temperature = 291(K)
C1
= 0.061 × 10−3 or 6.06 × 10−5 (m3)
A1
3
[14]
Page 47 of 80
4
5
6
7
C
[1]
D
[1]
C
[1]
(a)
energy = heat capacity × temperature change
C1
22 J
A1
(2)
(b)
E = mcθ
C1
0.13 K condone °C /
(allow e.c.f. from (i))
A1
(2)
[4]
8
(a)
(i)
pV / T = constant in any form
C1
correct substitution including absolute temperatures / 345K
C1
72°C not 345K
condone no unit and condone just °
A1
(3)
(ii)
pV = nRT
C1
correct substitution: n =
or
C1
3.8(5) × 10–3 (mol) or 3.8(4) × 10–3 (mol)
A1
e.c.f. for their (i)
(3)
(iii)
pV = Nm<c2 or p = ρ<c2
C1
3.0 × 105 m2 s–2 condone subsequent calculation of rms speed
A1
(2)
Page 48 of 80
(iv)
no heat transfer / ΔQ = 0 / no energy loss
B1
process too quick (for conduction to take place) /
glass is poor (thermal) conductor / the system is isolated
B1
(2)
(v)
molecules move faster / have more KE
B1
greater number of collisions (per second) (between molecules and wall)
not between molecules
B1
greater (rate of) change in momentum in each collision
B1
(3)
(b)
(i)
anticlockwise arrows correctly labelled - both arrows needed
B1
(3)
(ii)
work done on the gas (during compression)
B1
(1)
[15]
9
(a)
momentum before collision = momentum after collision (1)
provided no external force acts (1)
(2)
(b)
(i)
p = mυ (1)
10 × 10–3 × 200 = 2.(0) (1)
(ii)
kg m s–1 (N s) (1)
total mass after collision = 0.40 kg (1)
0.40υ = 2.0 gives υ = 5.(0) m s–1 (1)
(allow e.c.f. from (i))
(4)
(c)
(i)
kinetic energy = ½mv2
(1) (= 200 J)
(ii)
kinetic energy =
(iii)
ΔQ = 200 – 5 = 195 (J) = mcΔθ (1)
Δθ =
(1)
= 78 K (1)
(= 5.0 J)
(allow e.c.f. for incorrect ΔQ)
(5)
Page 49 of 80
(d)
kinetic energy lost (= potential energy gained) = mgh (1)
1.3 m (1)
(2)
[13]
10
(a)
energy required to heat the ice up
C1
2100 J needed to raise / extracted to lower temperature
of 1 kg by 1 deg (K or °C)
A1
2
(b)
(i)
either water @ 18 to water @ 0 = 75600 J or ice @
0 to ice at –5 = 10500 J
M1
water @ 0 to ice @ 0 = 330000 J
M1
total = 416100 J
A1
3
(ii)
cand. bi × 1.5 or cand. bi/300
C1
power = 2080 W
[0.4 MJ yields 2 kW condone 1 sf; J s–1]
A1
2
[7]
11
(a)
35 × 103 × 4200 × 24
Cl
= 3.53 × 109 W
Al
Page 50 of 80
(b)
3.53 × 109 / 2.4 × 106
[ecf; ans to (a) / 2.4 × 106]
Cl
= 1.47 × 103 kg s–1 [allow kg]
Al
(c)
800 MW / sensible power
Cl
= 0.8 / (3.53 + 0.8) = 0.185 or 18.5 %
[ecf from ai]
Al
[6]
12
(a)
½mv2 or substitution ignoring powers of 10
C1
3.75 × 1010 J
A1
(b)
Q = mcΔθ
C1
1785 seen
C1
2.36 × 109 J
A1
(c)
(i)
W = Fs or correctly substituted values
C1
2.48 × 106 N
condone effect of change of g.p.e.
A1
(ii)
force = rate of change of momentum or use of an equation of motion
C1
0.33 s
A1
(iii)
P = W / t or P = Fv or substitution ignoring powers of 10
C1
1.88 (or 1.86) × 108 W e.c.f. from (c)(ii)
A1
[11]
13
(a)
(i)
quantity of energy supplied to unit mass (1)
which raises temperature by 1°C [or 1K] (1)
Page 51 of 80
(ii)
quantity of energy required to change state of unit mass (1)
solid to liquid [or ice to water] (1)
without change of temperature (1)
(max 4)
(b)
(i)
Q (= mcΔθ ) = 0.15 × 1200 × (58 – 18) = 7200 (J) (1)
P=
(ii)
= 24 W (1)
Q = 24 × 7 × 60 = 10080 (J) (1)
0.15l = 10080 gives l = 67200 J kg–1 (1)
(iii)
24 × 4 × 60 = 0.15 × sL × (94 – 58) (1)
gives sL = 1070 J kg–1 K–1 (1)
(6)
[10]
14
(a)
(i)
no net flow of (thermal) energy (between two or more bodies) (1)
bodies at same temperature (1)
(ii)
(kinetic) energy is exchanged in molecular collisions (1)
until average kinetic energy of all molecules is the same (1)
max 3
Page 52 of 80
(b)
(i)
(1)
= 1340m s–1 (1)
(ii)
average k.e. of nitrogen molecules = average k.e. of helium molecules (1)
=
×
(1340)2 = 5.97× 10–21 J (1)
alternative schemes for (ii):
average k.e. =
=
kT (1)
× 1.38 × 10–23 ×290
= 6.00 × 10–21 J (1)
or
average k.e. =
=
(1)
×
= 6.00 × 10–21 J (1)
(iii)
use of p =
(1)
pHe =
or equivalent [or, at same temperature, p ∝ no. of molecules]
× 120 = 80 kPa (1)
6
[9]
Page 53 of 80
15
(a)
(i)
energy/heat input needed to change liquid into gas/vapour
when at its boiling point/without change of temperature
M1
energy per unit mass/1 kg
A1
(ii)
idea that more energy has to be supplied to separate
molecules than to break solid bond
or
for vaporisation work is done against atmospheric pressure
or
Idea that there is a greater change in PE in L-G than S-L
B1
3
(b)
(i)
ml = Mc∆θ or energy gain by water = 89250 (J)
m × 2.3 × 106 = 0.25 × 4200 × 85
C1
m = 0.0388 kg
A1
total mass = 0.289 (0.29) kg (0.25 + their m)
B1
(ii)
energy from steam is needed to raise temperature of
the cup
or
energy/heat will be lost to the surroundings/cup/tube
during the heating
B1
4
[7]
Page 54 of 80
16
(a)
1/C = 1/500 + 1/1000 or
C1
330 (333) µF
A1
2
(b)
(i)
Q = VC or Q = 0.25 × 9
C1
2.3 or 2.25 C (c.a.o. unit essential)
A1
(ii)
energy = ½ CV2 or 0.5 × 0.25 × 92 or ½ QV used
C1
10(.1) J (allow e.c.f. for Q)
A1
(iii)
V = Vo e−t/RC
C1
7
0.1 = 9 e−t/(8.5 x 0.25)
C1
9.6 (9.56) s
A1
(c)
(i)
Q = mc∆θ or mass = volume × density
C1
correct substitution 10.1 = (2.2 × 10−7 × 8900 × 400 × ∆θ)
C1
12 (12.3) K or °C ecf for energy from (b) (ii)
A1
5
Page 55 of 80
(ii)
some energy raises temperature of the thermometer
B1
energy/heat lost to (raise temperature of) surroundings
B1
[14]
17
(a)
(use of Ek = ½mv2 gives) Ek =
× 95 × 8.02 (1)
= 3040 J (1)
2
(b)
(i)
ΔQ = 0.60 × 3040 = 1824 (J) (1)
(allow C.E. for Ek from (a))
(use of ΔQ = mc Δθ gives)
1824 = 0.12 × 1200 Δθ (1)
Δθ = 13 K (1)
(12.7 K)
(allow C.E. for ΔQ)
(ii)
no heat is lost to the surroundings (1)
4
[6]
18
(a)
(use of
= Pt gives)
0.725 × c × (100 – 20) (1) = 2000 × 120 (1)
c = 41 00 (1) J kg–1 (1) (4140 J kg–1)
4
QWC 2
(b)
(i)
(use of mL = Pt gives) 94 × 10–3 L = 2000 × 105 (1)
L = 2.2 × 106 J kg–1 (1)
(ii)
no evaporation (before water heated to boiling point)
no heat lost (to the surroundings)
heater 100% efficient any two (1) (1)
4
[8]
Page 56 of 80
19
= 3.5 ×10–25 kg (1)
(a)
mass of one atom =
(b)
energy supplied = 23 × 103 ×3.5 ×10–25 (1)
= 8.1 × 10–21 J (1) (8.05 × 10–21 J
(allow C.E. for value from (i))
(c)
(use of ½ mv2 = EK gives)
½ × 3.5 × 10–25 × v2 = 8.1 × 10–21
= 220 m s–1 (1)
(215 m s–1)
(EK = 8.05 × 10–21 gives v = 214 m s–1) (allow C.E. for value of EK from (ii))
[5]
20
(a)
17 K
2
(b)
t = 15 s
2
[4]
21
(a)
(i)
(P = VIt gives) P = 2500 × 360 (1)
= 9.0 × 105 kJ (1)
(ii)
(Q = mcΔt gives) Q = 3 × 4200 × 40 (1)
= 5 0 × 105 J (1)
(4)
(b)
heat is lost to the surroundings
the dishwasher is heated
evaporation of the water
any two (1) (1)
(2)
[6]
22
(a)
adequate scale (1)
points plotted correctly (1)
best fit line (at least a point to right and left of line) (1)
3
Page 57 of 80
(b)
use of triangle for at least half line (1)
gradient
= 0.056 ± 0.004 (°C / s) (1)
2
gives 48 = c × (1.0) × 0.056 (1)
(c)
c = 860 ± 60 J kg-1K-1 (or J kg-1 °C-1) (1)
2
(d)
(use of Eth = ml gives) 48 × 200 = 32 ×10-3 × l (1)
l = 3.0 × 105 J kg-1 (1)
sensible assumption, e.g. no heat lost to surroundings or temperature
does not change or heat is transferred to ice (1)
3
[10]
23
(a)
(i)
(use of E = Pt gives) E = 3000 × 320 = 960 kJ (1)
(ii)
(use of Q = mcΔθ gives) Q = 2.4 × 4200 (100 - 16) (1)
= 850 kJ (1)
(iii)
energy needed to heat the kettle material (1)
[or heat loss to surroundings]
4
(b)
(i)
(use of I =
I=
gives)
= 13 (A) (1)
= 18 Ω (1)
(use of V = IR gives) R
(17.7 Ω)
(allow C.E. for value of I)
[or correct use of R =
(ii)
A=π
to give correct R]
(m2) (1)
(use of ρ =
(= 3.32 ×10–7(m2))
gives)
(1)
= 2.3 × 10-5 Ω m (1)
(2.35 × 10-5 Ωm)
(use of R = 18 Ω gives ρ = 2.4 × 10-5 Ω m)
(allow C.E. for value of R from (i) and value of A)
5
[9]
Page 58 of 80
24
(a)
(i)
(use of ΔQ = mcΔθ gives) energy lost by water
= 0.20 × 4200 × 20 (1)
= 1.7 × 104 J (1) (1.68 × 104 J)
(ii)
rate of loss of energy =
= 28 (W) (1)
(allow C.E. for value of energy lost in (i))
3
(b)
(i)
(use of ΔQ = ml gives) (28 × t) = 0.20 × 3.3 × 105 (1)
t = 2.4 × 103 s (1) (2.36 × 103 s)
(allow C.E. for value of rate of loss of energy in (a)(ii)
(ii)
e.g. constant rate of heat loss (1)
ice remains at 0°C (1)
max 3
[6]
25
(a)
(i)
(in 1 s), E = 0.045 × 4200 × (47 – 15) (1)
= 6050 J
(ii)
P
= 6.0 kW (1)
3
Page 59 of 80
(b)
(i)
(use of P = VI gives) I =
= 26 A (1)
(26.3 A)
(allow C.E. for value of P from (a))
(ii)
radius = 1.2 × 10-3 (m) (1)
cross-sectional area = π(1.2 × 10-3)2 (or 4.5 × 10-6(m2)) (1)
(1)
=
(1)
= 3.8 × 10-3 Ω m-1
(allow C.E. for value of A)
= 0.1 (V m–1) (per wire)
(iii)
two wires per cable gives pd per metre = 2 × 0.1 (1)
(= 0.20 V m–1) (1)
(iv)
maximum length =
= 30 m (1)
9
[12]
26
(a)
(use of ∆Q = mc∆T gives) ∆Q = 0.45 × 4200 × (35 − 15) (1)
= 3.8 × 104 J (3.78 × 104 J) (1)
2
(b)
(i)
3.8 × 104 J (1)
(allow C.E. for incorrect value of ∆Q from (a))
(ii)
(mc∆T = ∆Q gives) 0.12 × 390 × ∆T = 3.8 × 104 (1)
∆T = 812 K (1)
(use of ∆Q = 3.78 gives ∆T = 808 K
(allow C.E. for incorrect value of ∆Q from (i))
(iii)
(812 + 35) = 847 ºC (1)
(use of 808 gives 843 ºC)
(allow C.E. from (ii))
4
[6]
Page 60 of 80
27
(a)
reasons:
α particle has much more mass/momentum than β particle
α particle has twice as much charge as a β particle
α particle travels much slower than a β particle any two (1) (1)
2
QWC 1
(b)
(i)
energy absorbed per sec (= energy released per sec)
= 3.2 × 109 × 5.2 × 106 ×1.6 ×10–19 (1)
= 2.7 ×10–3 (J) (1) (2.66 × 10–3 (J))
(ii)
temperature rise in 1 minute
=
(for numerator) (1) (for denominator) (1)
= 0.90 K (or °C) (1)
(allow C.E. for incorrect value in (i))
5
[7]
28
(a)
(b)
(i)
energy = 800 × 60 = 48 × 103J (1)
(ii)
(use of Q = mc θ gives) 48 × 103 = 60 × 3900 ×
θ = 0.21 K (1)
(0.205 K)
(allow C.E. for value of energy from (i))
θ (1)
3
Q = ml gives 500 × 60 (1) = m × 2.3 × 106 (1)
m = 0.013 kg (1)
3
(c)
not generating as much heat internally (1)
still losing heat (at the same rate)
[or still sweating] (1)
hence temperature will drop (1)
lmax 2
[8]
29
(a)
(i)
mass each sec [=(vol / sec) × density] = 5.2 × 10–5 × 1000 (1)
= 0.052 kg (s–1) (1)
(ii)
power (= energy supplied per sec = mcΔθ) = 0.052 × 4200 × (42 - 10) (1)
= 7.0 × 103W (1)
(6.99 × 103W)
(allow C.E. for value of mass each sec from (i)
4
Page 61 of 80
(b)
h = ½gt2 gives the time to reach the floor (1)
= 0.64s (1)
(0.639 s)
range = (horizontal) speed of projection × time = 2.5 × 0.64 = 1.6 m (1)
(allow C.E. for value of t)
3
[7]
30
(a)
(i)
(use of ΔQ = mcΔθ gives)
Q = 30 × 1000 × 15 (1)
5
= 4.5 × 10 J (1)
(ii)
P × t = 4.5 × 105 (1)
t=
= 225 s (1)
(allow C.E. for value of Q from (i)
4
(b)
heat is lost to surroundings or other objects in room or to
heater itself (1)
more (thermal) energy required from heater (1)
[or because convection currents cause uneven heating]
[or rate of heat transfer decreases as temperature increases]
2
[6]
31
(a)
(i)
neutron
B1
1
(ii)
p = 36
B1
n = 144
B1
2
Page 62 of 80
(b)
(i)
total energy produced =
MJ each second
C1
number of reaction = 4.2 × 1019 per second
A1
2
(ii)
1 kg contains (1000/235) × 6.02 × 1023 atoms of uranium
C1
total number of fissions = (1000/235) × 6.02 × 1023 × 2 × 104
(5.1 × 1028)
C1
time = total fissions available/number per second or 1.2 × 109s
C1
38.7(39) years
A1
4
(iii)
too few neutrons produced to maintain the chain reaction
B1
probability of a neutron colliding with a uranium nucleus too low
B1
more absorption of neutrons in non–fission capture
B1
2
(c)
pressure = 150 × 105 (Pa) or F = PA
C1
force on 1 cm 2 = 1500N
A1
2
Page 63 of 80
(d)
energy removed each second
E=
MJ = 1.25 × 109 J or E = mc∆θ
C1
1.25 × 109 = m 5000 × 40
C1
mass per second = 6250 kg
C1
volume per second = 8.6(8.56) m3
A1
4
(e)
control rods
neutrons are absorbed
B1
by the nucleus of the boron/atoms
B1
moderator
neutrons are slowed down
B1
when colliding with the protons/hydrogen nucleus
B1
4
[21]
32
(a)
(use of ΔQ = m c ΔT)
30 × 98 = 0.100 × c × 14
c = 2100 (J kg–1 K–1)
2
Page 64 of 80
(b)
(use of ΔQ = m l + m c ΔT)
500 × 98 = 0.100 × 3.3 × 105
+ 0.100 × 4200 × ΔT
(ΔT = 38 °C)
T = 38°C
3
(c)
the temperature would be higher
as the ice/water spends more time below 25°C
or heat travels in the direction from hot to cold
or ice/water first gains heat then loses heat
any one line
2
[7]
33
(a)
(use of ΔQ = mcT gives)
ΔQ1 = 1.5 × 4200 × 18 (1)
= 1.134 × 105(J) (1)
ΔQ2 = 1.5 × 3.4 × 105 = 5.1 × 105(J) (1)
total energy released (= 1.134 × 105+ 5.1 × 105)
= 6.2 × 105J (1)
(6.23 × 105J)
4
(b)
(ice) requires energy to melt [or mention of latent heat] (1)
stays at 0 °C (for longer) (or cools for longer) (1)
(or extracts more energy from the drink)
2
[6]
34
(a)
(i)
heat water to 100 °C, energy (= 190 × 4200 × 79) = 63 (MJ) (1)
vapourise water, energy
(=190 × 2.3 × l06) = 440(MJ) (1)
(437MJ)
energy transferred (per sec) = (437 + 63) MJ (1)
(= 500 MJ)
Page 65 of 80
(ii)
mass of rocks (= 4.0 × 106 × 3200)
= 1.3 × 1010(kg) (1)
(1.28 × 1010)
temperature fall of ΔT in one day, energy removed
(= 1.28 ×1010 × 850 × ΔT) = 1.1 × 1013 ΔT (1)
(1.09 x 1013 AT)
(allow C.E. for value of mass of rocks)
energy transfer in one day (= 500 × 106 × 3600 × 24)
= 4.3 × 1013 (J) (1)
in one day
K (1)
7
(b)
number of nuclei in 1 kg of 238 U =
(1)
activity of lkg of 238U =
(1)
(1)
energy released per sec per kg of 238 U
= 1.2(6) × 107 × 4.2 × 1.6 × 10–13(J) (1)
(8.47 × 10–6(J))
mass of 238Uneeded =
= 5.9(0) × 1013kg (1)
5
[12]
35
(a)
using Q = mcΔθ
= 3.00 × 440 × (84-27) (1)
7.5 × 104 (J) (1)
2
Page 66 of 80
(b)
using Q = ml
= 1.20 × 2.5 × 104
= 3.0 × 104 (J) (1)
1
(c)
(heat supplied by lead changing state + heat supplied by cooling lead =
heat gained by iron)
3.0 × 104 + heat supplied by cooling lead = 7.5 × 104 (1)
heat supplied by cooling lead = 4.5 × 104 = mcΔθ
c = 4.5 × 104/(1.2 × (327 – 84) (1)
c = 154 (J kg–1 K–1) (1)
3
(d)
any one idea (1)
no allowance has been made for heat loss to the surroundings
or the specific heats may not be a constant over the range
of temperatures calculated
1
[7]
36
37
38
B
[1]
A
[1]
(a)
the energy required to change the state of a unit mass of water to steam / gas ✓
when at its boiling point temperature / 100°C / without a change in temperature) ✓
allow 1 kg in place of unit
allow liquid to vapour / gas without reference to water
don't allow ‘evaporation’ in first mark
2
(b)
(i)
thermal energy given by copper block ( = mcΔT)
= 0.047 × 390 × (990 – 100)
= 1.6 × 104 (J) ✓
2 sig figs ✓
can gain full marks without showing working
a negative answer is not given credit
sig fig mark stands alone
2
Page 67 of 80
(ii)
thermal energy gained by water and copper container
( = mcΔTwater + mcΔTcopper)
= 0.050 × 4200 × (100 – 84) + 0.020 × 390 × (100 – 84)
or
= 3500 (J) ✓ (3485 J)
available heat energy ( = 1.6 × 104 – 3500) = 1.3 × 104 (J) ✓
allow both 12000 J and 13000 J
allow CE from (i)
working must be shown for a CE
take care in awarding full marks for the final answer – missing out
the copper container may result in the correct answer but not be
worth any marks because of a physics error
(3485 is a mark in itself)
ignore sign of final answer in CE
(many CE’s should result in a negative answer)
2
(iii)
(using Q = ml)
m = 1.3 × 104 / 2.3 × 106
= 0.0057 (kg) ✓
Allow 0.006 but not 0.0060 (kg)
allow CE from (ii)
answers between 0.0052 → 0.0057 kg resulting from use of 12000
and 13000 J
1
[7]
39
(i)
(heat supplied by glass = heat gained by cola)
(use of mg cg ΔTg =mc cc ΔTc)
1st mark for RHS or LHS of substituted equation
0.250 × 840 × (30.0 – Tf) = 0.200 × 4190 × (Tf – 3.0)
2nd mark for 8.4°C
(210 × 30 – 210 tf = 838 Tf – 838 × 3)
Tf = 8.4(1)
Alternatives:
8°C is substituted into equation (on either side shown will get mark)
resulting in 4620J~4190J
or
8°C substituted into LHS
(produces ΔT = 5.5°C and hence)
= 8.5°C ~ 8°C
8°C substituted into RHS
(produces ΔT = 20°C and hence)
= 10°C ~ 8°C
2
Page 68 of 80
(ii)
(heat gained by ice = heat lost by glass + heat lost by cola)
NB correct answer does not necessarily get full marks
(heat gained by ice = mcΔT + ml)
heat gained by ice = m × 4190 × 3.0 + m × 3.34 × 105
(heat gained by ice = m × 346600)
3rd mark is only given if the previous 2 marks are awarded
heat lost by glass + heat lost by cola
= 0.250 × 840 × (8.41 – 3.0) + 0.200 × 4190 × (8.41 – 3.0)
(= 5670 J)
(especially look for m × 4190 × 3.0)
the first two marks are given for the formation of the substituted
equation not the calculated values
m (=5670 / 346600) = 0.016 (kg)
if 8oC is used the final answer is 0.015 kg
or (using cola returning to its original temperature)
(heat supplied by glass = heat gained by ice)
(heat gained by glass = 0.250 × 840 × (30.0 – 3.0))
heat gained by glass = 5670 (J)
(heat used by ice = mcΔT + ml)
heat used by ice = m(4190 × 3.0 + 3.34 × 105)
(= m(346600))
m (=5670 / 346600) = 0.016 (kg)
3
[5]
Page 69 of 80
Examiner reports
1
2
This question was performed well by a majority of students. The explanation of a specific heat
capacity in part (a) was very straightforward. The calculation in part (b) was done well by all but
the weakest students even though it contained parts dealing with both specific heat capacity and
latent heat. It was in choosing an incorrect number of significant figures that students lost the
most marks.
(a)
This was usually completed successfully. A minority gave the deceleration rather than the
decelerating force as the answer.
(b)
The majority decided to use the work done = Fs approach rather than the approach using
Δ(½mv2) / t and there were a large number of correct answers to the question. Although
the majority realised that they had to use work done = Fs, many failed to determine s
correctly with a significant number arriving at a stopping distance of 75 m. These
candidates used s = ut + ½at2 but used +4.0 m s–2 for a instead of –4.0 m s–2.
(c)
The majority appreciated that all that was necessary was to divide their answer to (b) by 2.5
s so that most were able to gain the two marks for this part allowing any error carried
forward.
(d)
(i)
This was often completed successfully. A common error was to use the value for rate
of energy dissipation from part (c), instead of the total energy from part (b). The
majority who were on the right track either divided the energy by 4 or multiplied the
mass by 4 when using Q = mcΔθ. Calculating Δθ for one brake and then dividing by
4 was an unconvincing approach.
(ii)
Many stated that the temperature rise of the brakes would be lower but were unable
to give a convincing explanation. There were many who thought that friction when the
brakes were applied had not been taken into account and that ‘energy would be lost
to friction’. Others referred to energy dissipated due to friction between the tyres and
the road but thought that this would increase the temperature rise of the brakes.
Page 70 of 80
3
(a)
(i)
There was a significant number of candidates who attempted this part by using W =
PΔV or W = PV. They made statements that because the two work values were the
same then there was no change in the internal energy. Other candidates thought that
PV = temperature.
Only the higher achieving candidates were able to produce convincing work that was
correct and well-laid out with due regard to the fact that this was a “Show that...” style
question.
(ii)
The application of the first law of thermodynamics to the adiabatic expansion was well
understood and effectively communicated by most candidates. Some candidates
were unaware that this expansion was adiabatic and neglected to address this in
their explanation.
(iii)
Over 50% of candidates achieved all 3 marks in this part. However, in many cases
the working was unconvincing with candidates opting for
workings. A few candidates did not realise that
without any supporting
should have been
expressed as a decimal fraction in keeping with standard procedure for numerical
answers.
(b)
Many candidates attempted to use an approximation using W = Pmean × ΔV. This was
equivalent to approximating the area under the curve to that of a trapezium. Such
approaches were acceptable for 1 mark but required a statement identifying the answer as
an over-estimate in order to access both marks.
The usual method for estimating an area was not seen as often as expected and even
when seen it was done in an unconvincing manner. Again, with a “Show that...” style
question working had to be clearly communicated; many candidates were unable to do this
effectively and consequently dropped marks.
(c)
Just under half of the candidates were able to deal with this multi-stage calculation and
obtain all 3 marks. A common error was for candidates to convert the temperature increase
of 18 °C to 291 K and substituting this value into Q = mcΔθ . Other common mistakes were
an inability to deal with the per-second aspect of the question and also quoting or
rearranging ρ =
7
8
incorrectly.
(a)
Many candidates confused heat capacity with specific heat capacity. Otherwise, this
calculation was well done.
(b)
Being more familiar to candidates, this calculation was done better than that in part (a).
Those who calculated the energy incorrectly in (a) were not further penalised in this part.
(a)
(i)
Most candidates quoted the correct equation. The majority also carried on to
complete the calculation correctly but a significant number used centigrade
temperature instead of absolute temperature.
Page 71 of 80
(b)
9
(ii)
This was done correctly by most candidates. Some of those who had previously used
centigrade temperatures corrected their mistakes here.
(iii)
Although the correct process was carried by most candidates, the use of incorrect
data was common. Unit penalties and significant figure penalties were common on
this part of the question.
(iv)
Many candidates explained what is meant by the word adiabatic but fewer could go
on to say why the process described in the question would be approximately
adiabatic.
(v)
Candidates tended to mention either the more frequent collisions with the wall of the
syringe or the fact that the molecules moved faster. Relatively few mentioned both
factors. Even fewer went further and explained how these factors would affect the
rate of change of momentum of the molecules and, hence, the pressure exerted on
the walls of the syringe.
(i)
This was generally well done.
(ii)
This was also answered correctly by many candidates but some were insufficiently
clear about the direction of the process.
This question was very well answered with many candidates scoring maximum or nearly
maximum marks. Most candidates knew in part (a) that the momentum before a collision
equalled the momentum after the collision, but rather few gave the condition that no external
force must act on the system. Part (b) was almost always correct, although some candidates did
not add the masses of the bullet and block.
Parts (c)(i) and (c)(ii) were usually correct. In section (iii) some candidates failed to subtract the
remaining kinetic energy of 5.0 J from the initial kinetic energy of 200 J, or alternatively used 5.0
J for the internal energy. Part (d) was most often correct, but a number of candidates used ʋ2 =
u2 + 2as and scored no marks.
10
(a)
Precise and clear explanations of the specific heat capacity statement were quite rare.
There was often ambiguity in wording that was inappropriate at A2 level.
(b)
(i)
The majority of candidates were able to carry out and structure this three-part
calculation well to yield the correct answer. Examiners were lenient with those who
presented their solution poorly, but candidates need to recognise that it is difficult to
award full credit if information about the intermediate stages in the answer is missing.
(ii)
This was a simple calculation and was done well by all those who realised how
straightforward it was. However, a minority felt the need to re-work only the latent
heat energy part of the previous calculation and lost marks accordingly.
Page 72 of 80
11
12
This question was answered well with many high scores of 5 or 6.
(a)
Some candidates carried out the evaluation using the temperature difference with 273
added to it. Significant figure and unit error penalties were very common.
(b)
Well done even by those who failed in (a).
(c)
A very common error was to forget to add the water power value into the total power
involved in the system. Candidates who did this could still gain some credit however for a
partially correct efficiency calculation.
(a)
This part was generally well answered; the most common mistake was to quote the formula
correctly but to forget to square the speed in the calculation. Weaker candidates forgot to
convert km s−1 into ms−1.
(b)
Again this part was generally well answered. A minority of candidates added 273 onto the
change of temperature before substituting their value into the specific heat capacity
equation.
(c)
(i)
A minority of candidates calculated the weight of the meteorite (using, F = mg).
Candidates approaching this part using an equation of motion frequently calculated
the maximum force rather than the average force. A considerable number of
candidates used the value for time, calculated in part (ii) and substituted this into the
relationship between force and rate of change of momentum.
(ii)
This part was generally well answered although many candidates suffered a
significant figure penalty for quoting the time either as □ s or as
(iii)
13
s.
A significant number of candidates made frequent use of P = Fv, but very few
recognised that this meant that F needed to be the average force and v the average
velocity in this example.
It seemed that some centres had overlooked, or not had sufficient time to cover, the topic of heat
calculations. Definitions of specific heat capacity and specific latent heat in part (a) were often
very good and were awarded all four marks. A common misconception was that heat must be
supplied to water in order to change it to ice.
The calculations in part (b) were sometimes missed out completely or often misunderstood, but
occasionally completed perfectly. Many candidates who were able to justify the 24W in part (b)(i)
were then stumped by parts (b)(ii) and (b)(iii). In both of these parts the incorrect use of 7200J
was very common. The answer to part (b)(iii) was often penalised because it was quoted to an
excessive number of significant figures.
Page 73 of 80
14
The anticipated emphasis in answers to part (a)(i) was on how thermal equilibrium is manifested,
whilst in part (a)(ii) it was on a molecular explanation. In general, higher marks would have been
awarded had the candidates read the question more carefully. Many answers simply fudged the
two different aspects. Most candidates seem to have gained only the most basic understanding
of molecular behaviour, a problem which pervaded the whole question. In particular, candidates
should realise that the molecules in any system have a range of energies and that temperature is
a measure of their mean kinetic energy. Thermal equilibrium may be satisfactorily explained by
considering two systems in contact. When in equilibrium there is no net flow of energy between
them because both systems are at the same temperature. However, there is a microscopic flow
of energy in each direction between them. It is untrue that “every molecule has the same kinetic
energy” when equilibrium has been established, as was stated in most answers.
Very few candidates were able to calculate the r.m.s. speed of the helium molecules in part (b)(i).
The main problems, also seen in other similar questions in the recent past, were confusion
between molar and molecular quantities and an inability to differentiate between mean square
values and r.m.s. values. Many candidates recovered in part (b)(ii) to gain a couple of marks by
using 3k T / 2 or its molar counterpart. Only a handful of candidates spotted that the average
kinetic energy of a nitrogen molecule would be the same as that of a helium molecule, and that
the result from part (b)(i) could be combined with ½ mʋ2 to arrive at the answer. There were
many correct numerical answers to part (b)(iii) but most were arrived at instinctively; justification
of (2 / 3) of 120KPa was expected for one of the two marks.
15
(a)
(b)
16
(i)
There were many good definitions, but many seemed unfamiliar with the idea of
providing a formal definition/explanation. Many answers either forgot the ‘per kg’ or
‘at the boiling point’ aspect, so gained one of the two available marks. Poorer
answers ignored the word specific completely, so that answers referred only to the
energy to change from liquid to vapour.
(ii)
This was not done well and relatively few good answers were seen. Few seemed to
have come across the idea of the energy needed to expand the vapour or gas
against atmospheric pressure when boiling. Many wrote about it requiring more
energy to break liquid bonds than to make solid bonds, apparently not appreciating
that fusion is a melting process not a solidifying process. Answers that provided
explanations in terms of more energy needed to break liquid bonds than to break
solid bonds, without reference to the large increase in separation, were not
convincing.
(i)
This was not done well and, surprisingly, few completely correct answers were seen.
Many of those who found the mass of steam needed correctly then failed to add the
0.25 kg. Most knew the formulae to use but were at a loss to know what to do with
the data. Often the 0.25 kg was used in the calculation of ml, and this was then
equated to mcΔθ (giving an answer of 1.6 kg). Others simply used the 0.25 kg in ml
and mcΔθ making no progress to the answer.
(ii)
Many candidates gave a satisfactory response. Those who did not often wrote about
the condensed steam which had already been accounted for in (i).
The vast majority of candidates found this a very accessible question and marks were generally
very high.
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(a)
The correct formula was used by most candidates. The main errors arising were failure to
do the final reciprocal, inappropriate significant figures and units.
(b)
(i)
This was usually successfully completed.
(ii)
This was also usually correct.
(iii)
Although there were many correct answers some confused V and Vo, others used
the value of charge from (i) and some were unable to do the calculation having
substituted correctly.
(i)
The majority knew which formula to use but a significant proportion did not know how
to calculate the mass and some used the value for charge from (b) (i) for the energy
(confusing Q in Q = VC and Q = mc∆θ).
(ii)
The majority gained credit for appreciating losses to the surroundings but few referred
to the effect of the heat capacity of the thermometer which is the other most
significant reason for the lower temperature. Some suggested that the lower
temperature was
because not all the heat energy supplied went to the
thermometer.
(c)
17
18
The last question in the paper was generally done well and candidates across the ability range
were able to perform the various calculations successfully. A minority were confused by the need
to use only 60% of the kinetic energy, but because marks were awarded for consequential errors,
this did not prove to be too much of a penalty.
A large number of candidates found this question difficult. Many seemed aware of the correct
equation in part (a) but made mistakes such as using power instead of energy, converting
temperatures to Kelvin and using an incorrect value for the mass of water.
The latent heat calculation in part (b) produced even fewer correct responses. Many candidates
included a change of temperature in their calculations or used an incorrect value for the mass of
steam. Part b(ii) realised better answers although stating only one assumption was quite
common.
19
Most candidates made progress in parts (a) and (b) and gained some credit. The most common
error was due to the failure to recognise that specific latent heat was given in kJ kg –1 rather than
J kg–1 . These candidates knew how to proceed in part (iii) but some used the molar mass
instead of the mass of the atom or used the specific latent heat instead of the energy supplied to
each atom.
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20
21
22
In part (a) almost all students knew the correct equation to use and only the less able students
made errors. The first of these was to use the mass of water in the heating chamber rather than
the rate of flow of water. The second error, which was less common, was to try to convert
between Kelvin and Celsius by adding 273 to the answer. Again in part (b) it was only the less
able students who had any difficulty. The problem was that they could not cope with being given
the rate of supply of energy. Overall the question was done well.
The calculations did not cause too many problems, but a minority of candidates were confused
between the energy provided by the heating element and the energy required to heat the water
and often the answers were interchanged. Part (b) proved to be more difficult and only a few
candidates scored both marks available.
Plotting the graph in part (a) was, by and large, well done and the majority of candidates were
able to gain the allocated three marks. Again, calculating the gradient of the graph in part (b) was
satisfactorily done although a minority of candidates found the inverse of the gradient.
Parts (c) and (d) proved to be very good discriminators. The better candidates tackled the
calculations impressively, but weaker candidates seemed unsure as to where to begin. The unit
for specific heat capacity did cause a lot of problems to even the best candidates.
23
The majority of candidates were able to calculate the electrical energy supplied to the kettle in
part (i) and the heat energy supplied to the water in part (ii). Significant figure penalties were
imposed in part (ii). In part (iii), some candidates considered that the kettle cable became heated
because of its resistance and many missed the point that there was heat loss from the kettle to
the surroundings.
In part (b) many candidates scored well, particularly in part (i). Candidates often lost a mark in
part (ii) through an incorrect calculation of the cross-sectional area or by using an incorrect
formula for the cross-sectional area. Weak candidates often penalised themselves heavily as a
result of being unable to rearrange the resistivity equation correctly.
24
This question was well answered and candidates were consistently able to extract and use
appropriate formulae in their calculations. In part (a), a minority of candidates used 293 K as the
change in temperature rather than 20 K, but this was less prevalent than in previous years.
There were problems with stating the appropriate assumptions in part (b). Many candidates
stated incorrectly that no heat was lost to the surroundings.
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25
Most candidates were awarded full marks in part (a) although some candidates added 273 to the
temperature difference. A few candidates lost the mark in part (ii) because they failed to describe
the correct relationship between power and energy.
In part (b)(i), a significant number of candidates obtained the correct value of current but then
divided, or multiplied, by
, clearly unaware that the required answer had already been
obtained. In part (ii), many candidates scored full marks with a clear and carefully expressed
calculation, but a few candidates did lose one or more marks as a result of failure to use the
correct value for the radius or the correct expression for the cross-sectional area. A significant
number of candidates were awarded only one mark in part (iii) because they had not obtained
the correct value of resistance per metre in part (ii) or else they failed to appreciate that the cable
contained two wires. In general, candidates who scored well in part (ii) usually scored both marks
in part (iii). In part (iv), most candidates gained the available mark, including weaker candidates
who were unable to make progress in the earlier parts of part (b).
26
27
Many candidates scored full marks in this question. Less able candidates created the usual
confusion between the Kelvin and Celsius scale, but this was less of an issue than has been the
case in the past. Candidates are clearly benefiting from practice with thermal energy questions.
Many answers in part (a) lacked sufficient detail to gain more than one or two marks. Although
candidates knew that the mass of an α particle was greater than that of a β particle, few stated
that its mass or momentum was much greater. Again, candidates knew that an α particle had
more charge than a β particle but few stated it had twice as much charge. Few candidates
realised that an α particle travels much slower than a β particle with the same kinetic energy;
some even claimed that it travelled faster because it had more momentum. In addition,
candidates often wrote at length about the relative penetrating powers of the two types of
radiation.
In part (b) (i), many candidates gave a correct calculation although some candidates made
arithmetical errors, often in the conversion of MeV to J, or in rounding the answer incorrectly. In
part (ii), the underlying principles behind the calculation were known but some candidates failed
to realise that the increase in temperature in one minute was required.
28
Questions involving thermal energy have caused problems for candidates in previous papers, but
this question was well answered. It also proved to be a good discriminator. Weaker candidates
did have problems, however, in structuring their calculations in a logical way, leading to errors in
the final answers.
The explanation in part (c) was answered well by good candidates who, in some cases, provided
considerable detail which was worthy of more than the two marks allocated.
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29
A large majority of the candidates gained full credit in part (a) (i) and were able to calculate the
power correctly. However, the temperature difference confused other candidates, not realising
that the temperature difference in °C is the same as in Kelvins. Some candidates lost the final
mark as a result of a significant figure error or giving the power in joules and not watts or joules
per second. Arithmetical errors leading to grossly unrealistic answers were made by a few
candidates.
In part (b), the time of descent needed to be calculated from the height drop in order to determine
the horizontal distance travelled. Many candidates were able to do this correctly. Other
candidates failed to gain any credit as a result of attempting to bring the speed of projection into
the calculations for time of descent and the horizontal distance.
30
31
This question was generally answered well with only a minority of candidates failing to complete
the calculation. The explanation as to why actual heating might take longer generated a variety of
explanations some of which were quite imaginative.
The neutron was identified by most students in part (a) (i).
There were many correct answers to part (a) (ii) but less able students did not take account of
the two ‘X‘ particles on the right hand side of the equation so obtained n= 145.
Most made some progress with part (b) (i). Obtaining the total energy produced by the reactor
proved difficult for less able students.
There were a good proportion of correct answers in part (b) (ii). Some progress was made by a
majority of the students.
Few were able to make a sensible comment in part (b) (iii).
Most were able to gain one of the two available marks in part (c) but the conversion from cm2 to
m2 was the downfall of many students.
Whilst many were able to gain credit in part (d) for making some progress, relatively few could
correctly complete the problem.
Most showed an appreciation of the different roles of the moderator in part (e) and the control
rods but the majority gave inadequate detail of the processes.
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32
Most candidates performed well in part (a).
In part (b) the less able candidates tended to score only one mark because they could not form
the energy balance equation when both changes of temperature and changes of state were
taking place.
Part (c) caught a majority of candidates out. Even grade A students were tempted to roll out the
usual answer, ‘the temperature would be less because heat is lost to the surroundings’. This
statement scored no marks.
33
The calculation in part (a) was done well and candidates seemed to be quite happy calculating
thermal energies. Some were still confused by the temperature scales and added 273 to the
change in temperature, not realising that a difference in two temperatures on the Celsius scale
will be in Kelvin.
Part (b) generated some very interesting responses. It seems to be a common misapprehension
that when ice melts it gives out energy and this cooled the cans of drinks. Candidates seem
much happier discussing heat losses when something cools rather than when it warms up. Less
able candidates discussed ice transferring ‘coldness’, and energy changes during a change of
state is clearly a topic that is not well understood.
34
The calculation, in part (a) (i), of the energy needed to heat the water and to turn it to steam at
100°C was usually done correctly. Some failed by choosing an incorrect temperature change. In
part (ii), many candidates obtained full marks although some correctly worked out the
temperature change per second, but did not proceed to calculate the temperature change in one
day. A small minority of candidates used an incorrect density formula or an incorrect value of the
specific heat capacity.
In part (b) most candidates knew how to convert MeV into joules correctly and used the given
mass number correctly. They also knew how to calculate the total activity of the rocks from the
energy released per second and the energy released per decay. Many of these candidates
correctly used their activity value and the decay constant, worked out from the half-life, to
calculate the total number of atoms and their mass. Some lost a mark in their calculation of the
decay constant as a result of not converting the half-life into seconds. A significant number of
candidates considered their answer for the total activity to be the total number of atoms and
consequently lost two marks because they failed to use the total activity and the decay constant
to calculate the total number of atoms.
Page 79 of 80
35
Almost all candidates knew which equation to use in part (a) and only a small minority used the
wrong temperature change.
In part (b) most candidates obtained full marks.
Part (c) turned out to be a very good discriminator. About one third of candidates were not using
the heat energy released by the lead, as it cooled, in their calculation. These candidates either
used their answer to (a) or (b) or the sum of the two. In addition another 10% calculated the
incorrect temperature change.
Part (d) was answered well on the whole. The most common error by candidates was to not say
where the heat energy might go in their answer. Candidates simply said that heat is lost.
38
A majority of candidates only scored one mark in part (a). These candidates either forgot to
indicate a unit mass or, as in a majority of cases, they omitted the phrase, 'without a change in
temperature', or equivalent. A few had problems in appreciating whether energy was required or
whether energy was given out. It was very noticeable that at the lower ability end candidates
have a poor vocabulary associated with this area of physics. Phrases like, 'to change water to a
gas without changing state', or 'condense water into steam', and others showed a lack of
distinction between boiling and evaporation.
The calculation of part (b)(i) did not hold many difficulties for the bulk of the candidates but the
significant figure issue did. In part (b)(ii) most candidates were relatively clear how to tackle this
question. It was in the detail that errors were made. The most significant was to forget about the
copper can, which also gained energy to reach the final temperature. Also at the lower ability end
there were many opportunities to make arithmetic errors.
The scores were much better for part (b)(iii) albeit from an error carried forward from part (b)(ii) in
many cases. So the use of the latent heat equation is not difficult to grasp for a majority of
candidates. The main error was from rounding off incorrectly or making errors in powers of 10
when converting to SI units.
39
Candidates found (i) quite difficult for a number of reasons. Some started correctly by equating
heat supplied to glass equals heat gained by cola but then they could not make the final
temperature the subject of the resulting equation. Others substituted the temperature the wrong
way round and used (3 − Tf), which was negative and fudged the arithmetic. As in a previous
question candidates did not explain their approach which made it difficult to award partial marks.
It was interesting to see some candidates who jumped in too quickly and made an initial mess of
the calculation fared better on additional pages when they thought more carefully over the
problem.
Part (ii) was also very discriminating. Only the best candidates scored full marks. Good
candidates who just missed full marks usually forgot about the 3 degree rise in temperature of
the ice after it had melted. Most other candidates were aware of the mcΔT and ml equations but
then made all manner of different errors.
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