Thermal Physics Unit 5
166 minutes
153 marks
Q1.
(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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(ii)
State two assumptions made in your calculation.
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(4)
(Total 8 marks)
Q2.
(a) A cylinder of fixed volume contains 15 mol of an ideal gas at a pressure of 500 kPa
and a temperature of 290 K.
(i)
Show that the volume of the cylinder is 7.2 × 10–2 m3.
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(ii)
Calculate the average kinetic energy of a gas molecule in the cylinder.
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(4)
(b)
A quantity of gas is removed from the cylinder and the pressure of the remaining gas falls
to 420 kPa. If the temperature of the gas is unchanged, calculate the amount, in mol, of
gas remaining in the cylinder.
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(2)
(c)
Explain in terms of the kinetic theory why the pressure of the gas in the cylinder falls when
gas is removed from the cylinder.
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(4)
(Total 10 marks)
Q3.
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)
(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)
Q4.
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)
(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)
Q5.
(a) The air in a room of volume 27.0 m3 is at a temperature of 22 °C and a pressure of
105 kPa.
Calculate
(i)
the temperature, in K, of the air,
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(ii)
the number of moles of air in the room,
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(iii)
the number of gas molecules in the room.
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(5)
(b)
The temperature of an ideal gas in a sealed container falls. State, with a reason, what
happens to the
(i)
mean square speed of the gas molecules,
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(ii)
pressure of the gas.
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(4)
(Total 9 marks)
Q6.
The figure below shows a tube containing small particles of lead. When the tube is inverted
the particles of lead fall freely through a vertical height equal to the length of the tube.
(a)
Describe the energy changes that take place in the lead particles during one inversion of
the tube.
You may be awarded marks for the quality of written communication in your answer.
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(3)
(b)
The tube is made from an insulating material and is used in an experiment to determine
the specific heat capacity of lead. The following results are obtained.
mass of lead:
50length of tube:
K
0.025 kgnumber of inversions:
1.2 mchange in temperature of the lead:
4.5
Calculate
(i)
the change in potential energy of the lead as it falls after one inversion down the
tube,
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(ii)
the total change in potential energy after 50 inversions,
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(iii)
the specific heat capacity of the lead.
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(4)
(Total 7 marks)
Q7.
The graph shows how the pressure of an ideal gas varies with its volume when the mass
and temperature of the gas are constant.
(a)
On the same axes, sketch two additional curves A and B, if the following changes are
made.
(i)
The same mass of gas at a lower constant temperature (label this A).
(ii)
A greater mass of gas at the original constant temperature (label this B).
(2)
(b)
A cylinder of volume 0.20 m3 contains an ideal gas at a pressure of 130 kPa and a
temperature of 290 K. Calculate
(i)
the amount of gas, in moles, in the cylinder,
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(ii)
the average kinetic energy of a molecule of gas in the cylinder,
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(iii)
the average kinetic energy of the molecules in the cylinder.
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(5)
(Total 7 marks)
Q8.
The number of molecules in one cubic metre of air decreases as altitude increases. The
table shows how the pressure and temperature of air compare at sea-level and at an altitude of
10 000 m.
(a)
altitude
pressure/Pa
temperature/K
sea-level
1.0 × 105
300
10 000 m
2.2 × 104
270
Calculate the number of moles of air in a cubic metre of air at
(i)
sea-level,
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(ii)
10 000 m.
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(3)
(b)
In air, 23% of the molecules are oxygen molecules. Calculate the number of extra oxygen
molecules there are per cubic metre at sea-level compared with a cubic metre of air at an
altitude of 10 000 m.
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(2)
(Total 5 marks)
Q9.
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–1specific heat capacity of copper =
J kg–1 K–1
(a)
390
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)
Q10.
(a) The volume of air in a fully expanded pair of human lungs is 5.0 × 10–3 m3. The
pressure of the air in the lungs is 1.0 × 105 Pa and its temperature is 310 K.
Calculate
(i)
the number of moles of air in the lungs,
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(ii)
the average kinetic energy of an air molecule in the lungs.
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(4)
(b)
Air is a mixture of oxygen and nitrogen molecules. The mass of an oxygen molecule is
greater than the mass of a nitrogen molecule. State and explain the effect this has on the
mean square speeds of the oxygen and nitrogen molecules in the lungs.
You may be awarded additional marks to those shown in brackets for the quality of written
communication in your answer.
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(3)
(Total 7 marks)
Q11.
(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
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(4)
(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.
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(2)
(Total 6 marks)
Q12.
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–1Specific heat capacity of
copper = 390 J kg–1 K–1
(a)
(i)
The temperature of the water rises from 15ºC to 35ºC. Calculate the thermal energy
gained by the water.
thermal energy gained = ....................................
(ii)
Calculate the temperature reached by the copper in the flame. Assume no heat is
lost when the copper is transferred.
temperature = ....................................
(4)
(b)
When the lump of copper entered the water, some of the water was turned to steam.
(i)
The specific latent heat of vaporisation of steam is 2.25 MJ kg–1. What further
measurement would need to be made to calculate the energy used to produce this
steam?
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(ii)
Without further calculation, describe how this further measurement should be used
to obtain a more accurate value of the flame temperature.
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(3)
(Total 7 marks)
Q13.Some liquids in open bottles deteriorate exposure to air. The diagram below shows one device
used to reduce this deterioration. It consists of a rubber valve that is inserted into the neck of
the bottle together with a pump that is used to remove some of the air in the bottle through this
rubber valve. On an up-stroke of the pump, air enters the pump chamber from the bottle. On the
down-stroke, the rubber valve closes and the air in the chamber is expelled to the atmosphere
through another valve (not shown) in the handle.
(a)
There is 3.5 × 10–4 m3 of air space in the bottle and the volume of the pump chamber
changes from zero at the beginning of the up-stroke to 6.5 × 10–5 m3 at the end of the upstroke. The initial pressure of the air in the bottle is that of the atmosphere with a value of
99 kPa.
Assuming the process is at constant temperature, calculate the pressure in the bottle after
one up-stroke of the pump.
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(3)
(b)
Calculate the number of molecules of air originally in the air space in the bottle at a
temperature of 18 °C.
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(3)
(c)
Explain how the kinetic theory of an ideal gas predicts the existence of a gas pressure
inside the bottle. Go on to explain why this pressure decreases when some of the air is
removed from the bottle.
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(5)
(Total 11 marks)
Q14.
Figure 1
Figure 1 shows a cylinder, fitted with a gas-tight piston, containing an ideal gas at a constant
temperature of 290 K. When the pressure, p, in the cylinder is 20 × 104 Pa the volume, V, is
0.5 × 10–3 m3.
Figure 2 shows this data plotted.
Figure 2
(a)
By plotting two or three additional points draw a graph, on the axes given in Figure 2, to
show the relationship between pressure and volume as the piston is slowly pulled out.The
temperature of the gas remains constant.
(3)
(b)
(i)
Calculate the number of gas molecules in the cylinder.
answer = ...................................... molecules
(2)
(ii)
Calculate the total kinetic energy of the gas molecules.
answer = .................................................... J
(3)
(c)
State four assumptions made in the molecular kinetic theory model of an ideal gas.
(i)
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(ii)
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(iii)
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(iv)
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(4)
(Total 12 marks)
Q15.
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)
(d)
State one reason why the answer to part (c) is only an approximation.
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(1)
(Total 7 marks)
Q16.
A fixed mass of ideal gas at a low temperature is trapped in a container at
constantpressure. The gas is then heated and the volume of the container changes so that the
pressure stays at 1.00 × 105 Pa. When the gas reaches a temperature of 0 °C the volume is
2.20 × 10–3m3.
(a)
Draw a graph on the axes below to show how the volume of the gas varies
withtemperature in °C.
(2)
(b)
Calculate the number of moles of gas present in the container.
answer = .................................moles
(2)
(c)
Calculate the average kinetic energy of a molecule when this gas is at a
temperature of 50.0 °C. Give your answer to an appropriate number of
significant figures.
answer = .........................................J
(3)
(d)
Calculate the total internal energy of the gas at a temperature of 50.0 °C.
answer = .........................................J
(1)
(e)
By considering the motion of the molecules explain how a gas exerts a pressure and why
the volume of the container must change if the pressure is to remain constant as the
temperature increases.
The quality of your written communication will be assessed in this question.
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(6)
(Total 14 marks)
Q17.
The pressure inside a bicycle tyre of volume 1.90 × 10–3 m3 is 3.20 × 105 Pa when the
temperature is 285 K.
(i)
Calculate the number of moles of air in the tyre.
answer = ................................... mol
(1)
(ii)
After the bicycle has been ridden the temperature of the air in the tyre is 295 K.
Calculate the new pressure in the tyre assuming the volume is unchanged.
Give your answer to an appropriate number of significant figures.
answer = ...................................... Pa
(3)
(b)
Describe one way in which the motion of the molecules of air inside the bicycle tyre is
similar and one way in which it is different at the two temperatures.
similar ............................................................................................................
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different ..........................................................................................................
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(2)
(Total 6 marks)
Q18.
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)
Q19.(a)
Outline what is meant by an ideal gas.
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(2)
(b)
An ideal gas at a temperature of 22 °C is trapped in a metal cylinder of volume 0.20 m3 at
a pressure of 1.6 × 106 Pa.
(i)
Calculate the number of moles of gas contained in the cylinder.
number of moles ....................................... mol
(2)
(ii)
The gas has a molar mass of 4.3 × 10−2 kg mol−1.
Calculate the density of the gas in the cylinder.
State an appropriate unit for your answer.
density ......................... unit ..............
(3)
(iii)
The cylinder is taken to high altitude where the temperature is −50 °C and the
pressure is 3.6 × 104 Pa. A valve on the cylinder is opened to allow gas to escape.
Calculate the mass of gas remaining in the cylinder when it reaches equilibrium with
its surroundings.
Give your answer to an appropriate number of significant figures.
mass .......................................... kg
(3)
(Total 10 marks)
Q20.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)
(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−1specific latent heat of fusion of ice =
3.34 × 105 J kg−1
mass .......................................... kg
(3)
(Total 5 marks)
M1.
(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]
M2.
(a)
(i)
p V = nR T (1)
V=
(ii)
(1) (gives V = 7.2 × 10–2m3)
(use of Ek =
kT gives) Ek =
× 1.38 × 10–23 × 290 (1)
= 6.0 × 10–21 (J) (1)
4
(b)
(use of pV = nRT gives)
[or use p
n]
(1)
n = 13 moles (1) (12.5 moles)
2
(c)
pressure is due to molecular bombardment [or moving molecules] (1)
when gas is removed there are fewer molecules in the cylinder
[or density decreases] (1)
(rate of) bombardment decreases (1)
molecules exert forces on wall (1)
is constant (1)
[or pV =
Nm (c2) (1)
V and m constant (1)
(c2) constant since T constant (1)
p
N (1)]
[or p =
p(c2) (1)
explanation of ρ decreasing (1)
(c2) constant since T constant (1)
p (c2) ρ (1)]
max 4
[10]
M3.
(a)
(ii)
(i)
(use of ΔQ = mcΔθ gives) energy lost by water
= 0.20 × 4200 × 20 (1)
= 1.7 × 104 J (1) (1.68 × 104 J)
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]
M4.
(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]
M5.
(a)
(i)
T (=273 + 22) = 295 (K) (1)
(ii)
pV = nRT (1)
105 × 103 × 27 = n × 8.31 × 295 (1)
n = 1160 (moles) (1)
(1156 moles)
(allow C.E. for T (in K) from (i)
(iii)
N = 1156 × 6.02 × 1023 = 7.0 × 1026 (1)
(6.96 × 1026)
5
(b)
(i)
decreases (1)
because temperature depends on mean square speed (or
[or depends on mean Ek] (1)
(ii)
)
decreases (1)as number of collisions (per second) falls (1)rate of change of
momentum decreases (1)
[or if using pV = nRTdecreases (1)as V constant (1)as n constant (1)][or if using p =
1/3ρ
decrease (1)as ρ is constant (1)as
is constant (1)]
max 4
[9]
M6.
(a) lead particles fall and lose potential energy (as tube is inverted) (1)
this is converted to kinetic energy (1)
kinetic energy converted into heat energy on impact (1)
3
QWC 1
(b)
(i)
Ep (= mgh) = 0.025 × 9.81 × 1.2 = 0.29(4) J (1)
(ii)
total change of energy (= 50 × 0.294) = 15 J (1) (14.7 J)
(use of 0.29 gives 14.(5) J)
(allow C.E. for value of Ep from (i))
(iii)
(use of ∆Q = mc∆θ gives) 14.7 = 0.025 ×c × 4.5 (1)
c = 131 J kg–1 K–1 (1) (130.7 J kg–1 K–1)
(use of 15 from (ii) gives 133 J kg–1 K–1)
(allow C.E. for value of energy from (ii))
4
[7]
M7.
(a)
(ii)
(i)
curve A below original, curve B above original (1)
both curves correct shape (1)
2
(b)
(i)
(use of pV = nRT gives) 130 × 103 × 0.20 = n × 8.31 × 290 (1)
n = 11 (mol) (1) (10.8 mol)
(ii)
(use of Ek =
kT gives) Ek =
× 1.38 × 10–23 × 290 (1)
= 6.0 × 10–21 J (1)
(iii)
(no. of molecules) N = 6.02 × 1023 × 10.8 (= 6.5 × 1024)total k.e. = 6.5 × 1024 × 6.0 ×
10–21 = 3.9 × 104 J (1)(allow C.E. for value of n and Ek from (i) and (ii))(use of n = 11
(mol) gives total k.e. = 3.9 (7) × 104 J)
5
[7]
M8.
(a)
(i)
(use of
gives)
(1)
= 40(.1) moles (1)
(ii)
= 9.8(1) moles (1)
3
(b)
(total) = (40 × 6 × 1023) − (9.8 × 6 × 1023) = 1.8(1) ×1025 (1)
(allow C.E. for incorrect values of n from (a))
(oxygen molecules) = 0.23 × 1.8 ×1025 = 4.2 × 1024 (1)
2
[5]
M9.
(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]
M10.
(a)
(i)
(use of pV = nRT gives)
(1)
=0.194 moles (1)
(ii)
average kinetic energy = 3/2 × 1.38 × 10-23 × 310 (1)
= 6.4 × 10–21J (1)
4
(b)
same temperature (1)
(hence) same average kinetic energy (1)
(since) kinetic energy is mv2 (1)
nitrogen molecules must have a higher mean square speed (1)
3
[7]
M11.
(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]
M12.
thermal energy gained by water = 0.45 × 4200 × (35 – 15)= 3.78 × 104 J (1)
(a)
(i)
(ii)
(thermal energy loss by copper = thermal energy gained bywater gives)
0.12 × 390 × ΔT = 3.78 × 104 (1)
ΔT =
= 808 K
flame temperature (= 808 + 35ºC) = 843ºC or 1116 K (1)
3
(b)
(i)
measure the total mass of the water, beaker and iron lump
(to find the mass of water lost) (1)
mass of water lost due to conversion to steam,
m = mass measured in (b) (i) - initial mass of water, beaker
and iron (1)
add the thermal energy due to steam produced, mL, to the
thermal energy gained by the water (1)
calculated flame temperature would be greater (1)
4
[7]
M13.(a)
use of pV = constant or p1V 1 = p2V2 (1)
p = 99 × 3.50/4.15 (1)
= 83.5 kPa (1)
3
(b)
no. of moles = 99 [103] × 3.5 × 10–4/8.31 × 291 (1)
= 1.4(3) × 10–2 moles (1)
no. of molecules (= 1.4(3) × 102 × 6.02 × 1023)
= 8.6 × 1021 (1)
3
(c)
molecules/particles have momentum (1)
momentum change at wall (1)
momentum change at wall/collision at wall leads to force (1)[allow impulse arguments]
less air so fewer molecules (1)
so change in momentum per second/rate of change is less[or per unit per time] (1)
pressure is proportional to number of molecules (per unit volume) (1)
max 5
[11]
M14.
(a)
3
curve with decreasing negative gradient that passes through the given point
which does not touch the x axis (1)
designated points
pressure/104 Pa
volume/10–3 m3
10
1.0
5.0
2.0
4.0
2.5
2.5
4.0
2 of the designated points (1)(1) (one mark each)
(b)
(i)
N = PV/kT = 5 × 104 × 2 × 10–3/1.38 × 10–23 × 290 (1)
[or alternative use of PV = nRT5 × 104 × 2.0 × 10–3/8.31 × 290 = 0.0415 moles]= 2.50
× 1022 molecules (1)
2
(ii)
(mean) kinetic energy of a molecule
=
(1) (= 6.00 × 10–21 J)
(total kinetic energy = mean kinetic energy × N)
= 6.00 × 10–21 × 2.50 × 1022 (1)
= 150 (J) (1)
3
(c)
all molecules/atoms are identical
molecules/atoms are in random motion
Newtonian mechanics apply
gas contains a large number of molecules
the volume of gas molecules is negligible (compared to the volume
occupied by the gas) or reference to point masses
no force act between molecules except during collisions or the
speed/velocity is constant between collisions or motion is in a
straight line between collisions
collisions are elastic or kinetic energy is conserved
and of negligible duration
any 4 (1)(1)(1)(1)
max 4
[12]
M15.
(a)
using Q = mcΔθ
= 3.00 × 440 × (84-27) (1)
7.5 × 104 (J) (1)
2
(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]
M16.
(a) graph passes through given point 2.2 × 10–3 m3 at 0 °C straight
line with positive gradient
(straight) line to aim or pass through –273 °C at zero volume
2
(b)
(use of n = P V/R T)
1.00 × 105 × 2.20 × 10–3/8.31 × 273
n = 0.0970 (moles)
2
(c)
(use of mean kinetic energy = 3/2 K T)
= 3/2 × 1.38 × 10–23 × 323
6.69 × 10–21 (J)
3 sfs
3
(d)
total internal energy = 6.69 × 10–21 × 0.0970 × 6.02 × 1023 = 390 (J)
1
(e)
The candidate’s writing should be legible and the spelling,punctuation and
grammar should be sufficiently accurate for themeaning to be clear.
The candidate’s answer will be assessed holistically. The answerwill be assigned to one
of three levels according to the following criteria.
High Level (Good to excellent): 5 or 6 marks
The information conveyed by the answer is clearly organised,logical and coherent, using
appropriate specialist vocabularycorrectly. The form and style of writing is appropriate to
answerthe question.
The candidate provides a comprehensive and coherent sequenceof ideas linking the
motion of molecules to the pressure they exerton a container. At least three of the first
four points listed below must begiven in a logical order. The description should also show
awarenessof how a balance is maintained between the increase in speed andshortening
of the time interval between collisions with the wall to maintaina constant pressure.To be
in this band, reference must be made to force being the rate ofchange of momentum or
how, in detail, the volume compensates for theincrease in temperature.
Intermediate Level (Modest to adequate): 3 or 4 marks
The information conveyed by the answer may be less well organised
and not fully coherent. There is less use of specialist vocabulary, or
specialist vocabulary may be used incorrectly. The form and style of
writing is less appropriate.
The candidate provides a comprehensive list of ideas linking the
motion of molecules to the pressure they exert on a container. At least
three of the first four points listed below are given. The candidate also
knows than the mean square speed of molecules is proportional to
temperature. Using this knowledge, an attempt is made to explain how
the pressure is constant.
Low Level (Poor to limited): 1 or 2 marks
The information conveyed by the answer is poorly organised and may
not be relevant or coherent. There is little correct use of specialist vocabulary.
The form and style of writing may be only partly appropriate.
The candidate attempts the question and refers to at least two of the
points listed below.
Incorrect, inappropriate of no response: 0 marks
No answer or answer refers to unrelated, incorrect or inappropriate
physics.
Statements expected in a competent answer should include some ofthe following
marking points.
molecules are in rapid random motion/many molecules are involved
molecules change their momentum or accelerate on collision withthe walls
reference to Newton’s 2nd law either F = ma or F = rate of change ofmomentum
reference to Newton’s 3rd law between molecule and wall
relate pressure to force P = F/A
mean square speed of molecules is proportional to temperature
as temperature increases so does change of momentum or change invelocity
compensated for by longer time between collisions as the temperatureincreases
as the volume increases the surface area increases which reduces thepressure
max 6
[14]
M17.
(a)
(i)
n = PV/RT = 3.2 × 105 × 1.9 × 10-3/8.31 × 285
n = 0.26 mol
(0.257 mol)
1
(ii)
3.31 × 105 Pa
3 sig figs
(allow 3.30-3.35 × 105 Pa)
sig fig mark stands alone even with incorrect answer
3
(b)
similar
-( rapid) random motion
- range of speeds
different
- mean kinetic energy
- root mean square speed
- frequency of collisions
2
[6]
M18.
(a)
17 K
2
(b)
t = 15 s
2
[4]
M19.(a)
molecules have negligible volume
collisions are elastic
the gas cannot be liquified
there are no interactions between molecules (except during collisions)
the gas obeys the (ideal) gas law / obeys Boyles law etc.
at all temperatures/pressures
any two lines
a gas laws may be given as a formula
2
(b)
(i)
n (= PV / RT) = 1.60 × 106 × 0.200 / (8.31 × (273 + 22))
= 130 or 131 mol
(130.5 mol)
2
(ii)
mass = 130.5 × 0.043 = 5.6 (kg)
allow ecf from bi
(5.61kg)
density (= mass / volume) = 5.61 / 0.200 = 28
(28.1 kg m−3)kg m−3
a numerical answer without working can gain the first two marks
3
(iii)
(V2 = P1 V1 T2 / P2 T1)
V2 = 1.6 × 106 × .200 × (273 – 50) / 3.6 × 104 × (273 + 22) or 6.7(2) (m3)
allow ecf from bii
[reminder must see bii]
look out for
mass remaining = 5.61 × 0.20 / 6.72 = 0.17 (kg)
(0.167 kg)
or
n = (PV / RT = 3.6 × 104 × 0.200 / (8.31 × (273 − 50)) = 3.88(5) (mol)
mass remaining = 3.885 × 4.3 × 10−2 = 0.17 (kg)
2 sig figs
any 2 sf answer gets the mark
3
[10]
M20.(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
(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]
E1.
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.
E2.
There were significant variations between candidates. Most were able to correctly apply the
equation of state for an ideal gas but were less confident when it came to calculating in part (a)
the average kinetic energy of a gas molecule in the cylinder. It was evident that a significant
number of candidates did not know the appropriate formula. In part (c) explaining, in terms of
the kinetic theory, why the pressure of the gas fell was well answered. The only common error
was assuming that the kinetic energy of molecules fell even though the temperature remained
constant.
E3.
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.
E4.
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.
E5.
Again candidates found this a demanding question and a significant minority were not even
able to convert the temperature from Celsius to Kelvin in part (a). In the same section, the
calculation of number of moles of gas and the number of gas molecules also proved difficult and
significant figure errors were common.
Part (b) realised better answers, although many candidates did not seem to appreciate that
temperature is related to the mean kinetic energy and not to the kinetic energy of an individual
molecule.
E6.
This question was answered well, although part (a) did generate some badly structured
responses. Most candidates realised that potential energy was being converted to kinetic
energy but were less clear what happened to the kinetic energy on impact. The most common
answer was that it changed back into gravitational potential energy.
Part (b) worked well, although the unit for specific heat capacity still causes problems. It was a
shame that many candidates, having successfully calculated specific heat capacity, then lost
both marks due to a significant figure penalty and a unit penalty.
E7.
Candidates found this question quite demanding and although in part (a) most were able to
draw the two graphs correctly, the answers were spoilt due to a lack of care in the sketches.
The calculations in part (b) were more discriminating and arithmetic errors were common. The
correction to part (b) (iii) did not seem to cause problems, although only the more able
candidates were able to come up with a correct response. A significant proportion of candidates
answered the question correctly but then lost the final mark due to a significant figure penalty.
E8.
E9.
Again many candidates scored high marks in this question. There was a tendency to incur a
significant figure penalty for either quoting an answer to too many significant figures or rounding
down to one significant figure. Part (b) caused problems for less able candidates who, although
correctly determining the change in number of moles, were then either unable to calculate 23%
of the number or else did not multiply the change in number of moles by Avogadro’s number.
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.
E10.
The calculation in part (a) was answered well and candidates used the equation of state
confidently. Part (b) produced more varied answers. Most candidates realised that the oxygen
molecules would have a lower mean square speed, but the explanations as to why this was the
case were a little less robust. Candidates often failed to appreciate the fact that the
temperatures of both gases were the same and consequently they must have the same mean
kinetic energy. There was also a tendency to write about speed rather than mean square speed,
and kinetic energy rather than mean kinetic energy.
E11.
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.
E14.
Part (a) proved difficult for less able candidates. Some drew straight lines and others tried
to force the curve to intercept the volume axis. The less able candidates sometimes marked
correct points on the grid but did not draw a line. It seemed that some less able candidates
followed the wrong order in tackling this part. They drew the curve before they marked points on
the grid. As a result the points were just randomly placed on the curve they had drawn.
Part (b) (i) was done well by most. Candidates who used the alternative
equation PV = nRToften stopped when they had found the number of moles of gas. Part (b) (ii)
was much more discriminating with less than 50% of candidates obtaining the correct answer.
Many candidates did not have a clue whereas others could find the mean kinetic energy but
then did not follow this up by finding the total kinetic energy.
Although part (c) looks like a basic question it did discriminate well. It was only the more able
candidates who scored full marks. Many did not know what the question was getting at and
guessed. Sometimes these candidates did score the mark associated with molecules moving in
random motion. In other cases candidates did not complete their statements fully. For example,
stating ‘atoms travel in straight lines’, rather than, ‘atoms travel in straight lines between
collisions’.
E15.
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.
E16.
The graph in part (a) was done well by most, but the less able candidates were not careful
in reading the temperature scale and did not place the x-axis intercept at absolute zero. In some
cases they had drawn a curve that had no intercept on the x-axis.
Parts (c) (d) and (e) were tacked well by more able candidates. The less able could only
manage to do part (b) but then started either to substitute the wrong data, eg temperature in °C,
or quote incorrect equations in the parts that followed. It was appreciated that not enough space
was given to answer.
Part (e) allowed almost all candidates to score some marks, but the scores tended to be
grouped in the following way. Less able candidates scored a couple of marks by discussing
movement of molecules but did not go any further because of their poor use of physics in using
phrases such as, ‘the molecules have more energy and so hit each other harder giving more
pressure’.
Some candidates started to use Newton’s second law more effectively and referred to pressure
in a more scientific manner.
The more able candidates could explain how increasing the volume allowed the pressure to
remain constant as the temperature increased in terms of molecular motion.
E17.
Part (a)(i) was an easy introductory question, which most students got correct. Part (a)(ii)
was also successfully attempted in a majority of scripts. Use of the ideal gas equation again
was more popular than using pressure is proportional to temperature. A small percentage of
papers gave answers to only 2 significant figures rather than the 3 required. A majority of
students only scored one mark out of two for part (b). They correctly referred to the random
motion but failed to refer to a mean when giving some quantity, such as kinetic energy, that
increases with temperature.
E18.
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.
E19.A majority of candidates referred to obeying a gas law in answer to part (a). A second marking
point was often missed out, wrong or vague. This is illustrated in the two answers that follow: ‘It
has properties of a gas such as Brownian Motion’, and ‘The gas obeys the assumptions of the
kinetic theory’.
Parts (b)(i)+(ii) were done well by most. Only a few did not convert the temperature to Kelvin
before performing a calculation. Again very few did not know the unit for density.
In part (b)(iii) more than half the candidates could perform the calculation but a significant
number of those did not quote the answer to 2 significant figures. Of those missing out on the
calculation many did score the first mark but then went wrong by using the wrong density or by
not finding the proportion of the gas still in the container.
E20.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.