Combined Science Double Award 9204 | Paper 3 Mock 01 | Physics
Question 1 Report
A technician prepares a container for used fuel from a research reactor. Fig. 1 shows the transport cask with a steel shell, thick concrete and a water-filled inner space. The fuel continues to emit ionising radiation after it has been removed from the reactor. A warning meter beside the cask reads 6.0 mSv h-1 without its outer concrete lid and 0.15 mSv h-1 when the lid is fitted.
(a) Which type of radiation is most likely to need thick concrete shielding: alpha, beta or gamma? [2]
(b) Describe two ways in which the cask design reduces the risk to workers. [3]
(c) Calculate by what factor the dose rate is reduced when the lid is fitted. [2]
(d) Use the information in Fig. 1 to give one reason why water is kept around the used fuel. [3]
- Gamma radiation needs thick concrete shielding because it is the most penetrating of the three types listed and needs dense, thick material to reduce it effectively. [2]
- The thick concrete absorbs or reduces gamma radiation; the steel provides additional shielding and containment; the water absorbs radiation; and the sealed cask prevents radioactive material being released. Any three correctly linked design features are credited. [3]
- \[\text{reduction factor}=\frac{6.0\text{ mSv h}^{-1}}{0.15\text{ mSv h}^{-1}}=40\]The dose rate is reduced by a factor of 40. [2]
- Water around the fuel absorbs radiation, reducing the radiation dose outside the cask. It can also remove heat from fuel that remains radioactive. [3]
Answer Details
- Gamma radiation needs thick concrete shielding because it is the most penetrating of the three types listed and needs dense, thick material to reduce it effectively. [2]
- The thick concrete absorbs or reduces gamma radiation; the steel provides additional shielding and containment; the water absorbs radiation; and the sealed cask prevents radioactive material being released. Any three correctly linked design features are credited. [3]
- \[\text{reduction factor}=\frac{6.0\text{ mSv h}^{-1}}{0.15\text{ mSv h}^{-1}}=40\]The dose rate is reduced by a factor of 40. [2]
- Water around the fuel absorbs radiation, reducing the radiation dose outside the cask. It can also remove heat from fuel that remains radioactive. [3]
Question 2 Report
Fig. 1 shows a simple transit observation. A distant star produces a steady light signal. A planet passes in front of the star once every 3.0 days, causing a small fall in the light received by a space telescope. The telescope circuit records the signal as a repeating dip. The planet cannot be seen directly because the star is much brighter.
(a) What causes each dip in the light signal? [1]
(b) Calculate the number of transits expected in 30 days. [2]
(c) Describe why a larger planet produces a deeper dip in the received light. [2]
(d) Explain why this method is useful for finding planets orbiting distant stars. [3]
(e) Give two measurements, other than light level, that astronomers could use to investigate an orbiting planet. [2]
(a) Each dip occurs when the planet passes in front of the star and blocks some of its light [1]. This is called a transit.
(b) The planet transits once every 3.0 days:
\[\frac{30\text{ days}}{3.0\text{ days per transit}}=10\]
So there are 10 transits [2].
(c) A larger planet covers a larger area of the star [1]. It therefore blocks more light and causes a greater reduction in received light intensity [1].
(d) Planets emit very little visible light [1], and are difficult to see beside a much brighter star [1]. However, repeated dips provide evidence that an object is orbiting the star and allow its orbital period to be found [1].
(e) Two acceptable measurements are the star’s radial velocity and the wavelength shift of the star’s light [2]. Transit duration, planet mass and planet radius are also accepted.
Answer Details
(a) Each dip occurs when the planet passes in front of the star and blocks some of its light [1]. This is called a transit.
(b) The planet transits once every 3.0 days:
\[\frac{30\text{ days}}{3.0\text{ days per transit}}=10\]
So there are 10 transits [2].
(c) A larger planet covers a larger area of the star [1]. It therefore blocks more light and causes a greater reduction in received light intensity [1].
(d) Planets emit very little visible light [1], and are difficult to see beside a much brighter star [1]. However, repeated dips provide evidence that an object is orbiting the star and allow its orbital period to be found [1].
(e) Two acceptable measurements are the star’s radial velocity and the wavelength shift of the star’s light [2]. Transit duration, planet mass and planet radius are also accepted.
Question 3 Report
A research balloon samples high-altitude air where atoms can be ionised by radiation from space. Fig. 1 shows an oxygen atom before and after it gains two electrons. The charged oxygen ion can affect instruments that measure electrical current in the thin air. Each diagram gives the number of protons, and the shells show the electron arrangement. The balloon team needs to calculate the charge correctly before calibrating its detector.
(a) Calculate the charge on the oxygen ion. [1]
(b) What is the formula of this ion? [1]
(c) Which particles have moved into the oxygen atom? [1]
(d) Describe why the ion has a different charge from the atom. [2]
(e) Give the number of protons in the oxygen ion. [1]
(a) The oxygen ion has 8 protons but, after gaining two electrons, 10 electrons. It has two extra negative charges, so its charge is \(-2\). [1]
(b) The formula is \(\mathrm{O^{2-}}\). [1]
(c) The particles that moved into the atom are electrons. [1]
(d) The ion has gained two electrons. It therefore has two more electrons than protons, giving an overall negative charge. [2]
(e) Ion formation changes electrons, not the nucleus. The oxygen ion still has 8 protons. [1]
Answer Details
(a) The oxygen ion has 8 protons but, after gaining two electrons, 10 electrons. It has two extra negative charges, so its charge is \(-2\). [1]
(b) The formula is \(\mathrm{O^{2-}}\). [1]
(c) The particles that moved into the atom are electrons. [1]
(d) The ion has gained two electrons. It therefore has two more electrons than protons, giving an overall negative charge. [2]
(e) Ion formation changes electrons, not the nucleus. The oxygen ion still has 8 protons. [1]
Question 4 Report
During a museum renovation, a conservator examines a faded painting under an angled glass cover. Fig. 1 shows a lamp, the glass, and the conservator's eye. The lamp produces a bright reflected patch that hides some detail in the picture. The glass surface is smooth, and the incident ray and normal are shown. The conservator wants to move the lamp, not the painting, so that reflected light misses the eye and the colours can be inspected.
(a) Calculate the angle of reflection if the angle of incidence is 28 degrees. [1]
(b) Draw the reflected ray that causes the bright patch. [2]
(c) Which direction should the lamp be moved to reduce the reflection into the eye: towards the normal or away from the normal? [1]
(d) Describe why the glare is produced by the glass. [2]
(a) By the law of reflection, the angle of reflection equals the angle of incidence. The angle of reflection is 28°. [1]
(b) The reflected ray is drawn on the opposite side of the normal, at 28° to the normal, and directed towards the eye. [2]
(c) The lamp should be moved away from the normal to change the reflected ray so that it misses the eye. [1]
(d) Smooth glass produces regular reflection. [1] The reflected rays can enter the eye and make a bright image of the lamp, causing glare. [1]
Answer Details
(a) By the law of reflection, the angle of reflection equals the angle of incidence. The angle of reflection is 28°. [1]
(b) The reflected ray is drawn on the opposite side of the normal, at 28° to the normal, and directed towards the eye. [2]
(c) The lamp should be moved away from the normal to change the reflected ray so that it misses the eye. [1]
(d) Smooth glass produces regular reflection. [1] The reflected rays can enter the eye and make a bright image of the lamp, causing glare. [1]
Question 5 Report
Fig. 1 shows the optical path in a compact periscope fitted to an inspection mast on a harbour wall. A worker uses it to look over a barrier at a moored boat without climbing the wall. Two parallel plane mirrors are fixed inside the box. Light from the boat enters the top opening, changes direction at mirror A, then reaches mirror B. The worker sees the outgoing ray at the lower opening. Both mirrors are at 45° to the vertical sides of the box. This is light radiation, not a sound wave, and the electrical fan used to cool the box has no effect on the ray path.
(a) Describe the changes in direction of the light ray between the boat and the worker. [3]
(b) Calculate the angle of incidence at mirror A. [2]
(c) Draw a solid line over each dashed section to show the path of the reflected light. [2]
(d) What are two properties of the image of the boat seen through this periscope? [2]
(e) Give two reasons why the mirrors should be smooth and shiny. [2]
(a) Light travels horizontally into the top of the periscope. [1] Mirror A reflects it downwards. [1] Mirror B then reflects it horizontally towards the worker. [1]
(b) The angle of incidence at mirror A is 45°. [1] This angle is measured between the incident ray and the normal, not between the ray and the mirror surface. [1]
(c) The reflected path is shown by the solid downward line from A to B and the solid horizontal line from B towards the worker. [2]
(d) The image is virtual [1] and upright. [1] It is also the same size as the object.
(e) The mirrors should be smooth so that they give regular reflection. [1] They should be shiny so that they reflect a high proportion of the light, giving a brighter, clearer image with less scattering. [1]
Answer Details
(a) Light travels horizontally into the top of the periscope. [1] Mirror A reflects it downwards. [1] Mirror B then reflects it horizontally towards the worker. [1]
(b) The angle of incidence at mirror A is 45°. [1] This angle is measured between the incident ray and the normal, not between the ray and the mirror surface. [1]
(c) The reflected path is shown by the solid downward line from A to B and the solid horizontal line from B towards the worker. [2]
(d) The image is virtual [1] and upright. [1] It is also the same size as the object.
(e) The mirrors should be smooth so that they give regular reflection. [1] They should be shiny so that they reflect a high proportion of the light, giving a brighter, clearer image with less scattering. [1]
Question 6 Report
Fig. 1 shows a delivery cyclist using brakes while travelling down a hill. The bicycle and rider have a mass of 90 kg. At the bottom, the rider applies the brakes and comes to rest. The brake discs become much hotter and a small amount of sound is heard. The cyclist wants to reduce the heating of the brakes on a long descent.
(a) What energy store decreases as the bicycle slows down? [1]
(b) Describe the useful and wasted energy transfers when the brakes are applied. [2]
(c) Which action would reduce the rate at which energy is transferred to the brake discs? [1]
(d) Give one reason why brake discs need to conduct thermal energy well. [1]
A thermostat in an electric incubator uses a bimetallic strip. Fig. 1 shows two bonded metal layers, P and Q, next to an electrical contact. Metal P expands more than metal Q for the same temperature increase. When the strip bends, the circuit to the heater opens and the heater stops transferring energy to the air.
(a) Which metal layer becomes longer by the greater amount when heated? [1]
(b) Describe how the strip bends when the temperature increases. [2]
(c) What happens to the circuit when the strip touches the contact? [1]
(d) Give one reason why a thermostat saves energy. [1]
Bicycle brakes
- (a) The kinetic energy store decreases as the bicycle slows down. [1]
- (b) Friction at the brakes transfers kinetic energy mainly to the thermal energy stores of the brake discs and surrounding air. [1] Some energy is also transferred as sound, which is wasted. [1]
- (c) The rider could brake more gently, apply the brakes for a shorter time, or use a lower gear earlier. [1] Each reduces the rate at which energy is transferred to the discs.
- (d) Brake discs need to conduct thermal energy well so that energy is transferred away quickly and the brakes do not overheat. [1]
Bimetallic-strip thermostat
- (a) P becomes longer by the greater amount when heated, because the question states that P expands more than Q. [1]
- (b) P expands more than Q. [1] Therefore the strip bends towards Q, with P on the outside of the curve. [1] The outside of a bend must be the longer layer.
- (c) When the strip touches the contact, the circuit is completed and current flows to the heater. [1]
- (d) A thermostat saves energy because it switches the heater off at the required temperature, preventing unnecessary energy transfer. [1]
Answer Details
Bicycle brakes
- (a) The kinetic energy store decreases as the bicycle slows down. [1]
- (b) Friction at the brakes transfers kinetic energy mainly to the thermal energy stores of the brake discs and surrounding air. [1] Some energy is also transferred as sound, which is wasted. [1]
- (c) The rider could brake more gently, apply the brakes for a shorter time, or use a lower gear earlier. [1] Each reduces the rate at which energy is transferred to the discs.
- (d) Brake discs need to conduct thermal energy well so that energy is transferred away quickly and the brakes do not overheat. [1]
Bimetallic-strip thermostat
- (a) P becomes longer by the greater amount when heated, because the question states that P expands more than Q. [1]
- (b) P expands more than Q. [1] Therefore the strip bends towards Q, with P on the outside of the curve. [1] The outside of a bend must be the longer layer.
- (c) When the strip touches the contact, the circuit is completed and current flows to the heater. [1]
- (d) A thermostat saves energy because it switches the heater off at the required temperature, preventing unnecessary energy transfer. [1]
Question 7 Report
Fig. 1 shows the path of alpha particles near a thin gold foil in a museum reconstruction of a scattering experiment. The source produces a narrow beam. The detector can be moved around the foil to count particles at different angles. Students observe that the beam is not attracted towards the foil. Their teacher asks them to use the paths to infer the charge and position of particles within an atom.
(a) What is the charge on an alpha particle? [1]
(b) Describe why some alpha particles were deflected as they passed the foil. [2]
(c) Which part of the atom caused the deflection? [1]
(d) Give the relative mass of an electron. [1]
(e) Calculate the number of neutrons in a gold-197 atom with proton number 79. [1]
(a) An alpha particle has relative charge \(+2\). [1]
(b) Alpha particles are positively charged. When they pass close to the positively charged nucleus, they are repelled, so their paths are deflected. [2]
(c) The part of the atom that causes the deflection is the nucleus. [1]
(d) The relative mass of an electron is \(\frac{1}{1836}\), often treated as approximately zero compared with protons and neutrons. [1]
(e) Neutrons = mass number − proton number:
\[197-79=118\]
A gold-197 atom has 118 neutrons. [1]
Examination reminder: Repulsion shows that the nucleus is positive. Alpha particles are not attracted because they are also positive.
Answer Details
(a) An alpha particle has relative charge \(+2\). [1]
(b) Alpha particles are positively charged. When they pass close to the positively charged nucleus, they are repelled, so their paths are deflected. [2]
(c) The part of the atom that causes the deflection is the nucleus. [1]
(d) The relative mass of an electron is \(\frac{1}{1836}\), often treated as approximately zero compared with protons and neutrons. [1]
(e) Neutrons = mass number − proton number:
\[197-79=118\]
A gold-197 atom has 118 neutrons. [1]
Examination reminder: Repulsion shows that the nucleus is positive. Alpha particles are not attracted because they are also positive.
Question 8 Report
A radio astronomy club listens for pulses from a compact star. Fig. 1 shows a neutron star with a magnetic axis that is not aligned with its rotation axis. A narrow beam of radiation sweeps through space as the star spins. The club receives a pulse only when the beam points towards Earth, rather like a rotating lighthouse.
(a) What is the usual name for this regularly pulsing object? [1]
(b) Describe why the radio signal is not detected continuously. [2]
(c) Which event can form a neutron star? [1]
(d) Give one reason why the neutron star has a very high density. [1]
(a) This regularly pulsing object is called a pulsar. [1]
(b) The radiation beam is narrow. As the neutron star rotates, the beam moves towards Earth and then away from Earth, so the radio signal is not detected continuously. [2]
(c) A neutron star can form in a supernova. [1]
(d) It has very high density because a large mass is compressed into a very small volume. [1]
Answer Details
(a) This regularly pulsing object is called a pulsar. [1]
(b) The radiation beam is narrow. As the neutron star rotates, the beam moves towards Earth and then away from Earth, so the radio signal is not detected continuously. [2]
(c) A neutron star can form in a supernova. [1]
(d) It has very high density because a large mass is compressed into a very small volume. [1]
Question 9 Report
Fig. 1 shows a simplified magnetic field surrounding a rapidly rotating star. A science team uses this model to predict the path of charged particles in the thin air-free region above the star. The arrows show the direction of the magnetic field. A proton enters the region at right angles to the field lines with a velocity of 2.4 × 106 m/s. At this point, the magnetic flux density is 0.30 T. The charge on a proton is 1.6 × 10−19 C.
(a) What is the name of the region around the star in which a magnetic force can act on the proton? [1]
(b) Describe the direction of the magnetic field at the position of the proton in Fig. 1. [2]
(c) Calculate the magnetic force on the proton. Use the equation:
force = magnetic flux density × charge × velocity
Give your answer in N. [3]
(d) Use ideas about forces to describe the proton’s path as it travels through a uniform magnetic field. [2]
A moving charged particle in a magnetic field experiences a force when it moves at right angles to the field. That force changes its direction, rather than simply increasing its speed.
- (a) The region is a magnetic field. [1]
- (b) At the proton, the field is towards the star, to the left [1], and is perpendicular to the proton's velocity. [1]
- (c) \[F=Bqv=0.30\times1.6\times10^{-19}\times2.4\times10^6\]
\[F=1.152\times10^{-13}\ \text{N}\] Therefore \(1.15\times10^{-13}\ \text{N}\), or \(1.2\times10^{-13}\ \text{N}\), is accepted. [3] - (d) The magnetic force acts at right angles to the proton's motion [1], so the proton follows a curved or circular path. [1]
Answer Details
A moving charged particle in a magnetic field experiences a force when it moves at right angles to the field. That force changes its direction, rather than simply increasing its speed.
- (a) The region is a magnetic field. [1]
- (b) At the proton, the field is towards the star, to the left [1], and is perpendicular to the proton's velocity. [1]
- (c) \[F=Bqv=0.30\times1.6\times10^{-19}\times2.4\times10^6\]
\[F=1.152\times10^{-13}\ \text{N}\] Therefore \(1.15\times10^{-13}\ \text{N}\), or \(1.2\times10^{-13}\ \text{N}\), is accepted. [3] - (d) The magnetic force acts at right angles to the proton's motion [1], so the proton follows a curved or circular path. [1]
Question 10 Report
During a field survey, students use the circuit in Fig. 1 to identify a buried metal cap. A search coil is connected to a sensitive meter. The coil is moved steadily over the ground. The meter needle deflects only as the coil passes the cap.
(a) What observation shows that a potential difference has been induced? [1]
(b) Which action produces no meter deflection? [1]
(c) Give one way to make the induced potential difference larger. [1]
(d) Describe why the meter returns to zero after the coil has passed the cap. [2]
Two students build the alarm circuit in Fig. 1 for a greenhouse door. A reed switch is fixed to the frame and a magnet is fixed to the door. The buzzer sounds only when the door is open. The cell provides energy to the circuit.
(a) Which component changes electrical energy into sound energy? [1]
(b) What happens to the reed switch when the magnet is close to it? [1]
(c) Give one reason this circuit is useful for an alarm. [1]
(d) Draw the circuit symbol for a cell. [1]
Buried metal cap circuit
- The observation is that the meter needle deflects. A deflection shows that current is flowing, so an induced potential difference has been produced. [1]
- Holding the coil still produces no deflection. Induction requires a changing magnetic field through the coil; a stationary coil has no changing flux. [1]
- Move the coil faster, or use a coil with more turns. Either change increases the induced potential difference. [1]
- After the coil has passed the cap, the magnetic field through it stops changing. Therefore, no potential difference or current is induced, and the meter returns to zero. [2]
Greenhouse door alarm
- The buzzer changes electrical energy into sound energy. [1]
- When the magnet is close, the reed switch closes and completes the circuit. [1]
- The circuit detects that the door has moved/opened and gives a sound warning, so it is useful as an alarm. [1]
- A cell is represented by two parallel lines of unequal length:
The longer line is the positive terminal. [1]
Answer Details
Buried metal cap circuit
- The observation is that the meter needle deflects. A deflection shows that current is flowing, so an induced potential difference has been produced. [1]
- Holding the coil still produces no deflection. Induction requires a changing magnetic field through the coil; a stationary coil has no changing flux. [1]
- Move the coil faster, or use a coil with more turns. Either change increases the induced potential difference. [1]
- After the coil has passed the cap, the magnetic field through it stops changing. Therefore, no potential difference or current is induced, and the meter returns to zero. [2]
Greenhouse door alarm
- The buzzer changes electrical energy into sound energy. [1]
- When the magnet is close, the reed switch closes and completes the circuit. [1]
- The circuit detects that the door has moved/opened and gives a sound warning, so it is useful as an alarm. [1]
- A cell is represented by two parallel lines of unequal length:
The longer line is the positive terminal. [1]
Question 11 Report
Fig. 1 is a force diagram for an automated library trolley as it travels across a smooth concrete floor. The trolley carries boxes of returned books. Its battery-powered motor provides a force of 180 N towards the north. A side roller against a guide rail produces a force of 240 N towards the east. The students assume that friction and air resistance are too small to affect the calculation. The two forces act at right angles.
(a) Calculate the size of the resultant force on the trolley. Give your answer in N. [3]
(b) Describe the direction of the resultant force, using north and east. [1]
(c) Use the resultant force to state what happens to the trolley's velocity while these forces act. [2]
(d) Draw a labelled arrow on Fig. 1 to show a third force that would make the resultant force zero. [2]
(e) Explain why the trolley can still move at a steady velocity when the resultant force is zero. [2]
(a) The forces are at right angles, so use Pythagoras' theorem:
\[R^2=180^2+240^2=32400+57600=90000\]
\[R=\sqrt{90000}=300\text{ N}\]
The resultant force is 300 N. [3]
(b) The resultant acts towards the north-east. [1]
(c) Since there is a non-zero resultant force, the trolley accelerates and its velocity changes in the north-east direction. [2]
(d) To make the resultant zero, a third force must be equal and opposite to the 300 N north-east resultant: 300 N towards the south-west.
[2]
(e) A zero resultant force means zero acceleration, so there is no change in velocity. An object already moving therefore continues at constant velocity. [2]
Answer Details
(a) The forces are at right angles, so use Pythagoras' theorem:
\[R^2=180^2+240^2=32400+57600=90000\]
\[R=\sqrt{90000}=300\text{ N}\]
The resultant force is 300 N. [3]
(b) The resultant acts towards the north-east. [1]
(c) Since there is a non-zero resultant force, the trolley accelerates and its velocity changes in the north-east direction. [2]
(d) To make the resultant zero, a third force must be equal and opposite to the 300 N north-east resultant: 300 N towards the south-west.
[2]
(e) A zero resultant force means zero acceleration, so there is no change in velocity. An object already moving therefore continues at constant velocity. [2]
Question 12 Report
| lunch box | insulation material | temperature after 30 min / degrees C |
|---|---|---|
| A | none | 46 |
| B | foam | 58 |
| C | foil-lined foam | 61 |
| D | vacuum wall | 68 |
This apparatus is used to test a toy launcher. Fig. 1 shows a spring compressed by 4.0 cm before it pushes a small cart along a track. The cart then climbs a ramp and stops at point P. Friction between the cart wheels and the track is small but not zero. The students repeat the test using the same cart and a greater compression.
(a) What energy store increases when the spring is compressed? [1]
(b) Describe the main energy transfers as the cart moves up the ramp. [2]
(c) Which position gives the cart its greatest gravitational potential energy? [1]
The table below gives the temperature inside four identical lunch boxes after they have been outdoors for 30 minutes. Each box begins with soup at 75 degrees C. Box D has a vacuum layer between two walls. The students use the results to compare how well each design reduces energy transfer to cold air.
| lunch box | insulation material | temperature after 30 min / degrees C |
|---|---|---|
| A | none | 46 |
| B | foam | 58 |
| C | foil-lined foam | 61 |
| D | vacuum wall | 68 |
(a) Which lunch box has the smallest temperature decrease? [1]
(b) Calculate the temperature decrease for box C. [1]
(c) Describe how foam reduces energy transfer by conduction. [2]
(d) Explain why a vacuum layer reduces conduction and convection. [1]
Spring launcher
- (a) Compressing the spring increases its elastic potential energy store. [1]
- (b) The spring's elastic potential energy is transferred to the cart's kinetic energy store. [1] As the cart climbs, kinetic energy is transferred to its gravitational potential energy store. [1] A small amount is also dissipated by friction, but the question asks for the main transfers.
- (c) The cart has its greatest gravitational potential energy at point P, its highest position. [1]
Lunch boxes
- (a) Box D has the smallest temperature decrease: it finishes at 68 °C, the highest final temperature. [1]
- (b) \[ \text{temperature decrease}=75-61=14\ ^\circ\text{C} \] 14 °C. [1]
- (c) Foam contains trapped air. [1] Air is a poor conductor, so particles transfer energy slowly through the foam. [1]
- (d) A vacuum contains no particles, so there are no particles to conduct energy or to form convection currents. [1]
Answer Details
Spring launcher
- (a) Compressing the spring increases its elastic potential energy store. [1]
- (b) The spring's elastic potential energy is transferred to the cart's kinetic energy store. [1] As the cart climbs, kinetic energy is transferred to its gravitational potential energy store. [1] A small amount is also dissipated by friction, but the question asks for the main transfers.
- (c) The cart has its greatest gravitational potential energy at point P, its highest position. [1]
Lunch boxes
- (a) Box D has the smallest temperature decrease: it finishes at 68 °C, the highest final temperature. [1]
- (b) \[ \text{temperature decrease}=75-61=14\ ^\circ\text{C} \] 14 °C. [1]
- (c) Foam contains trapped air. [1] Air is a poor conductor, so particles transfer energy slowly through the foam. [1]
- (d) A vacuum contains no particles, so there are no particles to conduct energy or to form convection currents. [1]
