155 free, high-quality past-paper-style questions on Thermal energy transfers for IB Physics SL, each with a full mark scheme. Covers: Density & the molecular model; Temperature scales; Mechanisms of thermal transfer; and more. For when you have finished the past papers, or want to practise one topic at a time.B.1 at Higher Level →Read the B.1 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.763 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1B-1A-01Density & the molecular modelPaper 1AEasy1 mark
A block of metal has a mass of 0.500 kg and a volume of 2.5 × 10−4 m³.
A sample of water is at 300 K. Its temperature then rises by 20 °C.
What is its initial temperature in degrees Celsius, and what is the rise in kelvin?
Initial temperature / °CRise / K
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Step 1Subtract 273 to convert a Kelvin temperature to Celsius: 300 − 273 = 27 °C. A kelvin and a degree Celsius are the same size, so a change of 20 °C is a change of 20 K.
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Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
A273 has been added to the change as well. Only a temperature itself is converted; a change is the same number on both scales.
B273 has been added instead of subtracted. Kelvin values are larger than Celsius ones, so converting K to °C means subtracting.
CThis leaves the temperature unconverted. 300 °C and 300 K are very different temperatures.
DCorrect: 300 − 273 = 27 °C, and because the two scales have the same size of degree a rise of 20 °C is a rise of 20 K.
3B-1A-10Mechanisms of thermal transferPaper 1AEasy1 mark
An electric heater stands on the floor of a closed room. A convection current is set up in the air of the room.
Which statement explains why the air just above the heater rises?
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Step 1Air near the heater is warmed and expands, so its density decreases. The denser, cooler air around it sinks and pushes the less dense warm air upwards, setting up a convection current.
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Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
A'Heat rises' is a misconception. Energy is carried upwards only because the warm air itself moves, and it moves because of a difference in density.
BCorrect: the warm air is less dense than the cool air around it, so it is pushed upwards while the denser air sinks to take its place.
CThe mass of each molecule does not change on heating. It is the density of the air — mass per unit volume — that falls, because the molecules spread further apart.
DThe molecules themselves do not expand. The air expands because its molecules move faster and spread further apart.
4B-1A-23Internal energy & the molecular modelPaper 1AEasy1 mark
Which statement about the internal energy of a substance is correct?
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Step 1Internal energy is the sum of the total random kinetic energy of the molecules and the total intermolecular potential energy arising from the forces between them.
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: internal energy has both a kinetic part, which the temperature measures, and a potential part, which depends on the separations.
BThis leaves out the potential energy stored in the separations between molecules, which is what changes during melting and boiling.
CTemperature measures the average kinetic energy per molecule. Two samples at the same temperature can have very different internal energies if they differ in mass or in phase.
DInternal energy is about the random motion of the molecules. Motion of the body as a whole is ordinary kinetic energy, not internal energy.
8B-1A-45Density & the molecular modelPaper 1AEasy1 mark
A substance can exist as a solid, as a liquid and as a gas.
Which statement describes the molecules of the substance when it is a liquid?
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Step 1In a liquid the molecules are close together (similar separation to the solid, so similar density), the intermolecular forces are still significant, but the molecules are not locked in place: they move randomly and slide past one another, which is why a liquid flows.
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Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AThis describes a solid. In a liquid the molecules are not held at fixed positions, which is why a liquid takes the shape of its container.
BThis describes a gas. The density of a liquid is close to that of its solid, so its molecules are about as close together as in the solid.
CCorrect: the separation is similar to that in the solid, but the molecules are not fixed in place and can move past one another.
DThe intermolecular forces in a liquid are not negligible: they hold the liquid together with a definite volume. Only in an ideal gas are the forces neglected.
Step 1A black body is a perfect absorber of all incident radiation. It emits a continuous spectrum whose shape, peak wavelength (Wien's law) and total power per unit area (σT⁴) depend only on its absolute temperature.
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Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AA black body looks black only when it is cool, because it reflects nothing. When it is hot enough it glows visibly: a black body at 5800 K, like the Sun, has its peak in the visible region.
BThe spectrum is continuous, covering a range of wavelengths. Only the peak wavelength λmax is single-valued, and that is what becomes shorter as T rises.
CThe first half is right, but the spectrum of a black body is the same for every material at a given temperature. That is what makes the black body a useful model.
DCorrect: a black body is a perfect absorber, and its continuous emission spectrum is fixed by its temperature alone.
10B-1A-64Temperature & direction of transferPaper 1AEasy1 mark
A small steel rivet at 450 °C is dropped into a tank containing 200 kg of water at 15 °C. The internal energy of the water in the tank is far greater than the internal energy of the rivet.
Which statement about the thermal energy transfer once the rivet is in the water is correct?
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Step 1The direction of the resultant thermal energy transfer is set by the temperature difference, not by the amounts of internal energy: energy goes from the rivet (450 °C) to the water (15 °C) until both are at the same temperature, just above 15 °C.
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Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AThe amount of internal energy does not decide the direction. The water has more internal energy only because it has far more molecules; its molecules have a smaller average kinetic energy than those of the rivet.
BEquilibrium means equal temperatures, not equal internal energies. A 200 kg tank and a small rivet at the same temperature still have very different internal energies.
CCorrect: the resultant transfer is from the hotter body to the colder body, whatever their internal energies.
DThe transfer stops when the rivet reaches the temperature of the water, about 15 °C. The rivet still has internal energy then: its particles still have random kinetic energy and intermolecular potential energy.
The front glass wall of a tropical aquarium is 0.60 m wide, 0.40 m high and 8.0 mm thick. The thermal conductivity of the glass is 0.96 W m−1 K−1. The inner surface of the glass is at 24.5 °C and the outer surface is at 23.9 °C.
What is the rate of thermal energy transfer through the front glass?
Diagram not to scale
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Step 1A = 0.60 × 0.40 = 0.24 m², ΔT = 0.6 K and Δx = 8.0 × 10−3 m.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2ΔQ/Δt = kAΔT/Δx = 0.96 × 0.24 × 0.6 / 8.0 × 10−3 = 17.3 W ≈ 17 W.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThe width and the height have been added (0.60 + 0.40 = 1.0 m²) instead of multiplied to give the area: 0.96 × 1.0 × 0.6 / 0.0080 = 72 W.
BCorrect: 0.96 × 0.24 × 0.6 / 0.0080 = 17 W.
CThe thickness has been left in millimetres: 0.96 × 0.24 × 0.6 / 8.0 = 1.7 × 10−2 W.
DThe thickness has been multiplied instead of divided: 0.96 × 0.24 × 0.6 × 0.0080 = 1.1 × 10−3 W.
A titanium hip implant has been inside a patient’s body for several months. The implant and the bone in contact with it are both at 37 °C.
Which statement describes the particles where the implant touches the bone?
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Step 1Equal temperatures mean equal average kinetic energies of the particles. The particles at the boundary keep vibrating and interacting, so energy passes in both directions, but on average as much passes from the implant to the bone as from the bone to the implant: there is no resultant conduction.
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Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AParticles vibrate at every temperature above absolute zero. At 37 °C (310 K) their average kinetic energy is ³⁄₂kBT, which is far from zero.
BThe particles in contact still collide and interact. Energy passes both ways in these interactions; it is only the resultant transfer that is zero.
CThe implant and the bone are at the same temperature, so their particles have the same average kinetic energy. A good conductor transfers energy faster only when there is a temperature difference.
DCorrect: conduction depends on a difference in the average kinetic energy of the particles. With no temperature difference the energy exchanged in each direction balances, so there is no resultant transfer.
13IBP2S-009-01-01Internal energy & the molecular modelPaper 1AEasy1 mark
A sports physiotherapist places an ice pack on an athlete's sprained ankle. The ice in the pack melts at a constant 0 °C. Which describes the changes in the molecules of the ice while it melts?
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Temperature measures the mean random kinetic energy of the molecules. The temperature stays at 0 °C, so the mean kinetic energy is unchanged. The energy absorbed from the ankle breaks the bonds holding the molecules in the lattice. This increases their intermolecular potential energy, so the internal energy rises. Options A and C wrongly link absorbed energy to a temperature rise. Option D reverses the potential energy change, which describes freezing.
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Answer B
Answer: B
Common error: During a phase change at constant temperature, all the energy supplied goes into potential energy. Do not say the molecules 'speed up'.
A laptop processor is cooled by a copper heat sink rather than a plastic block of the same shape. What is the main reason copper conducts thermal energy much better than plastic?
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In a metal, free electrons gain kinetic energy at the hot region and move quickly through the lattice. They collide with ions elsewhere and pass on that energy. This adds to the slower transfer by lattice vibrations, which is the only mechanism in plastic. Option B describes bulk movement of matter, which is convection and cannot happen in a solid. Option C is false, because copper's specific heat capacity is low, and it does not determine conduction rate anyway. Option D wrongly attributes conduction to atomic mass.
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Answer A
Answer: A
Exam tip: When explaining conduction in metals, name both mechanisms: lattice vibrations and free-electron transport. State that free electrons are the dominant one.
15IBP2S-009-01-05Mechanisms of thermal transferPaper 1AEasy1 mark
A communications satellite orbits far above the atmosphere. Which mechanism(s) can transfer thermal energy from the Sun to the satellite?
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The space between the Sun and the satellite is almost a perfect vacuum. Conduction needs particles to collide or vibrate, and convection needs a fluid to flow. Neither can happen without a medium. Electromagnetic radiation needs no medium, so radiation is the only mechanism.
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Answer C
Answer: C
Exam tip: Whenever a vacuum separates two objects, only radiation can transfer thermal energy between them. This explains spacecraft design and vacuum flasks.
A walk-in freezer has an insulating wall panel. The panel is replaced by one made of the same material with double the thickness and double the area. The temperature difference across the panel is unchanged. How does the rate of thermal energy transfer through the new panel compare with that through the original panel?
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The rate of conduction is ΔQ/Δt = kAΔT/Δx. Doubling A doubles the rate, and doubling Δx halves it. The two effects cancel, so the rate is unchanged. Options A and C each consider only one of the two changes. Option B wrongly treats both changes as reducing the rate.
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Answer D
Answer: D
Exam tip: Write down the proportionality ΔQ/Δt ∝ AΔT/Δx and apply each factor in turn. This avoids sign slips in ratio questions.
17IBP2S-009-04-01Temperature & direction of transferPaper 1AEasy1 mark
A small steel rivet at 400 °C is dropped into a large tank of water at 20 °C during a shipyard quenching process. The water in the tank has far more internal energy than the rivet. Which statement about the thermal energy transfer is correct?
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The direction of net thermal energy transfer depends only on temperature difference. Energy flows from the hotter rivet to the cooler water. It stops when both reach the same temperature, not the same internal energy. The total internal energy of each body is not what matters.
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Answer B
Answer: B
Common error: Do not mix up internal energy, which depends on the amount of material, with temperature, which depends on the mean kinetic energy of the particles. Only temperature decides the direction of transfer.
18IBP2S-009-04-02Temperature scalesPaper 1AEasy1 mark
A biological sample is stored in liquid nitrogen at 77 K. It is removed and left in a laboratory until it reaches 20 °C. What is the change in temperature of the sample?
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First convert 20 °C to kelvin: 20 + 273 = 293 K. The change in temperature is then 293 − 77 = 216 K. A change of 1 K is the same size as a change of 1 °C, but both temperatures must be on the same scale before you subtract.
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Answer B
Answer: B
Common error: Subtracting 20 from 77 mixes the two scales and gives 57 K, which is wrong. Put both temperatures on the same scale first.
19IBP2S-009-04-06Method of mixturesPaper 1AEasy1 mark
A nurse places a digital thermometer under a patient's tongue and waits before reading it. Why is the wait necessary?
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Thermal equilibrium is reached when the two bodies are at the same temperature. At that point there is no net transfer of thermal energy, so the reading equals the tongue's temperature. The thermometer is far smaller than the patient, so its internal energy never matches the patient's.
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Answer B
Answer: B
Common error: Heat is energy in transfer, not something an object stores. Avoid phrases such as 'the heat in the thermometer'.
20IBP2S-009-00-01Temperature scalesPaper 1AEasy1 mark
A hospital patient's body temperature rises from 37.0 °C to 39.5 °C during a fever. What is the change in the patient's absolute (thermodynamic) temperature?
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A kelvin is the same size as a degree Celsius, so a temperature difference has the same numerical value on both scales: 39.5 − 37.0 = 2.5 K. Adding or subtracting 273 is only needed when converting a single temperature, not a change.
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Answer A
Answer: A
Common error: Do not add 273 to a temperature change. Only individual temperatures are converted between °C and K.
A ski-boot liner is redesigned. The new liner has double the surface area and double the thickness of the old one. On a milder day, the temperature difference across it is half that of the original test. What is the new rate of thermal energy transfer by conduction compared with the original?
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The rate of conduction is \(\frac{Q}{t} = kA\frac{\Delta T}{\Delta x}\), with the same material so k is unchanged. The ratio of new to old is (2 × 0.5)/2 = 0.5, so the rate halves. 'One quarter' comes from ignoring the doubled area: 0.5/2. 'The same' comes from ignoring the halved temperature difference: 2/2. 'Double' comes from putting thickness in the numerator: 2 × 0.5 × 2.
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Answer B
Answer: B
Exam tip: Write the conduction equation as a ratio of new to old and substitute the scaling factors one at a time. Thickness goes in the denominator.
22IBP2S-009-00-05Temperature scalesPaper 1AEasy1 mark
The air in a sealed laboratory chamber starts at 27 °C. It is heated until its absolute temperature has doubled. What is its new temperature in degrees Celsius?
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27 °C = 300 K. Doubling the absolute temperature gives 600 K, which is 600 − 273 = 327 °C. 54 °C doubles the Celsius value, 600 °C forgets to convert back, and 873 °C adds 273 instead of subtracting.
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Answer B
Answer: B
Common error: 'Doubling the temperature' only has physical meaning on the kelvin scale. Convert to K first, then scale.
23IBP2S-009-03-01Temperature scalesPaper 1AEasy1 mark
A frozen tissue sample is stored at −196 °C. It is moved to a transport container and warms to −171 °C. What is the change in temperature of the sample in kelvin?
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A kelvin is the same size as a degree Celsius, so a temperature difference has the same number in both scales: −171 − (−196) = 25 K. The values 77 K and 102 K are the absolute temperatures at the start and the end, not the change. 298 K comes from wrongly adding 273 to a temperature difference.
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Answer A
Answer: A
Common error: Add 273 only when converting a single temperature. A temperature change is numerically the same in K and °C.
24IBP2S-009-03-02Temperature & molecular energyPaper 1AEasy1 mark
A deep-sea diving support vessel stores helium gas and xenon gas in separate cylinders kept at the same temperature. Which quantity is the same for the atoms in both cylinders?
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The average translational kinetic energy of a particle depends only on the absolute temperature: \(\bar{E}_k = \frac{3}{2}k_BT\). Equal temperatures therefore mean equal average kinetic energies. Xenon atoms have a much larger mass, so for the same kinetic energy they move more slowly and have more momentum. The density depends on the pressure and the molar mass, so it is not fixed by temperature alone.
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Answer B
Answer: B
Exam tip: Same temperature means same average kinetic energy per particle, not same speed. Heavier particles move more slowly.
25IBP2S-009-03-08Temperature & molecular energyPaper 1AEasy1 mark
The infrared detectors on a space telescope are cooled to a temperature close to absolute zero. Which statement describes absolute zero?
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Absolute zero (0 K) is the lowest possible temperature. At this temperature the random kinetic energy of the particles is at its minimum. Gases liquefy at many different temperatures, not at one shared temperature. Water freezes at 273 K. Density does not become zero, because the particles still have mass and occupy a volume.
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Answer B
Answer: B
Exam tip: Link the kelvin scale directly to particle kinetic energy. 0 K is where the kinetic energy is at its minimum.
26IBP2S-009-02-01Stefan–Boltzmann & WienPaper 1AEasy1 mark
The heating element of an electric grill glows when it is switched on. At its working temperature T (in kelvin), the radiation it emits has its peak at wavelength λ. When the grill is turned down, the peak wavelength becomes 1.5λ. Assume the element behaves as a black body. What is the new temperature of the element?
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Wien's displacement law gives λmaxT = constant, so peak wavelength is inversely proportional to absolute temperature. If λmax increases by a factor of 1.5, T decreases by a factor of 1.5, giving T/1.5 = 2T/3. Option B squares the factor. Options C and D treat wavelength as increasing with temperature.
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Answer A
Answer: A
Common error: Wien's law is a simple inverse proportion with no powers. The fourth power belongs only to the Stefan–Boltzmann law.
27IBP2S-009-02-02Stefan–Boltzmann & WienPaper 1AEasy1 mark
An LED-replacement project tests an old-style lamp filament. The filament is replaced by one with half the surface area that operates at twice the absolute temperature. The emissivity is unchanged. What is the ratio of the new radiated power to the original radiated power?
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From P = eσAT4, the new power ratio is (½) × 24 = 16/2 = 8. Option A assumes that halving the area cancels doubling the temperature, as if P were proportional to T. Option B uses T2 instead of T4. Option D doubles the area instead of halving it.
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Answer C
Answer: C
Exam tip: For ratio questions, write P ∝ AT⁴ and substitute the factor for each quantity separately before multiplying.
28IBP2S-009-02-06Stefan–Boltzmann & WienPaper 1AEasy1 mark
An engineer chooses a coating for the radiator fins of a motorcycle engine. Which statement correctly defines the emissivity of a surface?
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Emissivity compares the surface with a perfect emitter at the same temperature, so A is correct. It is a dimensionless number between 0 and 1. Option B describes reflectivity, which is the opposite of good emission. Option C equals eσ, not e, and has units. Option D confuses emissivity with Wien's law.
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Answer A
Answer: A
Exam tip: When defining emissivity, state 'at the same temperature'. Leaving this out makes the definition incomplete.
29B-1A-03Specific heat capacityPaper 1AMedium1 mark
An electric heater of power P warms a liquid of mass m for a time t. The temperature of the liquid rises by Δθ. Only 80 % of the energy supplied by the heater is transferred to the liquid.
What is the specific heat capacity of the liquid?
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Step 1Energy transferred to the liquid = 80 % of the energy supplied = 0.8Pt.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Q = mcΔθ, so c = Q / (mΔθ) = 0.8Pt / (mΔθ).
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis assumes that all of the heater's energy reaches the liquid; the 20 % lost to the surroundings has been ignored.
BCorrect: the liquid receives 0.8Pt, and c = Q / (mΔθ) = 0.8Pt / (mΔθ).
CThe 0.8 has been divided into the energy supplied instead of multiplying it, crediting the liquid with 1.25Pt — more than the heater supplied.
DThe equation has been inverted. More energy for the same rise means a larger c, so the energy belongs on top.
30B-1A-04Latent heat & phase changePaper 1AMedium1 mark
A block of ice at 0 °C is heated at a constant rate until all of the water formed from it has boiled away. No energy is lost to the surroundings.
The specific latent heat of vaporization of water is about seven times its specific latent heat of fusion.
Which sketch graph shows the variation with time t of the temperature θ of the sample?
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Step 1During each phase change the energy supplied increases the intermolecular potential energy while the mean kinetic energy, and so the temperature, stays constant: the graph must be horizontal at 0 °C and at 100 °C.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2At a constant heating rate the time for a phase change is proportional to mL. The same mass melts and boils, so the section at 100 °C lasts about seven times as long as the section at 0 °C.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe two horizontal sections are equal. The time for each phase change is proportional to mL, and L for vaporization is about seven times L for fusion.
BThe temperature keeps rising during melting and boiling. During a phase change the energy increases the intermolecular potential energy, not the mean kinetic energy, so the temperature is constant.
CCorrect: horizontal sections at 0 °C and 100 °C, with the boiling section about seven times as long as the melting section.
DThe lengths of the two horizontal sections are reversed. That would need the latent heat of fusion to be the larger of the two.
A hot metal block is placed in cold water in an insulated container. Neither the block nor the water changes phase. After some time thermal equilibrium is reached.
Which statement about the block and the water at thermal equilibrium is correct?
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Step 1The direction of the resultant energy transfer is set by the temperature difference. Energy passes from block to water until their temperatures are equal; then there is no resultant transfer.
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: temperature difference determines the direction of the resultant transfer, so once the temperatures are equal there is no resultant transfer.
BInternal energy depends on the mass, the material and the phase as well as on the temperature. Equilibrium means equal temperatures, not equal internal energies.
CEqual energies are transferred, not equal temperature changes. The temperature changes are in the inverse ratio of the thermal capacities mc, which are not equal in general.
DThe water gains energy and does not change phase, so its temperature must rise. The common final temperature lies between the two starting temperatures.
One end of a long glass rod is held in a flame. Some time later the other end of the rod has become warm.
Which statement describes how energy is conducted along the rod?
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Step 1Particles at the hot end have a greater mean kinetic energy. Through collisions (interactions) with neighbouring particles that have less kinetic energy they pass energy on, so energy travels along the rod although the particles stay in place.
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: conduction is the transfer of kinetic energy from particles with more to neighbouring particles with less, driven by the temperature difference along the rod.
BParticles in a solid vibrate about fixed positions and do not travel along the rod. Energy carried by moving matter is convection, which needs a fluid.
CParticles do not expand when heated; they vibrate with greater amplitude. Expansion of particles is not the mechanism of conduction.
DThere is no flow of 'cold'. The resultant energy transfer is from the hotter end to the cooler end.
34B-1A-08Stefan–Boltzmann & WienPaper 1AMedium1 mark
In each graph the solid line shows the variation with wavelength λ of the intensity I emitted by a black body at temperature T.
The body is then heated to a higher temperature. Which graph shows, as a dashed line, the emission of the body at the higher temperature?
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Step 1A hotter black body emits more at every wavelength, so the new curve lies above the old one everywhere (the total power, the area under the curve, grows as T⁴).
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Wien's law, λmaxT = 2.9 × 10−3 m K, puts the peak at a shorter wavelength.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe peak has moved to a longer wavelength. By Wien's law a higher temperature gives a shorter peak wavelength.
BThe peak has stayed at the same wavelength. λmax is inversely proportional to T, so the peak must move to a shorter wavelength.
CCorrect: the curve is higher at every wavelength and its peak is at a shorter wavelength.
DThe peak is lower than before. A hotter black body radiates more power at every wavelength, so the whole curve must rise.
35B-1A-25Specific heat capacityPaper 1AMedium1 mark
When an energy Q is supplied to block X, its temperature rises by ΔT. Block Y has twice the mass of X, and the specific heat capacity of Y is 1.5 times that of X.
The same energy Q is supplied to Y. No energy is lost. What is the rise in temperature of Y?
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Step 1ΔT = Q / (mc), so for the same Q the rise is inversely proportional to mc.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2mc for Y is 2 × 1.5 = 3 times that for X, so the rise is ΔT/3.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: Y's thermal capacity mc is 2 × 1.5 = 3 times larger, so the same energy gives a third of the rise.
BOnly the doubled mass has been taken into account. The larger specific heat capacity reduces the rise further.
COnly the specific heat capacity has been taken into account (1/1.5 = 2/3); the doubled mass has been ignored.
DThe proportionality has been inverted. A larger mc gives a smaller temperature rise for the same energy, not a larger one.
36B-1A-26Latent heat & phase changePaper 1AMedium1 mark
The water in an electric kettle is boiling at 100 °C. The heating element supplies energy at a rate of 2.0 kW, and all of this energy is transferred to the water.
Specific latent heat of fusion of water = 3.34 × 105 J kg−1. Specific latent heat of vaporization of water = 2.26 × 106 J kg−1.
Which row gives the mass of water turned into steam in 60 s, and the change in the intermolecular potential energy of the water that boils?
Mass turned into steamIntermolecular potential energy
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Step 1Energy supplied in 60 s: Q = Pt = 2000 × 60 = 1.2 × 105 J.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2m = Q/L = 1.2 × 105 / 2.26 × 106 = 0.053 kg ≈ 53 g. The temperature stays at 100 °C, so the mean kinetic energy is unchanged: the energy separates the molecules and increases their potential energy.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThe mass is right, but the energy supplied must go somewhere. At constant temperature the mean kinetic energy is unchanged, so the intermolecular potential energy increases as the molecules are pulled apart.
BCorrect: 1.2 × 10⁵ J boils 1.2 × 10⁵ / 2.26 × 10⁶ = 0.053 kg, and this energy increases the intermolecular potential energy at constant temperature.
CTwo errors: the latent heat of fusion has been used for a change from liquid to vapour, and the energy has not been assigned to the intermolecular potential energy, which increases.
DThe latent heat of fusion has been used (1.2 × 10⁵ / 3.34 × 10⁵ = 0.36 kg). Boiling is a change from liquid to vapour, so the latent heat of vaporization applies.
Two rods of the same length carry the same temperature difference between their ends. Rod X has four times the thermal conductivity of rod Y and twice the cross-sectional area.
The rate of thermal energy transfer along X is how many times that along Y?
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Step 1ΔQ/Δt = kAΔT/Δx. The length and the temperature difference are the same for both rods, so the rate is proportional to the product kA.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2k is four times larger and A is twice as large, and these are independent factors that multiply.
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Step 3Ratio = 4 × 2 = 8.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the rate is proportional to kA, so the ratio is 4 × 2 = 8.
BThe two factors have been added. In ΔQ/Δt = kAΔT/Δx they multiply.
CThis uses the conductivity ratio only, ignoring the difference in cross-sectional area.
DThis uses the area ratio only. The conductivity also differs, and both factors multiply the rate.
A metal block of mass M is heated to a temperature θh and then placed in an insulated cup containing a mass m of water at temperature θc. The final temperature of the block and the water is θf. The specific heat capacity of water is cw. Energy transferred to the cup and to the surroundings is negligible.
What is the specific heat capacity of the metal?
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Step 1Energy lost by the block = energy gained by the water: Mc(θh − θf) = mcw(θf − θc).
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2So c = mcw(θf − θc) / [M(θh − θf)].
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThe two temperature changes have been interchanged. The block falls from θh to θf; the water rises from θc to θf.
BThe two masses have been interchanged. The mass of water belongs with cw on the side of the energy gained by the water.
CThe fall in temperature of the block has been taken as θh − θc. The block only cools to the final temperature θf, not to θc.
DCorrect: equating the energy lost by the block to the energy gained by the water and rearranging for c gives this expression.
39B-1A-48Stefan–Boltzmann & WienPaper 1AMedium1 mark
The thermal radiation emitted by the outer surface of a mug of hot tea at 57 °C has its peak at a wavelength of 8.79 μm. The radiation emitted by the heating element of an oven has its peak at 2.93 μm. Both surfaces behave as black bodies.
What is the temperature of the heating element?
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Step 1Wien's law: λmaxT is constant, so T ∝ 1/λmax, with T in kelvin: Ttea = 57 + 273 = 330 K.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Telement = 330 × 8.79/2.93 = 3 × 330 = 990 K (or 2.9 × 10−3/2.93 × 10−6 = 990 K), which is 990 − 273 = 717 °C.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe wavelength ratio has been inverted: 330/3 = 110 K = −163 °C. A shorter peak wavelength means a hotter body.
BThe Celsius temperature has been multiplied by 3. Wien's law needs the absolute temperature, which is 330 K, not 57.
CCorrect: the peak wavelength is one third of that from the mug, so the absolute temperature is three times 330 K, 990 K, which is 717 °C.
D990 is the temperature in kelvin; it has not been converted to degrees Celsius.
40B-1A-65Density & the molecular modelPaper 1AMedium1 mark
Liquid X has density ρ and liquid Y has density 3ρ. Equal masses of X and Y are mixed. The liquids do not react, and the volume of the mixture is equal to the total volume of X and Y before mixing.
What is the density of the mixture?
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Step 1For a mass m of each liquid the total volume is m/ρ + m/3ρ = 4m/3ρ.
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AThis divides the mass of only one liquid, m, by the total volume. The mixture has a mass of 2m.
BCorrect: 2m ÷ (4m/3ρ) = 1.5ρ. The mixture contains three times as much volume of X as of Y, so its density is nearer to ρ than to 3ρ.
CThis is the mean of the two densities, which would be right for equal VOLUMES. Equal masses of X and Y have unequal volumes: X occupies three times the volume of Y.
DThe densities have been added. Density is total mass ÷ total volume; it is not additive.
41B-1A-66Temperature & molecular energyPaper 1AMedium1 mark
A sealed lamp contains a mixture of helium atoms and xenon atoms that has been at a steady temperature for a long time. The mass of a xenon atom is about 33 times the mass of a helium atom.
Which row compares the xenon atoms with the helium atoms?
Average kinetic energy of the xenon atomsAverage speed of the xenon atoms
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Step 1The two gases are in thermal equilibrium, so they are at the same temperature T. Ek = ³⁄₂kBT does not depend on the mass of the particle, so the average kinetic energies are equal.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2With equal ½mv², the heavier xenon atoms must move more slowly (about 1/√33 ≈ 0.17 of the speed of the helium atoms).
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AEqual kinetic energy does not mean equal speed. ½mv² is the same for both, so the atom with 33 times the mass must have the smaller speed.
BThis assumes that the atoms move at the same speed, so that the heavier ones carry more kinetic energy. In fact the temperature fixes the average kinetic energy, not the speed.
CThe speed is right, but the average kinetic energy depends only on the temperature, which is the same for both gases.
DCorrect: same temperature means the same average kinetic energy ³⁄₂kBT; with 33 times the mass, the xenon atoms must move more slowly.
42B-1A-67Latent heat & phase changePaper 1AMedium1 mark
Water is stored in a porous clay pot that stands in dry, moving air. Water seeps through the clay and evaporates from the outer surface of the pot. The water inside cools to a temperature below that of the air.
Which statement explains why evaporation lowers the temperature of the water that remains?
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Step 1The molecules have a range of kinetic energies. Only the fastest molecules at the surface have enough energy to escape against the intermolecular attraction, so those left behind have a smaller average kinetic energy — and temperature is a measure of the average kinetic energy.
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: evaporation removes the most energetic molecules, lowering the average kinetic energy, and so the temperature, of the remaining water.
BTemperature measures the average random kinetic energy, not the potential energy. A fall in potential energy alone would not lower the temperature.
CEvaporation takes place from the surface at any temperature. Boiling, which happens throughout the liquid, needs the boiling point.
DThe stem says the water ends up colder than the air. Energy then flows from the air into the water, so the air cannot be what cools it.
43B-1A-71Stefan–Boltzmann & WienPaper 1AMedium1 mark
A small electrically heated plate behaves as a black body. Energy is lost from the plate only by radiation, and the radiation it absorbs from the surroundings is negligible. The electrical power supplied to the plate is doubled.
When the temperature of the plate is steady again, what is (new kelvin temperature) / (original kelvin temperature)?
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Step 1In the steady state the power radiated equals the electrical power, so the radiated power doubles.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2L = σAT⁴ with the same A: T⁴ doubles, so T increases by a factor of 21/4 = 1.19.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: T ∝ L1/4, and 21/4 = 1.19.
BThis takes the square root (√2 = 1.41), as if the radiated power were proportional to T².
CThis assumes that the radiated power is proportional to T.
DThis raises 2 to the fourth power instead of taking the fourth root: it is the power ratio for a doubled temperature.
44B-1A-73Luminosity & apparent brightnessPaper 1AMedium1 mark
Star Y is 2.0 × 105 times as far from the Earth as the Sun is. The apparent brightness of Y at the Earth is 1.0 × 10−10 times the apparent brightness of the Sun.
What is the luminosity of Y? L☉ is the luminosity of the Sun.
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Step 1b = L/4πd², so L = 4πd²b and, comparing with the Sun, LY/L☉ = (bY/b☉) × (dY/d☉)².
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45B-1A-74Temperature & molecular energyPaper 1AMedium1 mark
A sample of an ideal gas is heated from 0 °C to 300 °C.
Which graph shows how the average kinetic energy Ek of the molecules of the gas varies with the temperature θ of the gas in degrees Celsius?
Sketch graphs — not to scale. The temperature axis runs from 0 °C to 300 °C in each graph.Show mark scheme
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Step 1Ek = ³⁄₂kBT with T = θ + 273, so Ek is a linear function of θ.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2At θ = 0 °C, T = 273 K and Ek is not zero: a straight line with a positive intercept (at 0 °C Ek is 273/573 ≈ 0.48 of its value at 300 °C).
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AA line through the origin treats the Celsius temperature as if it were the absolute temperature. At 0 °C the molecules still have kinetic energy; Ek would be zero only at −273 °C.
BThis curve makes Ek proportional to T². The average kinetic energy is proportional to the kelvin temperature itself.
CThis curve makes Ek proportional to √T, which is how the molecular SPEED depends on temperature, not the kinetic energy.
DCorrect: a straight line, because Ek ∝ (θ + 273), with a non-zero value at 0 °C.
46IBP2S-009-01-03Latent heat & phase changePaper 1AMedium1 mark
A marathon runner is cooled as sweat evaporates from their skin at a temperature well below 100 °C. Which statement best explains why evaporation cools the remaining sweat and the skin?
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Molecules in a liquid have a spread of kinetic energies. Only the fastest molecules near the surface have enough energy to overcome the intermolecular attractions and escape. When they leave, the mean kinetic energy of the remaining molecules falls, and so does the temperature. The liquid then absorbs energy from the skin. Option A is wrong because evaporation occurs at any temperature. Option B wrongly links temperature to potential energy. Option D describes run-off, not evaporation.
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Answer C
Answer: C
Common error: Temperature depends on mean kinetic energy, not potential energy. The cooling comes from losing the most energetic molecules.
47IBP2S-009-01-04Latent heat & phase changePaper 1AMedium1 mark
The superconducting magnet of a hospital MRI scanner is kept in liquid helium at its boiling point. Thermal energy leaks into the helium at a steady rate of 0.20 W. The specific latent heat of vaporisation of helium is 2.0 × 104 J kg−1. What mass of helium boils off in 5.0 × 104 s?
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The energy leaking in is Q = Pt = 0.20 × 5.0 × 104 = 1.0 × 104 J. The mass boiled off is m = Q/L = 1.0 × 104 / 2.0 × 104 = 0.50 kg. Each distractor comes from a specific slip: • A divides the power by L and forgets the time. • C inverts the ratio, giving L/(Pt). • D multiplies Pt by L.
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Answer B
Answer: B
Exam tip: Check the units of your answer: J ÷ (J kg⁻¹) gives kg. An answer of 10⁸ kg for a scanner should prompt a check.
48IBP2S-009-01-06Mechanisms of thermal transferPaper 1AMedium1 mark
Air-conditioning units that cool a room are usually mounted high on a wall, close to the ceiling. Which statement explains why this position cools the whole room effectively?
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The air that passes the unit cools, so its particles move more slowly and it contracts. Its density increases and it sinks. Warmer, less dense air near the floor is pushed upwards towards the unit, where it is cooled in turn. This circulating convection current eventually cools the whole room. Option A treats thermal energy as a substance that rises. Option B is wrong because cooling does not change molecular mass. Option C reverses the density change.
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Answer D
Answer: D
Common error: Do not write 'heat rises'. Convection happens because a warmer fluid is less dense and is displaced upwards by denser, cooler fluid.
49IBP2S-009-01-08Temperature & molecular energyPaper 1AMedium1 mark
On a hot day, the air in a car tyre warms from 20 °C to 40 °C. By what factor does the average random kinetic energy of the air molecules increase?
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The average kinetic energy is \(\bar{E}_k = \frac{3}{2}k_BT\), so it is proportional to the kelvin temperature. The factor is (40 + 273)/(20 + 273) = 313/293 ≈ 1.07. Option C uses the Celsius temperatures directly. Option D also squares that ratio, as if kinetic energy were proportional to T2. Option A is the ratio of the mean molecular speeds, √(313/293), not of the energies.
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Answer B
Answer: B
Common error: Always convert to kelvin before taking a ratio of temperatures. Doubling a Celsius temperature does not double the molecular kinetic energy.
50IBP2S-009-04-03Specific heat capacityPaper 1AMedium1 mark
Two brake discs, A and B, absorb the same amount of energy during an emergency stop. Disc B has twice the mass of disc A. The specific heat capacity of the material of B is one third of that of A. What is \(\frac{\Delta T_B}{\Delta T_A}\), the ratio of the temperature rise of B to that of A?
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From Q = mcΔT, the temperature rise is ΔT = Q/(mc), so ΔT is proportional to 1/(mc). For disc B, mc = 2m × c/3 = (2/3)mc. Therefore ΔTB/ΔTA = 1/(2/3) = 3/2.
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Answer C
Answer: C
Exam tip: For a ratio question, write ΔT = Q/(mc) and substitute the factors directly. Then check the direction: B has the smaller heat capacity, so it should heat up more.
51IBP2S-009-04-04Method of mixturesPaper 1AMedium1 mark
Two modules are docked in space and come into thermal contact. They are isolated from their surroundings. Module X has a heat capacity of 300 J K−1 and is at 80 °C. Module Y has a heat capacity of 100 J K−1 and is at 0 °C. What is their final common temperature?
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The energy lost by X equals the energy gained by Y: 300(80 − T) = 100(T − 0) 24000 = 400T, so T = 60 °C. The final temperature lies closer to that of X, which has the larger heat capacity.
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Answer C
Answer: C
Common error: A simple average (40 °C) is only correct when the heat capacities are equal. The answer 20 °C comes from swapping the heat capacities.
52IBP2S-009-04-05Specific heat capacityPaper 1AMedium1 mark
A sample of engine coolant of mass m is heated by an electric heater of constant power P. A data logger shows that the temperature rises at a constant rate R (in K s−1). Energy losses are negligible. What is the specific heat capacity of the coolant?
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In time t, the energy supplied is Pt. This equals mcΔT, where ΔT = Rt. So Pt = mcRt, which gives P = mcR. Therefore c = P/(mR).
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Answer C
Answer: C
Exam tip: Check by units: W/(kg × K s⁻¹) = J s⁻¹/(kg K s⁻¹) = J kg⁻¹ K⁻¹. The option P/R has no mass term, so it gives a heat capacity, not a specific heat capacity.
53IBP2S-009-04-07Temperature & molecular energyPaper 1AMedium1 mark
The absolute temperature of an ideal gas in a spacecraft propellant tank rises from 150 K to 300 K. What happens to the average random kinetic energy of the gas molecules?
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The average random kinetic energy of the molecules is \(\frac{3}{2}kT\), which is directly proportional to the absolute temperature. Doubling T from 150 K to 300 K doubles the mean kinetic energy. This does not depend on the volume.
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Answer B
Answer: B
Common error: The factor √2 applies to the root-mean-square speed, not to the kinetic energy.
54IBP2S-009-04-09Method of mixturesPaper 1AMedium1 mark
A processor dissipates 50 W into a copper heat sink of mass 0.25 kg. The specific heat capacity of copper is 400 J kg−1 K−1. Assume no energy leaves the heat sink. At what rate does the temperature of the heat sink rise?
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The rate of temperature rise is ΔT/Δt = P/(mc). P/(mc) = 50/(0.25 × 400) = 50/100 = 0.50 K s−1.
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Answer B
Answer: B
Common error: Inverting the expression gives 2.0 K s⁻¹, and multiplying everything together gives 5000 K s⁻¹. Check the units to catch both mistakes.
55IBP2S-009-00-03Luminosity & apparent brightnessPaper 1AMedium1 mark
A space probe travels from Earth's orbit (1.0 AU from the Sun) to Jupiter's orbit (5.2 AU from the Sun). Approximately what fraction of the apparent brightness of the Sun at Earth does the probe receive at Jupiter?
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Apparent brightness \(b = \frac{L}{4\pi d^2}\), so b ∝ 1/d². (5.2)² ≈ 27, giving about 1/27. The 1/5.2 option assumes b ∝ 1/d, 1/10 doubles the distance instead of squaring it, and 1/140 uses 1/d³.
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Answer C
Answer: C
Common error: Brightness follows an inverse-square law because the power spreads over the area of a sphere, 4πd², not over a distance.
56IBP2S-009-00-04Luminosity & apparent brightnessPaper 1AMedium1 mark
A red giant star has a radius three times that of star S and a surface temperature half that of star S. What is the ratio of the luminosity of the red giant to the luminosity of star S?
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\(L = 4\pi R^2\sigma T^4\), so L ∝ R²T⁴. Ratio = 3² × (1/2)⁴ = 9/16. 3/16 forgets to square R, 9/4 uses T² instead of T⁴, and 3/2 treats both as linear.
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Answer B
Answer: B
Exam tip: In Stefan–Boltzmann ratio questions, radius is squared and temperature is raised to the fourth power: check both powers before choosing.
An outside wall is made from a layer of brick (thermal conductivity 0.80 W m−1 K−1) and a layer of insulating foam (thermal conductivity 0.040 W m−1 K−1) of equal thickness and equal area, in contact with each other. Thermal energy is conducted through the wall at a steady rate. Which statement is correct?
Equal thickness and equal area
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In steady state, energy does not accumulate at the boundary, so the same power passes through both layers in series. With equal A and Δx, P = kAΔT/Δx gives ΔT ∝ 1/k, so ΔTfoam/ΔTbrick = 0.80/0.040 = 20.
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Answer B
Answer: B
Exam tip: For layers in series at steady state, the power is the same in every layer; the poorer conductor takes the larger share of the temperature drop.
58IBP2S-009-00-07Luminosity & apparent brightnessPaper 1AMedium1 mark
Two stars, X and Y, have the same apparent brightness when observed from Earth. The luminosity of X is 16 times the luminosity of Y. What is the ratio \(\frac{d_X}{d_Y}\) of their distances from Earth?
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b = L/(4πd²). Equal b gives LX/dX² = LY/dY², so dX/dY = √(LX/LY) = √16 = 4. The brighter-source star must be further away. 16 forgets the square root, 2 takes a fourth root (confusion with T⁴), and 1/4 inverts the ratio.
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Answer C
Answer: C
Common error: A more luminous star appearing equally bright must be further away, not closer: sanity-check the direction of your ratio.
59IBP2S-009-00-08Luminosity & apparent brightnessPaper 1AMedium1 mark
An astronomer wants to determine the radius of a distant star, treating it as a black body. Which set of measurements is sufficient?
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Apparent brightness and distance give the luminosity, L = 4πd2b. The peak wavelength gives the surface temperature by Wien's law, T = 2.9×10−3 m K / λmax. The radius then follows from L = 4πR2σT4, so R = √(L/(4πσT4)). Option A gives T but not L, because brightness is missing. Option C gives T and b but not L, because distance is missing. Option D gives L but not T, and mass does not help find the radius.
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Answer B
Answer: B
Exam tip: Finding a radius needs both L and T. L comes from b and d, and T comes from Wien's law.
60IBP2S-009-00-09Stefan–Boltzmann & WienPaper 1AMedium1 mark
Thermal conductivity is measured in W m−1 K−1. What is this unit expressed in SI base units?
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1 W = 1 J s−1 = 1 kg m2 s−3. Dividing by m and K gives kg m s−3 K−1. The second option forgets the m−1, the third subtracts two powers of metre, and the fourth treats the watt as having s−2 (like the newton or joule).
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Answer A
Answer: A
Common error: The watt is a joule per second, so it carries s⁻³, not s⁻².
61IBP2S-009-03-03Temperature & molecular energyPaper 1AMedium1 mark
Gas flowing through a heat exchanger is heated from 27 °C to 327 °C. What is the ratio \(\frac{\text{final average kinetic energy of a molecule}}{\text{initial average kinetic energy of a molecule}}\)?
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The average kinetic energy is proportional to the absolute temperature. Converting gives 27 °C = 300 K and 327 °C = 600 K, so the ratio is 600/300 = 2. The option 12 comes from using the Celsius values (327/27). The option √2 is the ratio of the molecular speeds, not the energies. The option 4 wrongly squares the temperature ratio.
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Answer B
Answer: B
Common error: Ratios involving molecular energy must use kelvin temperatures. Celsius ratios have no physical meaning.
62IBP2S-009-03-04Density & the molecular modelPaper 1AMedium1 mark
A 3D-printed lattice bracket for an aircraft is made from a titanium alloy of density 4.5 g cm−3. The bracket has a mass of 144 g and occupies an external volume of 80 cm3. What fraction of the external volume is empty space?
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The volume of metal is m/ρ = 144/4.5 = 32 cm3. The fraction filled with metal is 32/80 = 0.40, so the empty fraction is 1 − 0.40 = 0.60. The option 0.40 is the solid fraction, not the empty fraction. The option 1.8 g cm−3 is the average density of the whole bracket, not a fraction. The option 2.5 is 4.5/1.8, which is the inverse of the solid fraction.
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Answer B
Answer: B
Exam tip: Find the volume of material using the material's own density, then compare it with the overall volume.
63IBP2S-009-03-06Internal energy & the molecular modelPaper 1AMedium1 mark
Crushed ice at 0 °C in a sports injury pack melts while it absorbs energy from an athlete's ankle. The temperature of the pack stays at 0 °C. What happens to the average kinetic energy and the average potential energy of the water molecules during melting?
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The temperature is constant, so the average kinetic energy of the molecules does not change. The energy absorbed is used to overcome intermolecular bonds, which increases the intermolecular potential energy. The internal energy therefore rises without any change in temperature. Potential energy cannot decrease, because energy is being supplied to separate the molecules.
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Answer B
Answer: B
Exam tip: Constant temperature during a phase change means constant average kinetic energy. All the energy transferred changes the potential energy.
64IBP2S-009-03-09Internal energy & the molecular modelPaper 1AMedium1 mark
Two copper heat sinks in a computer have masses of 0.20 kg and 0.40 kg. Both are at 60 °C. Which statement is correct?
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Temperature measures the average kinetic energy per particle, so equal temperatures mean equal average kinetic energy. Internal energy is the total kinetic and potential energy of all the particles. The 0.40 kg heat sink is the same material in the same state but contains twice as many atoms, so its internal energy is twice as large.
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Answer B
Answer: B
Common error: Temperature is not the same as internal energy. Temperature is an average per particle, while internal energy is a total that depends on the amount of material.
65IBP2S-009-02-03Temperature & molecular energyPaper 1AMedium1 mark
The tip of a soldering iron is heated from 27 °C to 327 °C. The tip can be treated as a black body, and its area does not change. By what factor does the power radiated by the tip increase?
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The Stefan–Boltzmann law requires absolute temperature. The tip warms from 300 K to 600 K, so the temperature doubles and the power increases by 24 = 16. Option A assumes power is proportional to T. Option B uses T2. Option D uses Celsius temperatures: (327/27)4 ≈ 2.2×104.
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Answer C
Answer: C
Common error: Substituting Celsius temperatures into P = eσAT⁴ gives a wildly wrong answer. Convert to kelvin first.
66IBP2S-009-02-04Method of mixturesPaper 1AMedium1 mark
Hot water at 80 °C is poured into a ceramic mug at 20 °C. The mass of the water equals the mass of the mug. The specific heat capacity of the ceramic is one quarter of that of water. Assume no energy is transferred to the surroundings. What is the final common temperature?
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Energy lost by the water equals energy gained by the mug: m cw(80 − T) = m (cw/4)(T − 20). This simplifies to 4(80 − T) = T − 20, so 340 = 5T and T = 68 °C. Option B simply averages the temperatures, which ignores the different heat capacities. Option A places the factor ¼ on the water instead of the mug. Option D assumes the mug has no effect at all.
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Answer C
Answer: C
Exam tip: A check on the answer: the final temperature must lie closer to the starting temperature of the object with the larger heat capacity, mc.
67IBP2S-009-02-05Stefan–Boltzmann & WienPaper 1AMedium1 mark
Two stars X and Y have the same radius. The peak wavelength in the spectrum of X is half that of Y. What is the ratio of luminosity of X to luminosity of Y?
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By Wien's law, halving λmax means X has twice the surface temperature of Y. With equal radii, L = σ4πR2T4 gives L ∝ T4, so the ratio is 24 = 16. Option A treats luminosity as proportional to λ4 rather than T4. Options B and C use the wrong power of T.
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Answer D
Answer: D
Exam tip: Wien's law and the Stefan–Boltzmann law are often combined. Convert the wavelength information into a temperature ratio first.
68IBP2S-009-02-08Stefan–Boltzmann & WienPaper 1AMedium1 mark
A patient's exposed skin has area A, emissivity e and absolute temperature Ts. The hospital room walls are at absolute temperature Tr, where Tr < Ts. Which expression gives the net rate of energy loss by radiation from the skin?
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The skin emits eσATs4 and absorbs eσATr4 from the surroundings, so the net loss is eσA(Ts4 − Tr4). Option A raises the temperature difference to the fourth power, which is not the same thing. Option C ignores absorption from the room. Option D divides by e instead of multiplying.
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Answer B
Answer: B
Common error: (Ts − Tr)⁴ is not equal to Ts⁴ − Tr⁴. Raise each temperature to the fourth power separately, then subtract.
69IBP2S-009-02-09Method of mixturesPaper 1AMedium1 mark
A student measures the specific heat capacity of a metal block. She heats the block in boiling water, records the temperature of the boiling water, then carries the block to a calorimeter of cold water. The block cools noticeably in the air during the transfer. How does this affect the calculated specific heat capacity of the metal?
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The calculation uses cmetal = mwcwΔTw/(mmΔTm). The block's real starting temperature in the calorimeter is below the recorded value, so ΔTm is overestimated. Because ΔTm is in the denominator, cmetal is underestimated. Option D identifies the right cause but reverses its effect. Option B wrongly counts the energy lost to the air in the calculation. Option C ignores that the recorded data are wrong.
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Answer A
Answer: A
Exam tip: For systematic-error questions, write the equation used, decide which measured quantity is wrong and in which direction, then see where it appears in the equation.
71B-1A-24Density & the molecular modelPaper 1AHard1 mark
At 100 °C and atmospheric pressure the density of liquid water is 958 kg m−3 and the density of steam is 0.598 kg m−3.
What is the best estimate of the ratio (mean separation of the molecules in steam) / (mean separation of the molecules in liquid water)?
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Step 1The molecules have the same mass in both phases, so the volume occupied per molecule is inversely proportional to the density: steam has 958/0.598 ≈ 1600 times the volume per molecule.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Volume per molecule ∝ (separation)³, so the separation ratio is 16001/3 ≈ 12.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThe ratio has been inverted. The molecules in steam are much further apart than those in the liquid.
BCorrect: the volume per molecule is about 1600 times larger in steam, and the cube root of 1600 is about 12.
CA square root has been taken. Volume goes as the cube of the separation, so a cube root is needed.
DThis is the ratio of the densities, which gives the ratio of the volumes per molecule, not of the separations.
A lagged composite bar is made of a copper section and a stainless-steel section of equal length and equal cross-sectional area, joined end to end. The thermal conductivity of copper is about 25 times that of stainless steel. The free end of the copper is held at 100 °C, the free end of the steel at 0 °C, and the temperatures are steady.
Which graph shows how the temperature θ varies with the distance x along the bar from its hot end?
Sketch graphs — not to scaleShow mark scheme
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Notes
Step 1The bar is lagged, so in the steady state no energy is lost from the sides or stored at the joint: the rate of transfer ΔQ/Δt = kAΔT/Δx is the same through both sections.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2With equal A and lengths, the temperature gradient is inversely proportional to k: the steel has a gradient about 25 times steeper, so almost all of the 100 K drop is across it (the joint is at about 96 °C).
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: a straight, almost flat line through the copper and a straight, steep line through the steel, because the same rate of transfer needs a temperature gradient 25 times larger in the poorer conductor.
BThe gradients have been interchanged. The good conductor needs only a small temperature gradient to carry the same rate of transfer, so the steep section must be in the steel.
CA single straight line treats the bar as one material. The rate is the same in both sections, but the conductivities differ, so the gradients must differ.
DThis assumes that the rate of transfer falls as energy passes along the bar. The bar is lagged and the temperatures are steady, so the rate is the same at every point and within each section the graph is a straight line.
74B-1A-49Latent heat & phase changePaper 1AHard1 mark
A liquid in a container loses energy to its surroundings at a constant rate. The graph shows how its temperature θ varies with time. The liquid cools by Δθ in a time t1; its temperature is then constant for a time t2 while it freezes.
What is (specific latent heat of fusion) / (specific heat capacity of the liquid)?
Graph NOT drawn to scaleShow mark scheme
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Notes
Step 1With a constant rate of energy loss P: cooling the liquid, Pt1 = mcΔθ; freezing, Pt2 = mL.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Dividing: L/c = Δθt2/t1.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AΔθ has been left out, as if c were the energy per kilogram. Specific heat capacity is per kilogram per kelvin, so Δθ stays in the ratio.
BCorrect: mL = Pt2 and mcΔθ = Pt1, so L/c = Δθt2/t1.
CThe two times have been interchanged. The latent heat goes with the freezing time t2, which belongs in the numerator.
Ice at 0 °C is added to 0.300 kg of fruit juice at 25.0 °C in an insulated jug. When all of the ice has melted, the temperature of the mixture is 5.0 °C. No energy is exchanged with the jug or the surroundings.
Specific heat capacity of juice and of water = 4.18 × 103 J kg−1 K−1. Specific latent heat of fusion of ice = 3.34 × 105 J kg−1.
What mass of ice was added?
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Notes
Step 1Energy lost by the juice = 0.300 × 4180 × 20.0 = 2.51 × 104 J.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Each kilogram of ice needs 3.34 × 105 J to melt and a further 4180 × 5.0 = 20900 J for the meltwater to warm from 0 °C to 5.0 °C: 3.549 × 105 J kg−1 in all.
—
Step 3m = 2.51 × 104 / 3.549 × 105 = 0.0707 kg ≈ 71 g.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThe meltwater has been warmed through 25 K instead of 5 K: 2.51 × 104 / (3.34 × 105 + 4180 × 25) = 57 g. The meltwater starts at 0 °C and ends at the final temperature of 5.0 °C.
BCorrect: 2.51 × 104 J melts the ice AND warms the meltwater by 5.0 K, so m = 2.51 × 104 / 3.549 × 105 = 71 g.
CThe warming of the meltwater from 0 °C to 5.0 °C has been left out: 2.51 × 104 / 3.34 × 105 = 75 g.
DThe juice has been cooled to 0 °C instead of to 5.0 °C (fall of 25 K): 3.13 × 104 / 3.549 × 105 = 88 g.
A closed cool box has walls of expanded polystyrene. A second cool box is made of the same material, and every dimension of the second box, including the thickness of its walls, is twice the corresponding dimension of the first box.
The temperatures inside and outside the two boxes are the same. Energy enters each box only by conduction through its walls, and the walls are thin compared with the size of the box.
What is (rate of thermal energy transfer into the second box) / (rate of thermal energy transfer into the first box)?
Show mark scheme
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Notes
Step 1ΔQ/Δt = kAΔT/Δx, with the same k and ΔT for both boxes.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Doubling every length makes the wall area 2² = 4 times larger.
—
Step 3The walls are also twice as thick, which halves the temperature gradient ΔT/Δx: ratio = 4 × ½ = 2.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis scales the rate with the volume of the box (2³ = 8). Conduction depends on the area of the walls and on their thickness, not on the volume enclosed.
BThis allows for the larger area (×4) but ignores the thicker walls, which halve the temperature gradient.
CCorrect: the area is 4 times larger and the thickness 2 times larger, so the rate is 4/2 = 2 times larger.
DThis uses only the doubled thickness. The wall area is also larger, by a factor of 4.
77B-1A-70Mechanisms of thermal transferPaper 1AHard1 mark
On the Earth a candle flame is tall and pointed. On board an orbiting space station, where the station and everything in it are in free fall, a candle flame in still air is almost spherical and much dimmer.
Which statement explains this?
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Step 1Convection on the Earth happens because the denser, cooler air has more weight per unit volume and sinks, pushing the less dense hot gas upwards (buoyancy). In free fall there is no such resultant force, so there is no convection: the hot gases surround the wick and oxygen arrives only slowly, giving a small, round, dim flame.
✓ 1
Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: convection is driven by the buoyant force that arises from density differences in a gravitational field. In free fall that force is absent, so no convection current forms.
BThe hot gases ARE still less dense than the cooler air — they are at a higher temperature. What is missing is the buoyant force that would make the density difference produce motion.
CRadiation needs no medium and no gravity, so the flame still radiates. Radiation does not shape the flame.
DConvection is bulk flow driven by density differences under gravity; it has a preferred direction (up). With no buoyant force there is no convection at all, not convection in all directions.
78IBP2S-009-01-07Latent heat & phase changePaper 1AHard1 mark
A solar thermal store contains a solid wax. A heater supplies energy to the wax at a constant rate, and all of it is absorbed. The temperature of the solid rises by 20 K in 2.0 minutes. The wax then reaches its melting point and takes a further 10 minutes to melt completely. What is the value of \(\frac{L}{c}\), where L is the specific latent heat of fusion and c is the specific heat capacity of the solid wax?
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Notes
While the solid heats up: P × (2.0 min) = mc(20 K). While it melts: P × (10 min) = mL. Dividing the second equation by the first gives L/(c × 20 K) = 10/2.0 = 5, so L/c = 5 × 20 K = 100 K. Option B uses only the ratio of times and ignores the temperature rise. Option A divides the temperature rise by the time ratio instead of multiplying. Option D inverts the ratio and has the wrong unit.
✓ 1
Answer C
Answer: C
Exam tip: With a constant power supply, energy is proportional to time. You can compare stages without ever knowing P or m.
79IBP2S-009-01-10Method of mixturesPaper 1AHard1 mark
A cool box at a sports event contains 0.20 kg of water at 40 °C. Ice at 0 °C is added, and the system is insulated. What is the minimum mass of ice needed to bring all the water to 0 °C? Specific heat capacity of water = 4.2 × 103 J kg−1 K−1 Specific latent heat of fusion of ice = 3.36 × 105 J kg−1
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Notes
Energy released by the water cooling to 0 °C: 0.20 × 4.2 × 103 × 40 = 3.36 × 104 J. This energy melts the ice, and the melted ice stays at 0 °C, so m = 3.36 × 104 / 3.36 × 105 = 0.10 kg. Each distractor comes from a specific slip: • B wrongly also warms the melted ice to 40 °C. • C uses ΔT = 313 K, the kelvin temperature, instead of the temperature change. • D inverts the ratio.
✓ 1
Answer A
Answer: A
Common error: ΔT is a temperature change, so a change of 40 °C is a change of 40 K. Never use 313 K as ΔT.
80IBP2S-009-04-08Latent heat & phase changePaper 1AHard1 mark
An engineer measures the specific heat capacity of a battery cell for an electric vehicle. An electric heater supplies a measured amount of energy, and the engineer uses c = E/(mΔT). The cell loses a significant amount of energy to the surrounding air during heating. How does the calculated value compare with the true value?
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Notes
E is measured at the heater, so it is the full energy supplied. Some of that energy escapes, so the temperature rise ΔT is smaller than it would be with no losses. A smaller denominator in E/(mΔT) gives a calculated c that is too large.
✓ 1
Answer A
Answer: A
Exam tip: When you explain a systematic error, say which measured quantity is affected and in which direction. Then follow the change through the equation.
81IBP2S-009-04-10Method of mixturesPaper 1AHard1 mark
Two blocks, P and Q, are placed in thermal contact inside an insulated container. By the time they reach equilibrium, the temperature of P has fallen by 30 K and the temperature of Q has risen by 10 K. The masses of the blocks are not known. Which statement must be correct?
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Notes
The container is insulated, so the energy lost by P equals the energy gained by Q: CP × 30 = CQ × 10, so CQ = 3CP. The heat capacity of Q is three times that of P. Heat capacity is C = mc. Without the masses, nothing can be concluded about the specific heat capacities, so option A need not be true. Option C inverts the ratio: the block with the larger temperature change has the smaller heat capacity. Option D contradicts energy conservation, because the two energies are equal.
✓ 1
Answer B
Answer: B
Exam tip: Heat capacity (J K⁻¹) describes a whole object. Specific heat capacity (J kg⁻¹ K⁻¹) describes a material. Read the options carefully for which one is meant.
82IBP2S-009-00-10Temperature & molecular energyPaper 1AHard1 mark
A ceramic heating element in a toaster warms up from 27 °C to 327 °C. Its emissivity and surface area stay constant. By what factor does the power it radiates increase?
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Notes
Radiated power P = eσAT⁴ requires absolute temperature. 27 °C = 300 K and 327 °C = 600 K, so the factor is (600/300)⁴ = 2⁴ = 16. The factor 2 assumes P ∝ T, 4 assumes P ∝ T², and about 2×10⁴ comes from using Celsius values, (327/27)⁴.
✓ 1
Answer C
Answer: C
Common error: Always convert to kelvin before using the Stefan–Boltzmann law; ratios of Celsius temperatures are meaningless.
83IBP2S-009-03-07Temperature & molecular energyPaper 1AHard1 mark
The atmosphere of a distant exoplanet contains helium (molar mass 4 g mol−1) and methane (molar mass 16 g mol−1) at the same temperature. What is the ratio \(\frac{\text{root mean square speed of helium atoms}}{\text{root mean square speed of methane molecules}}\)?
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Notes
At the same temperature, both types of particle have the same average kinetic energy, \(\frac{1}{2}m\overline{v^2}\). This means \(v_{rms} \propto \frac{1}{\sqrt{m}}\). The ratio is therefore √(16/4) = 2, so helium atoms move faster. The option 4 forgets the square root. The option ½ inverts the ratio. The option 1 assumes that equal temperature means equal speed.
✓ 1
Answer C
Answer: C
Common error: Equal kinetic energy does not mean equal speed. Speed scales as one over the square root of the particle mass.
84IBP2S-009-03-10Internal energy & the molecular modelPaper 1AHard1 mark
A jeweller makes an alloy by melting together equal masses of two metals with densities ρ1 and ρ2. Assume that the volume of the alloy equals the sum of the volumes of the two metals. What is the density of the alloy?
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Notes
Let each mass be m. The total volume is m/ρ1 + m/ρ2 = m(ρ1+ρ2)/(ρ1ρ2). The density is the total mass divided by the total volume: 2m ÷ [m(ρ1+ρ2)/(ρ1ρ2)] = 2ρ1ρ2/(ρ1+ρ2). The simple mean is correct only for equal volumes. The last option forgets that the total mass is 2m.
✓ 1
Answer B
Answer: B
Common error: Averaging densities directly works only for equal volumes. For equal masses, add the volumes and divide the total mass by the total volume.
85IBP2S-009-02-07Luminosity & apparent brightnessPaper 1AHard1 mark
Two stars A and B have the same apparent brightness when observed from Earth. The surface temperature of A is twice that of B. A is twice as far from Earth as B. What is the ratio of the radius of A to the radius of B?
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Notes
Apparent brightness is b = L/(4πd2). Equal b with dA = 2dB requires LA = 4LB. Since L ∝ R2T4, 4 = (RA/RB)2 × 16, so (RA/RB)2 = 1/4 and RA/RB = 1/2. Option B forgets the square root. Option A ignores the distance. Option D inverts the ratio.
✓ 1
Answer C
Answer: C
Common error: Luminosity depends on R², not R. Take the square root at the end.
86IBP2S-009-02-10Method of mixturesPaper 1AHard1 mark
An athlete drops ice at 0 °C into an equal mass of sports drink at 60 °C in an insulated flask. Treat the drink as water: specific heat capacity 4200 J kg−1 K−1, specific latent heat of fusion of ice 3.3×105 J kg−1. Which describes the final state of the contents?
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Notes
Cooling the drink from 60 °C to 0 °C releases 4200 × 60 = 2.52×105 J per kg. Melting all the ice needs 3.3×105 J per kg, which is more. So only about 76% of the ice melts and the mixture stays at 0 °C. Option A ignores latent heat. Option B assumes the energies balance exactly. Option D comes from forcing an all-melted equation, which gives an impossible −9 °C.
✓ 1
Answer C
Answer: C
Exam tip: With ice, first compare the energy available from cooling the liquid to 0 °C with the energy needed to melt all the ice, before writing any mixing equation.
87B-1B-02Latent heat & phase changePaper 1BEasy9 marks
A beaker of water stands on a top-pan balance. An immersion heater in the water is supplied with a power of 150 W ± 2 %, measured with a joulemeter. When the water is boiling steadily, the balance reading m is recorded every 60 s.
t / s
0
60
120
180
240
300
360
m / g
486.3
482.8
478.8
475.2
471.2
467.8
464.0
Graph drawn to scale
(a)
State the absolute uncertainty in a reading of m.
(1)
(b)
Draw the line of best fit for the data.
(1)
(c)
Determine the rate at which water is boiled away, in kg s−1.
(2)
(d)
Calculate the specific latent heat of vaporization L of water obtained from this experiment.
(1)
(e)
The uncertainty in the rate in (c) is ±3 %. Determine the absolute uncertainty in L and write down L with its uncertainty.
(2)
(f)
The accepted value of L is 2.26 × 106 J kg−1. Discuss, with reference to the uncertainty, the difference between the accepted value and the value from this experiment, and suggest the systematic error responsible for it.
(2)
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Notes
Part (a)
±0.1 g
✓ 1
Accept ±0.05 g.
Part (b)
single straight line with the points scattered either side of it
✓ 1
Part (c)
gradient from a large triangle on the line, e.g. mass lost ÷ time
✓ 1
6.2 × 10−5 kg s−1
✓ 1
Accept 6.0 × 10−5 to 6.5 × 10−5 kg s−1. Sign may be omitted. Award [1] max if left in g s−1.
Value and uncertainty must have matching precision. Accept (2.4 ± 0.1) × 106.
Part (f)
the accepted value lies outside the range 2.29–2.53 × 106 J kg−1 so there is a systematic error «not only random error»
✓ 1
Allow ECF from (e).
energy is lost from the beaker/heater to the surroundings, so less water boils away than P alone would produce, the rate is too small and L is too large
✓ 1
The direction (too large) must be explained.
Answers: (c) 6.2 × 10−5 kg s−1 · (d) 2.41 × 106 J kg−1 · (e) (2.41 ± 0.12) × 106 J kg−1(the remaining parts are explanations — see the table above)
Steel ball bearings are left in boiling water for 10 minutes and then transferred quickly into water in an insulated polystyrene cup. The mixture is stirred and the highest temperature reached is recorded. Masses are measured with a balance of resolution 0.1 g and temperatures with a thermometer read to ±0.5 °C.
The specific heat capacity of water is 4180 J kg−1 K−1. Assume that no energy is transferred to the cup or to the air.
Quantity
Value
mass of steel balls
215.2 g
mass of water in cup
150.0 g
temperature of boiling water
99.5 °C
initial temperature of water in cup
18.5 °C
highest temperature of mixture
29.5 °C
Diagram not to scale
(a)
State the absolute uncertainty in the temperature rise of the water in the cup.
(1)
(b)
Calculate the specific heat capacity c of steel.
(2)
(c)
Determine the absolute uncertainty in c.
(2)
(d)
The accepted value for this steel is 490 J kg−1 K−1. Comment on the student’s value.
(1)
(e)
Identify the measurement that contributes most to the uncertainty in c, and suggest how its percentage uncertainty could be reduced.
(1)
(f)
The balls cool slightly in the air while being transferred. State and explain the effect of this on the calculated value of c.
(1)
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Part (a)
±1.0 °C / ±1.0 K
✓ 1
Two readings of ±0.5 °C are added. Accept ±1 °C.
Part (b)
msc(99.5 − 29.5) = mw × 4180 × (29.5 − 18.5)
✓ 1
Energy lost by the steel = energy gained by the water. Masses may be in g on both sides.
490 lies within the range 410–510 J kg−1 K−1, so the value is consistent with the accepted value
✓ 1
Allow ECF from (c).
Part (e)
the temperature rise of the water «≈ 9 %»; use a thermometer of finer resolution / a digital probe OR use less water «or more steel» so that its temperature rise is larger
✓ 1
Both the identification and a matching suggestion are needed.
Part (f)
the steel enters the cup below 99.5 °C, so its actual temperature fall is smaller than the value used; the calculated c is too small
✓ 1
Direction and reason both needed. OWTTE.
Answers: (a) ±1.0 °C · (b) 458 J kg−1 K−1 · (c) ±50 J kg−1 K−1(the remaining parts are explanations — see the table above)
89B-1B-21Specific heat capacityPaper 1BEasy11 marks
A student measures the specific heat capacity c of sunflower oil. An immersion heater of power P = 48.0 W ± 2 % is placed in a mass m of oil in a glass beaker. The oil is stirred, and the time t for its temperature to rise from 25.0 °C to 35.0 °C is measured. The thermometer reads to 0.1 °C. The experiment is repeated for different masses of oil, each time starting with the beaker and heater at 25.0 °C.
In this experiment energy is also needed to warm the beaker and the heater, so Pt = (mc + C)ΔT, where C is the energy needed to raise the temperature of the beaker and heater by 1 K and ΔT = 10.0 K. Each value of t has an uncertainty of ±2 s, shown by the error bars.
m / kg
0.080
0.120
0.160
0.200
0.240
0.280
t / s
69.0
84.2
102.6
118.4
136.8
152.4
Graph drawn to scale
(a)
State the absolute uncertainty in the temperature rise ΔT.
(1)
(b)
Draw the line of best fit for the data and extend it to m = 0.
(1)
(c)
Determine the gradient of your line.
(2)
(d)
Determine c.
(2)
(e)
The uncertainty in the gradient is ±4 %. Determine the absolute uncertainty in c and write down your value of c with its uncertainty.
(2)
(f)
A published value for sunflower oil is 1.97 × 103 J kg−1 K−1. Comment on your result.
(1)
(g)
Determine C using your graph.
(2)
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Part (a)
±0.2 K «two readings each ±0.1 °C»
✓ 1
Accept ±0.2 °C.
Part (b)
single straight line through all the error bars, extended to cut the t axis at a positive intercept
✓ 1
Do not accept a line forced through the origin.
Part (c)
gradient from a large triangle on the line
✓ 1
gradient = 422 s kg−1
✓ 1
Accept 401–443 s kg−1. Award [2] for a correct answer in the range.
Part (d)
t = (cΔT/P)m + CΔT/P, so gradient = cΔT/P AND c = P × gradient/ΔT
90B-1B-30Density & the molecular modelPaper 1BEasy11 marks
A student determines the density ρ of the steel used to make ball bearings. She measures the diameter d of each of six ball bearings with a micrometer screw gauge and the mass m of each ball with a balance of resolution 0.001 g.
For a sphere m = (πρ/6)d³, so a graph of m against d³ is expected to be a straight line through the origin. The table shows the results. The graph shows five of the six data points.
d / mm
d³ / mm³
m / g
6.35
256
1.053
7.94
501
2.048
9.52
863
3.541
11.11
5.617
12.70
2048
8.415
14.29
2918
11.952
Graph drawn to scale
(a)
(i)
State the absolute uncertainty in each measurement of d.
(1)
(ii)
Calculate the percentage uncertainty in d³ for the smallest ball bearing.
(1)
(b)
Calculate the missing value of d³ in the table and plot the point on the graph.
(1)
(c)
Draw the line of best fit for the data.
(1)
(d)
Determine the gradient of your line.
(2)
(e)
Determine ρ in kg m−3.
(2)
(f)
The manufacturer gives the density of the steel as 7.81 × 103 kg m−3. Comment on the student’s value, with reference to your answer to (a)(ii).
(1)
(g)
A second student uses a micrometer that reads +0.02 mm when its jaws are closed. He does not correct his readings for this. State and explain the effect on his value of ρ.
(2)
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Part (a)(i)
±0.01 mm
✓ 1
The resolution of a micrometer. Accept ±0.005 mm.
Part (a)(ii)
3 × 0.01/6.35 × 100 = 0.47 %
✓ 1
Accept 0.47–0.48 %. Award [0] for 0.16 % (power of 3 not applied).
Part (b)
1371 «mm³» AND point plotted correctly to within half a small square
✓ 1
Accept 1371.3.
Part (c)
single straight line through the points, passing close to the origin
✓ 1
Do not accept a line joining the points dot to dot.
Part (d)
gradient from a large triangle or two well-separated points on the line
✓ 1
4.10 × 10−3 g mm−3
✓ 1
Accept 4.0 × 10−3–4.2 × 10−3 g mm−3. Unit not required for this mark.
Part (e)
ρ = 6 × gradient/π
✓ 1
May be seen as gradient = πρ/6.
1 g mm−3 = 106 kg m−3 so ρ = 7.83 × 103 kg m−3
✓ 1
Allow ECF from (d). Accept 7.6 × 103–8.0 × 103 kg m−3. Award [1] max for a power-of-ten error.
Part (f)
the difference from the manufacturer’s value is about 0.2 %, smaller than the 0.47 % uncertainty in d³, so the value agrees with it
✓ 1
Allow ECF from (a)(ii) and (e). A comparison of the percentage difference with an uncertainty is needed.
Part (g)
every recorded d is too large, so every d³ is too large «points move to the right»
✓ 1
the gradient/m ÷ d³ is too small, so ρ is too small «an underestimate, by most for the smallest balls»
✓ 1
MP2 only scores if MP1 scores. OWTTE.
Answers: (a)(ii) 0.47 % · (b) 1371 mm³ · (d) 4.10 × 10−3 g mm−3 · (e) 7.83 × 103 kg m−3(the remaining parts are explanations — see the table above)
91B-1B-33Density & the molecular modelPaper 1BEasy11 marks
A student finds a published table of the density ρ of pure water at atmospheric pressure for temperatures θ from 0 °C to 20 °C. The table and the graph show the data.
θ / °C
0
2
4
6
8
10
ρ / kg m−3
999.84
999.94
999.97
999.94
999.85
999.70
θ / °C
12
14
16
18
20
ρ / kg m−3
999.50
999.24
998.94
998.60
998.21
Graph drawn to scale
(a)
State the resolution of the density values in the table.
(1)
(b)
Draw a smooth curve of best fit for the data.
(1)
(c)
State, in kelvin, the temperature at which the density of water is greatest.
(1)
(d)
Calculate the percentage increase in the volume of a fixed mass of water when it is warmed from 4 °C to 20 °C.
(2)
(e)
Each value of ρ has an uncertainty of ±0.01 kg m−3. Calculate the percentage uncertainty in the difference between the densities at 4 °C and at 6 °C, and comment on your answer.
(2)
(f)
In winter the water of a deep lake cools from the surface. Explain, using the data, why the deep water stays at about 4 °C while the surface water cools further and freezes.
(3)
(g)
Another student measures the density of water by weighing 100 cm³ of water in a measuring cylinder read to ±0.5 cm³. Suggest why this method could not show the variation in the table.
(1)
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Notes
Part (a)
0.01 kg m−3
✓ 1
Accept ±0.01 kg m−3.
Part (b)
single smooth curve with a maximum at about 4 °C, passing through or very close to all the points
✓ 1
Do not accept straight-line segments.
Part (c)
277 K
✓ 1
Accept 276–278 K. Accept 277.15 K.
Part (d)
V ∝ 1/ρ for a fixed mass, so V20/V4 = 999.97/998.21
✓ 1
= 1.00176, an increase of 0.18 %
✓ 1
Accept 0.17–0.18 %. Award [2] for CNA.
Part (e)
difference = 0.03 ± 0.02 kg m−3 so percentage uncertainty ≈ 67 %
✓ 1
The absolute uncertainties of the two values are added. Accept 67 %.
the uncertainty is a large fraction of the difference, so the data cannot fix the temperature of the maximum precisely «between about 2 °C and 6 °C» / the difference is barely significant
✓ 1
OWTTE. Accept a comparison with the difference between 4 °C and 20 °C, which has an uncertainty of only 1.1 %.
Part (f)
while the lake is above 4 °C, surface water that cools becomes denser, sinks and is replaced by warmer water from below «convection», until the whole lake is at about 4 °C
✓ 1
below 4 °C, cooling makes the surface water LESS dense, so it stays at the top and convection stops
✓ 1
the surface water continues to cool «by conduction to the cold air» and freezes, while the densest water, at 4 °C, remains at the bottom «ice also floats and insulates it»
✓ 1
OWTTE.
Part (g)
±0.5 cm³ in 100 cm³ is ±0.5 %, far larger than the variation in density «0.18 % over the range, 0.003 % near 4 °C»
✓ 1
Accept any comparison of the percentage uncertainty with the size of the variation.
Answers: (c) 277 K · (d) 0.18 % · (e) 67 % (the remaining parts are explanations — see the table above)
92B-1B-01Specific heat capacityPaper 1BMedium11 marks
A student determines the specific heat capacity c of brass. A brass cylinder of mass 0.745 ± 0.001 kg has two holes drilled in it: one holds an electric heater and the other a temperature probe connected to a data logger. The cylinder is wrapped in a layer of insulating foam.
The p.d. across the heater is constant at 11.9 ± 0.1 V and the current is constant at 1.52 ± 0.02 A. The heater is switched on at t = 0 and the temperature θ is recorded every 60 s. The table and the graph show the results.
t / s
0
60
120
180
240
300
360
420
480
540
600
θ / °C
19.6
21.0
23.4
26.8
30.0
33.5
37.1
40.3
43.9
47.4
50.7
Graph drawn to scale
(a)
(i)
Suggest why the temperature rises more slowly during the first 120 s than it does later.
(1)
(ii)
Draw the line of best fit for the data from t = 180 s onwards.
(1)
(b)
Determine the gradient of your line.
(2)
(c)
Determine c for brass.
(2)
(d)
The uncertainty in the gradient is ±3 %. Determine the absolute uncertainty in c and write down your answer to (c) with its uncertainty.
(3)
(e)
(i)
A data book gives c = 380 J kg−1 K−1 for brass. Comment on the student’s result.
(1)
(ii)
Suggest one change to the method that would reduce the systematic error you identified in (e)(i).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
energy is first needed to heat the heater «and the brass near it»/the probe takes time to reach the temperature of the cylinder/energy takes time to conduct from the heater to the probe
✓ 1
OWTTE. Do not accept “the heater is warming up” without reference to energy or conduction.
Part (a)(ii)
single straight line through the points from 180 s to 600 s, with points scattered either side
✓ 1
The line must not be forced through the origin or through the first points.
Part (b)
large triangle or two well-separated points read from the line
✓ 1
gradient = 0.0573 K s−1
✓ 1
Accept 0.055–0.060 K s−1. Accept °C s−1. Award [2] for a correct answer in the range.
Part (c)
P = VI = 11.9 × 1.52 = 18.1 W AND c = P / (m × gradient)
✓ 1
Must see P = VI and the rate form of Q = mcΔT.
c = 424 J kg−1 K−1
✓ 1
Allow ECF from (b). Accept 400–445 J kg−1 K−1.
Part (d)
percentage uncertainties: V 0.84 %, I 1.32 %, gradient 3 %
✓ 1
Mass contributes only 0.13 % and may be ignored.
total ≈ 5.3 % so Δc ≈ 22 J kg−1 K−1
✓ 1
Allow ECF from (c). Accept 5–6 %.
c = (420 ± 20) J kg−1 K−1
✓ 1
MP3 is for matching the precision of value and uncertainty. Accept (424 ± 22).
Part (e)(i)
380 lies outside the range 401–446 so the result is not consistent with the data-book value / there is a systematic error making c too large
✓ 1
Allow ECF from (d). Accept “energy is lost to the surroundings so the measured gradient is too small”.
Part (e)(ii)
thicker/better insulation OR start below room temperature and finish the same amount above it OR use a larger heater power/shorter time so that losses are a smaller fraction
✓ 1
The change must reduce energy exchange with the surroundings. Do not accept “repeat the experiment”.
Answers: (b) 0.0573 K s−1 · (c) 424 J kg−1 K−1 · (d) (420 ± 20) J kg−1 K−1(the remaining parts are explanations — see the table above)
93B-1B-13Luminosity & apparent brightnessPaper 1BMedium12 marks
A small filament lamp is treated as a point source that radiates uniformly in all directions. A radiation sensor measures the intensity b of the radiation at a distance d from the lamp. d is measured with a metre rule from a mark on the front edge of the lamp holder to the face of the sensor. The room is dark and the sensor is zeroed before the lamp is switched on. Each reading of b has an uncertainty of ±2 %.
The student plots b−½ against d. The graph shows six of the seven points.
d / m
b / W m−2
b−½ / W−½ m
0.100
12.3
0.285
0.150
5.99
0.409
0.200
3.65
0.523
0.250
2.39
0.300
1.73
0.760
0.400
0.983
1.009
0.500
0.651
1.239
Graph drawn to scale
(a)
Explain why a graph of b−½ against d is expected to be a straight line through the origin.
(2)
(b)
Calculate the missing value of b−½ and plot it on the graph.
(1)
(c)
State the percentage uncertainty in each value of b−½.
(1)
(d)
Draw the line of best fit and use it to determine the luminosity L of the lamp.
(4)
(e)
The line does not pass through the origin. Deduce what this shows about the measurement of d, and estimate the size of the effect.
(2)
(f)
Another student uses only the reading at d = 0.100 m to calculate L. Calculate her value and explain why it is smaller than yours.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
b = L/4πd2 so b−½ = √(4π/L) × d
✓ 1
√(4π/L) is constant, so b−½ ∝ d: a straight line through the origin with gradient √(4π/L)
✓ 1
Part (b)
0.647 «W−½ m» AND correctly plotted
✓ 1
Accept 0.65.
Part (c)
±1 %
✓ 1
Half the percentage uncertainty in b, because of the power −½.
Part (d)
single straight line of best fit through the points, not forced through the origin
✓ 1
gradient = 2.39 «W−½»
✓ 1
Accept 2.32–2.46.
L = 4π/gradient2
✓ 1
L = 2.2 W
✓ 1
Accept 2.1–2.3 W. Allow ECF.
Part (e)
the line meets the d axis at a negative value / has a positive intercept on the b−½ axis, so the true distance from the source is greater than d; systematic error «the filament is behind the mark»
A student measures the thermal conductivity k of copper. One end of a copper bar is heated by an electric heater; the other end is cooled by running water. The bar is surrounded by thick lagging. The p.d. across the heater is 12.0 ± 0.1 V and the current is 2.00 ± 0.02 A. The diameter of the bar, measured with vernier calipers, is 25.0 ± 0.1 mm.
When the temperatures are steady, thermocouples read the temperature θ at six distances x from the heated end. Each reading has an uncertainty of ±0.5 °C, shown by the error bars.
x / m
0.040
0.080
0.120
0.160
0.200
0.240
θ / °C
73.8
68.9
64.9
60.6
55.7
51.6
Graph drawn to scale
(a)
Explain why, if no energy were lost from the sides of the bar, the temperature gradient would be the same at every point along it.
(2)
(b)
Draw the line of best fit for the data.
(1)
(c)
Determine the temperature gradient along the bar.
(2)
(d)
Determine k, assuming that all the energy supplied by the heater is conducted along the bar.
(2)
(e)
The uncertainty in the temperature gradient is ±3 %. Determine the absolute uncertainty in k and write down k with its uncertainty.
(3)
(f)
The accepted value of k for copper is 390 W m−1 K−1. Comment on the student's result.
(1)
(g)
Suggest the cause of the difference in (f), and state its effect on the calculated value of k.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
in the steady state the temperature at each point is constant, so no energy is stored in any section: the rate of transfer into a section equals the rate out of it
✓ 1
OWTTE.
so ΔQ/Δt is the same at every cross-section, and with the same k and A, ΔT/Δx = (ΔQ/Δt)/(kA) is the same «a straight-line graph»
✓ 1
Part (b)
single straight line through all the error bars
✓ 1
Part (c)
gradient from a large triangle on the line «negative sign may be shown»
✓ 1
111 K m−1
✓ 1
Accept 106–115 K m−1. Accept °C m−1. Award [2] for CNA.
Part (d)
P = VI = 24.0 W AND A = π(12.5 × 10−3)² = 4.91 × 10−4 m²
✓ 1
Award [0] for this mark if the diameter is used as the radius.
k = P/(A × gradient) = 24.0/(4.91 × 10−4 × 111) = 440 W m−1 K−1
✓ 1
Allow ECF from (c). Accept 420–464.
Part (e)
percentage uncertainties: V 0.8 %, I 1.0 %, A 2 × 0.4 % = 0.8 %, gradient 3 %
✓ 1
The area term must be doubled because A ∝ d².
total ≈ 5.6 % so Δk ≈ 25 W m−1 K−1
✓ 1
Allow ECF from (d). Accept 5–6 %.
k = (440 ± 25) W m−1 K−1
✓ 1
MP3 is for matching the precision of value and uncertainty. Accept (440 ± 30) W m−1 K−1.
Part (f)
390 lies outside the range 415–465 W m−1 K−1, so the difference is not explained by the random uncertainties: there is a systematic error
✓ 1
Allow ECF from (e).
Part (g)
energy is lost from the heater/hot end to the surroundings «through the lagging», so less than 24.0 W is conducted along the bar; the gradient is smaller than it would be for 24.0 W, so k is overestimated
✓ 1
Direction (too large) and reason both needed. OWTTE.
Answers: (c) 111 K m−1 · (d) 440 W m−1 K−1 · (e) (440 ± 25) W m−1 K−1(the remaining parts are explanations — see the table above)
95B-1B-32Specific heat capacityPaper 1BMedium12 marks
A student uses a published database to investigate how the specific heat capacity c of a metal at room temperature depends on the molar mass M of the metal. She suggests that c = KMn, where K and n are constants, and tests the hypothesis that n = −1.
The table shows the data for six metals. The graph shows the points for four of them.
metal
molar mass M / g mol−1
c / J kg−1 K−1
lg (M / g mol−1)
lg (c / J kg−1 K−1)
aluminium
26.98
897
1.431
2.953
iron
55.85
449
1.747
2.652
copper
63.55
385
silver
107.87
235
2.033
2.371
tungsten
183.84
132
lead
207.2
129
2.316
2.111
Graph drawn to scale
(a)
Outline why a graph of lg c against lg M is used to test the relationship c = KMn.
(2)
(b)
(i)
Calculate the missing values of lg M and lg c for copper and for tungsten.
(1)
(ii)
Plot the two points on the graph and draw the line of best fit for the six points.
(2)
(c)
Determine the gradient of your line.
(2)
(d)
State whether the data support the hypothesis that n = −1.
(1)
(e)
The molar mass of platinum is 195.08 g mol−1. Use the graph to predict the specific heat capacity of platinum.
(2)
(f)
Explain, in terms of particles, why the specific heat capacity of lead is much smaller than that of aluminium.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
lg c = n lg M + lg K
✓ 1
Accept log or ln throughout.
this has the form y = mx + c, so the graph is a straight line «if the relationship holds» with gradient n «and intercept lg K»
✓ 1
OWTTE.
Part (b)(i)
copper: 1.803 and 2.585; tungsten: 2.264 and 2.121
✓ 1
All four needed. Accept 2 or 4 decimal places.
Part (b)(ii)
both points plotted correctly to within half a small square
✓ 1
Allow ECF from (b)(i).
single straight line of best fit with the points scattered either side
✓ 1
Part (c)
gradient from a large triangle on the line
✓ 1
gradient = −0.97
✓ 1
Accept −0.92 to −1.06. No unit. Negative sign needed.
Part (d)
yes: the gradient is close to −1 «within about 5 %» AND the points lie close to a straight line
✓ 1
Allow ECF from (c). The answer must refer to the value of the gradient.
Part (e)
lg M = 2.290 and lg c read from the line ≈ 2.12
✓ 1
ALTERNATIVE: cM ≈ constant ≈ 25 J mol−1 K−1 «from the table», so c ≈ 25 000/195 ✓.
c ≈ 130 J kg−1 K−1
✓ 1
Accept 120–140 J kg−1 K−1. Allow ECF from their line.
Part (f)
lead atoms have a much greater mass «7.7 times», so 1 kg of lead contains far fewer atoms than 1 kg of aluminium
✓ 1
Accept reference to the number of moles per kilogram.
raising the temperature by 1 K needs about the same energy for each atom «the average kinetic energy of an atom depends only on the temperature, not on the mass of the atom», so less energy is needed per kilogram of lead
✓ 1
OWTTE.
Answers: (b)(i) Cu 1.803, 2.585; W 2.264, 2.121 · (c) −0.97 · (e) 130 J kg−1 K−1(the remaining parts are explanations — see the table above)
96B-1B-34Luminosity & apparent brightnessPaper 1BMedium12 marks
A student estimates the luminosity of the Sun. A thin copper disc of diameter 40.0 ± 0.1 mm and mass 22.6 ± 0.1 g is painted matt black on its front face and mounted on a block of foam so that its front face is perpendicular to the Sun’s rays. A temperature sensor is attached to the back of the disc.
The disc is shaded until it reaches the air temperature of 21.0 °C. The shade is removed at t = 0 and the temperature θ is recorded every 30 s. Specific heat capacity of copper = 385 J kg−1 K−1. Assume that the disc absorbs all the solar radiation incident on its front face.
t / s
0
30
60
90
120
150
180
210
240
270
300
θ / °C
21.0
24.9
28.1
31.2
33.7
35.8
37.9
39.5
40.9
42.3
43.5
Graph drawn to scale
(a)
Explain why the rate at which the temperature of the disc rises decreases with time.
(1)
(b)
Draw the curve of best fit for the data.
(1)
(c)
Determine the initial rate of rise of the temperature of the disc.
(2)
(d)
Outline why the initial rate of rise, rather than a later one, is used to find the intensity of the solar radiation.
(1)
(e)
Calculate the intensity of the solar radiation incident on the disc.
(2)
(f)
The uncertainty in the initial rate of rise is ±10 %. Determine the absolute uncertainty in your answer to (e).
(2)
(g)
The mean distance of the Earth from the Sun is 1.50 × 1011 m. Estimate the luminosity of the Sun.
(1)
(h)
The accepted luminosity of the Sun is 3.83 × 1026 W. Comment on the student’s estimate and suggest one reason for the difference.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
as the disc becomes hotter than the air, the temperature difference increases, so the disc loses energy to the surroundings at a greater rate «leaving less to raise its temperature»
✓ 1
OWTTE.
Part (b)
smooth curve through the points, starting at (0, 21.0 °C), with a decreasing gradient
✓ 1
Do not accept a straight line through all points.
Part (c)
tangent drawn to the curve at t = 0 and its gradient found
✓ 1
0.137 K s−1
✓ 1
Accept 0.12–0.16 K s−1. Award [2] for a value in this range.
Part (d)
at t = 0 the disc is at the air temperature, so there is no resultant transfer to the surroundings: all the absorbed power raises the temperature of the disc
✓ 1
OWTTE.
Part (e)
absorbed power = mc × rate = 0.0226 × 385 × 0.137 = 1.19 W AND A = π(0.0200)² = 1.26 × 10−3 m²
✓ 1
Allow ECF from (c).
intensity = power/A = 950 W m−2
✓ 1
Accept 830–1110 W m−2. Award [1] max if the diameter is used as the radius.
Part (f)
percentage uncertainties: area 2 × 0.25 % = 0.5 %, mass 0.44 %, rate 10 %; total ≈ 11 %
✓ 1
The area term must be doubled.
Δb ≈ 100 W m−2, so b = (9.5 ± 1.0) × 102 W m−2
✓ 1
Allow ECF from (e). Accept ±90 to ±110 W m−2.
Part (g)
L = 4πd²b = 4π × (1.50 × 1011)² × 950 = 2.7 × 1026 W
✓ 1
Allow ECF from (e).
Part (h)
the estimate is about 30 % too small, much more than the ≈ 11 % uncertainty, so there is a systematic error
✓ 1
Allow ECF from (f) and (g).
part of the solar radiation is absorbed or scattered by the atmosphere before reaching the disc OR the painted surface reflects some of the radiation «absorbs less than all of it» OR the disc loses energy to the foam/sensor so the initial rate is too small
✓ 1
Any one. The reason must make the estimate too small.
Answers: (c) 0.137 K s−1 · (e) 950 W m−2 · (f) (9.5 ± 1.0) × 102 W m−2 · (g) 2.7 × 1026 W (the remaining parts are explanations — see the table above)
97B-1B-03Stefan–Boltzmann & WienPaper 1BHard11 marks
A student uses a published catalogue of eclipsing binary stars to test the Stefan–Boltzmann law. For these stars the radius R, the luminosity L and the surface temperature T have been found by independent methods. R is given in units of the solar radius and L in units of the solar luminosity.
If every star radiates as a black body, L = 4πσR2T4, and in these units lg(L/R2) = 4 lg T − 4 lg T☉, where T☉ is the surface temperature of the Sun.
For star D the catalogue does not give T but gives the peak wavelength of its spectrum, λmax = 4.12 × 10−7 m. The uncertainty in each value of lg(L/R2) is ±0.04; the uncertainty in lg T is too small to show. The graph shows the data for the other five stars.
Star
T / K
R / R☉
L / L☉
lg(T / K)
lg(L/R2)
A
4480
0.74
0.207
3.651
−0.42
B
5290
0.93
0.592
3.723
−0.16
C
6180
1.24
2.06
3.791
0.13
D
— (see text)
1.58
5.41
E
8350
1.83
14.1
3.922
0.62
F
9720
2.31
44.2
3.988
0.92
Graph drawn to scale
(a)
Show that the surface temperature of star D is about 7000 K.
(1)
(b)
Calculate the two missing values for star D in the table.
(1)
(c)
Plot the point for star D on the graph and draw the line of best fit for all six stars.
(2)
(d)
(i)
Determine the gradient of your line.
(2)
(ii)
By drawing lines of maximum and minimum gradient, determine the uncertainty in the gradient.
(2)
(e)
State whether the data support the Stefan–Boltzmann law for these stars.
(1)
(f)
Using your line, determine T☉.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
T = 2.9 × 10−3 / 4.12 × 10−7 = 7039 K
✓ 1
Must see the substitution into Wien’s law OR an answer to at least 3 s.f.
Part (b)
lg T = 3.848 AND lg(L/R2) = 0.34
✓ 1
Both needed. Accept 3.845–3.848 «from T = 7000 K» and 0.33–0.34.
Part (c)
star D plotted correctly to within half a small square
✓ 1
Allow ECF from (b).
single straight line passing through all six error bars
✓ 1
Part (d)(i)
gradient from a large triangle on the line
✓ 1
gradient = 4.0
✓ 1
Accept 3.8–4.2. No unit.
Part (d)(ii)
steepest and shallowest lines drawn through all the error bars
✓ 1
Expect gradients close to 4.2 and 3.8.
uncertainty = ½(max − min) ≈ ±0.2
✓ 1
Accept ±0.1 to ±0.3. Allow ECF.
Part (e)
yes: the law predicts a gradient of 4, which lies within the range of the gradient «AND all six error bars are crossed by one straight line»
✓ 1
Allow ECF from (d). Reference to the uncertainty is needed.
Part (f)
L/R2 = 1 «in solar units» for a star at T☉, so read lg T where the line crosses lg(L/R2) = 0, ≈ 3.76
✓ 1
Or use lg T☉ = −intercept/gradient.
T☉ = 103.761 ≈ 5800 K
✓ 1
Accept 5500–6100 K. Allow ECF.
Answers: (a) 7039 K · (b) 3.848 and 0.34 · (d)(i) 4.0 · (d)(ii) ±0.2 · (f) 5800 K (the remaining parts are explanations — see the table above)
98B-1B-12Latent heat & phase changePaper 1BHard12 marks
Crushed ice at 0 °C fills a funnel and surrounds an electric heater. Meltwater drips into a beaker. For each heater power P, once the dripping is steady, the meltwater is collected for 300 s and its mass m is measured. The ice also gains energy from the room, so some ice melts even when the heater is off.
From repeat runs, the uncertainty in each mass is ±1.5 g. The melting rate is R = m / 300 s. The graph shows R against P with error bars.
P / W
m / g
R / 10−5 kg s−1
12.0
13.7
4.57
20.0
19.3
6.43
28.0
27.4
36.0
34.9
11.6
44.0
40.3
13.4
52.0
48.2
16.1
Graph drawn to scale
(a)
Calculate the missing value of R for P = 28.0 W, giving your answer to an appropriate number of significant figures.
(1)
(b)
Draw the line of best fit for the data.
(1)
(c)
Determine the gradient of your line, giving its unit.
(2)
(d)
Show how the gradient is related to the specific latent heat of fusion L of ice, and hence determine L.
(2)
(e)
By drawing lines of maximum and minimum gradient, determine the absolute uncertainty in L. Hence state whether the result is consistent with the accepted value of 3.34 × 105 J kg−1.
(3)
(f)
State what is represented by the intercept of the line on the R axis.
(1)
(g)
Explain why the energy gained from the room does not affect the value of L found from the gradient.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
9.13 «× 10−5 kg s−1»
✓ 1
= 27.4 × 10−3 / 300. Must be given to 3 s.f., like the other values in the column.
Part (b)
single straight line through all the error bars, not forced through the origin
✓ 1
Part (c)
gradient from a large triangle
✓ 1
0.289 × 10−5 kg J−1 OR 2.9 × 10−6 kg s−1 W−1
✓ 1
Unit needed. Accept 0.275–0.304 × 10−5.
Part (d)
P + Proom = RL, so R = P/L + Proom/L and gradient = ΔR/ΔP = 1/L
✓ 1
OWTTE.
L = 1/gradient = 3.46 × 105 J kg−1
✓ 1
Allow ECF from (c). Accept 3.3 × 105–3.6 × 105.
Part (e)
max and min lines through all error bars «gradients ≈ 0.308 and 0.270 × 10−5» giving L between ≈ 3.2 × 105 and 3.7 × 105 J kg−1
✓ 1
ΔL = ½(3.7 × 105 − 3.2 × 105) ≈ 2 × 104 J kg−1
✓ 1
Accept 1 × 104 to 3 × 104. Allow ECF.
the accepted value lies within the range so the result is consistent
✓ 1
Allow ECF from their range.
Part (f)
the rate of melting due to energy from the room / when P = 0 «≈ 0.9 × 10−5 kg s−1»
✓ 1
OWTTE.
Part (g)
the energy «rate» gained from the room is the same for every value of P
✓ 1
OWTTE.
so it shifts the line up by a constant amount / changes only the intercept, not the gradient «ΔR/ΔP»
✓ 1
Answers: (a) 9.13 × 10−5 kg s−1 · (c) 0.289 × 10−5 kg J−1 · (d) 3.46 × 105 J kg−1 · (e) ±2 × 104 J kg−1(the remaining parts are explanations — see the table above)
A student measures the thermal conductivity k of acrylic. A square electric heater plate of side 10.0 ± 0.1 cm rests on a stack of acrylic sheets of the same area. The stack stands on a metal block kept at 0.0 °C by iced water, and a thermostat keeps the lower face of the heater at 50.0 °C. The temperature difference across the stack is 50.0 ± 1.0 K. The top and sides of the heater are insulated.
For stacks of 1 to 5 sheets the total thickness x is measured with vernier calipers, and the electrical power P supplied to the heater is measured once the temperatures are steady. The uncertainty in each P is ±0.5 W.
x / mm
(1/x) / m−1
P / W
2.9
345
33.6
6.0
167
16.6
9.1
110
11.8
12.0
83
8.8
15.1
66
7.6
Graph drawn to scale
(a)
Outline how the student could confirm that a steady state had been reached before recording P.
(1)
(b)
Draw the line of best fit for the data.
(1)
(c)
Determine k.
(3)
(d)
By drawing lines of maximum and minimum gradient, determine the percentage uncertainty in the gradient.
(2)
(e)
Determine the absolute uncertainty in k. Hence comment on whether the result agrees with the literature value of 0.19 W m−1 K−1.
(3)
(f)
The line does not pass through the origin. Suggest a reason for this and explain why it does not affect the value of k.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the power reading/the temperature of the cold block «and heater» no longer changes over several minutes
✓ 1
OWTTE.
Part (b)
single straight line through all error bars, not forced through the origin
✓ 1
Part (c)
gradient ≈ 0.094 W m
✓ 1
Accept 0.089–0.098 W m.
P = kAΔT × (1/x) «+ constant» so k = gradient / (AΔT)
✓ 1
k = 0.0937 / (0.0100 × 50.0) = 0.19 W m−1 K−1
✓ 1
Allow ECF from gradient. Unit needed.
Part (d)
max and min gradients ≈ 0.0969 and 0.0910 W m
✓ 1
percentage uncertainty = ½(max − min)/best ≈ 3 %
✓ 1
Accept 2–6 %. Allow ECF.
Part (e)
ΔA/A = 2 × 1 % = 2 % AND Δ(ΔT)/ΔT = 2 %
✓ 1
total ≈ 7 % so k = (0.187 ± 0.013) W m−1 K−1
✓ 1
Allow ECF from (c) and (d). Matching precision needed.
0.19 lies within the range so the result agrees
✓ 1
Allow ECF from their range.
Part (f)
energy is lost from the heater through the insulation «at the same rate for every stack» / the thermostat and wires use some power
✓ 1
OWTTE.
this adds a constant to P, changing the intercept but not the gradient, which alone is used for k
✓ 1
Answers: (c) 0.19 W m−1 K−1 · (d) ≈ 3 % · (e) (0.187 ± 0.013) W m−1 K−1(the remaining parts are explanations — see the table above)
100B-1B-22Stefan–Boltzmann & WienPaper 1BHard10 marks
A student tests Wien's displacement law. A small ceramic sample is heated in an electric furnace, and its temperature T is measured with a thermocouple to ±10 K. Radiation from the sample passes through a window to an infrared spectrometer, which records the spectrum and the peak wavelength λmax. Because the peak is broad, each value of λmax has an uncertainty of ±0.05 μm, shown by the error bars.
The ceramic may be treated as a black body. The graph shows λmax against 1/T for all six readings.
T / K
(1/T) / 10−4 K−1
λmax / μm
1150
8.70
2.55
1300
2.20
1450
6.90
2.02
1600
6.25
1.84
1750
1.64
1900
5.26
1.52
Graph drawn to scale
(a)
Calculate the two missing values of 1/T in the table.
(1)
(b)
Explain why, if Wien's law is obeyed, a graph of λmax against 1/T is a straight line through the origin.
(1)
(c)
Draw the line of best fit for the data.
(1)
(d)
Determine the gradient of your line in m K.
(2)
(e)
By drawing lines of maximum and minimum gradient, determine the absolute uncertainty in the gradient.
(2)
(f)
State whether the student's results are consistent with Wien's law.
(1)
(g)
Outline why no horizontal error bars are shown on the graph.
(1)
(h)
The emissivity of the ceramic is in fact about 0.85, the same at all wavelengths. State and explain whether this affects the values of λmax measured.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
7.69 AND 5.71 «× 10−4 K−1»
✓ 1
Both needed. Accept 3 or 4 s.f.
Part (b)
λmax = (2.9 × 10−3 m K) × (1/T), so λmax ∝ 1/T «form y = mx with zero intercept and gradient equal to Wien's constant»
✓ 1
OWTTE.
Part (c)
single straight line through all six error bars
✓ 1
Need not pass exactly through the origin.
Part (d)
gradient = 0.295 μm per 10−4 K−1 from a large triangle
✓ 1
Accept 0.287–0.303.
= 2.95 × 10−3 m K
✓ 1
Conversion of μm and 10−4 K−1 needed. Award [1] max for a power-of-ten error. Allow ECF.
Part (e)
steepest and shallowest lines through all the error bars «gradients ≈ 0.319 and 0.272 μm per 10−4 K−1»
✓ 1
uncertainty = ½(max − min) ≈ ±0.24 × 10−3 m K
✓ 1
Accept ±0.14 to ±0.35 × 10−3 m K. Allow ECF.
Part (f)
yes: 2.9 × 10−3 m K lies within the range 2.71–3.19 × 10−3 m K «AND a straight line through all the error bars passes close to the origin»
✓ 1
Allow ECF from (d) and (e). Reference to the uncertainty is needed.
Part (g)
the percentage uncertainty in 1/T equals that in T, at most 10/1150 ≈ 0.9 %, so the bars would be smaller than the plotted points «much smaller than the ≈ 2–3 % in λmax»
✓ 1
OWTTE.
Part (h)
no: a constant emissivity reduces the power at every wavelength by the same factor, so the shape of the spectrum and the position of its peak are unchanged
✓ 1
Do not accept a bare ‘no’.
Answers: (a) 7.69, 5.71 × 10−4 K−1 · (d) 2.95 × 10−3 m K · (e) ±0.24 × 10−3 m K (the remaining parts are explanations — see the table above)
A student estimates the thermal conductivity k of expanded polystyrene. A small cubical box is made from polystyrene sheet of thickness Δx = 10.0 ± 0.5 mm. The box is completely filled with 0.725 kg of hot water, which is stirred gently by a small stirrer that passes through the lid. The total area of the walls and lid, measured at the middle of their thickness, is A = 0.0600 m² ± 3 %. The room is at a constant 20.0 °C.
A data logger records the temperature θ of the water every 20 minutes for 4 hours. The table and the graph show the results. Assume that the energy lost by the water passes through the walls and lid by conduction, that the outer surface of the polystyrene is at room temperature and that the inner surface is at the temperature of the water. Specific heat capacity of water = 4180 J kg−1 K−1.
t / min
0
20
40
60
80
100
120
θ / °C
80.0
75.2
70.6
66.6
62.9
59.3
56.2
t / min
140
160
180
200
220
240
θ / °C
53.3
50.6
48.0
45.9
43.7
41.7
Graph drawn to scale
(a)
Draw the curve of best fit for the data.
(1)
(b)
Determine, by drawing a tangent, the rate of fall of the temperature of the water at t = 60 min. Give your answer in K s−1.
(2)
(c)
Show that the rate of thermal energy transfer from the water at t = 60 min is about 10 W.
(1)
(d)
Determine k.
(2)
(e)
The uncertainty in the rate of fall in (b) is ±10 %. Determine the absolute uncertainty in k and state whether the result is consistent with the manufacturer’s value of 0.033 W m−1 K−1.
(3)
(f)
Explain why the rate of fall of the temperature decreases during the experiment.
(1)
(g)
The lid does not fit tightly, and some water evaporates through the gap. State and explain the effect of this on the value of k found in (d).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
smooth curve through the points with a gradually decreasing gradient «points scattered either side»
✓ 1
Do not accept a straight line or dot-to-dot.
Part (b)
tangent drawn at t = 60 min and its gradient found from a large triangle
✓ 1
0.196 K min−1 ÷ 60 = 3.3 × 10−3 K s−1
✓ 1
Accept 2.9 × 10−3–3.7 × 10−3 K s−1. Award [1] max for an answer left in K min−1. Sign may be omitted.
Part (c)
ΔQ/Δt = mc × rate = 0.725 × 4180 × 3.27 × 10−3 = 9.91 W
✓ 1
Must see full substitution OR answer to at least 3 s.f. Allow ECF from (b).
Part (d)
θ at 60 min ≈ 66.6 °C so ΔT = 46.6 K AND k = (ΔQ/Δt)Δx/(AΔT)
✓ 1
Accept θ read from the curve in the range 66.2–67.0 °C.
k = 9.91 × 0.0100/(0.0600 × 46.6) = 0.035 W m−1 K−1
✓ 1
Allow ECF from (c). Accept 0.031–0.040 W m−1 K−1. Unit needed for this mark.
Part (e)
percentage uncertainties: Δx 5 %, A 3 %, rate 10 % «ΔT and m negligible»
✓ 1
total ≈ 18 % so Δk ≈ 0.006 W m−1 K−1
✓ 1
Allow ECF from (d). Accept k = (0.035 ± 0.006) W m−1 K−1.
0.033 lies within the range 0.029–0.042 so the result is consistent
✓ 1
Allow ECF from their range. A comparison with the range is needed.
Part (f)
as the water cools the temperature difference «gradient» across the walls decreases, so the rate of conduction/rate of energy loss decreases
✓ 1
OWTTE. Do not accept “the water has less energy”.
Part (g)
evaporation removes extra energy from the water, so the measured rate of energy loss is greater than the rate of conduction through the walls
✓ 1
k is overestimated / too large
✓ 1
MP2 only scores if MP1 scores.
Answers: (b) 3.3 × 10−3 K s−1 · (c) 9.91 W · (d) 0.035 W m−1 K−1 · (e) (0.035 ± 0.006) W m−1 K−1(the remaining parts are explanations — see the table above)
102B-2-01Specific heat capacityPaper 2Easy12 marks
A car of mass 1250 kg travels along a level road at 24.0 m s−1. The driver brakes and the car comes to rest. During the stop, 85 % of the kinetic energy of the car is transferred to the four steel brake discs, which have a total mass of 28.0 kg.
Specific heat capacity of steel = 450 J kg−1 K−1.
(a)
(i)
Calculate the kinetic energy of the car before it brakes.
(1)
(ii)
Show that the temperature of the discs rises by about 24 K.
(2)
(iii)
Before the stop the discs were at 35 °C. State their temperature after the stop, in kelvin.
(1)
(b)
(i)
Immediately after the stop, the discs are hotter than the surrounding air. State the direction of the net thermal energy transfer between the discs and the air, and what determines it.
(1)
(ii)
Outline two mechanisms by which thermal energy is transferred from the discs to the surroundings.
(2)
(c)
Later, the car travels down a long hill.
(i)
It moves at a constant speed and loses 120 m in height. The driver uses only the brakes to keep the speed constant. Assume that 85 % of the gravitational potential energy lost by the car is transferred to the discs and ignore air resistance. Calculate the energy transferred to the discs during the descent.
(2)
(ii)
The discs are at 35 °C at the top of the hill. Determine whether they would reach 150 °C by the bottom of the hill if they lost no energy to the surroundings.
(2)
(iii)
In practice the discs reach a lower temperature than your answer to (c)(ii) suggests. Suggest one reason why.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Ek = ½ × 1250 × 24.02 = 3.6 × 105 J
✓ 1
Part (a)(ii)
energy to the discs = 0.85 × 3.6 × 105 = 3.06 × 105 J
✓ 1
ΔT = Q/(mc) = 3.06 × 105 / (28.0 × 450) = 24.3 K
✓ 1
Must see full substitution OR answer to at least 3 s.f. Award [1] max if the 85 % is omitted (gives 28.6 K).
Part (a)(iii)
35 + 24.3 = 59.3 °C, so T = 332 K
✓ 1
Accept 332–333 K. Allow ECF from (a)(ii); accept 59 + 273.
Part (b)(i)
from the discs to the air, because the discs are at the higher temperature
✓ 1
Both needed. Accept 'the temperature difference determines the direction: hotter to colder'.
Part (b)(ii)
a first mechanism with a brief correct outline, e.g. convection: air next to the disc is heated, becomes less dense and rises / is carried away and replaced by cooler air
✓ 1
Names alone (e.g. 'convection and radiation') score [1] max.
a second, different mechanism outlined, e.g. radiation: the hot disc emits infrared electromagnetic radiation; OR conduction: energy passes by particle collisions to the hub/pads/wheel in contact
✓ 1
Part (c)(i)
ΔEp = mgΔh = 1250 × 9.81 × 120 = 1.47 × 106 J
✓ 1
energy to discs = 0.85 × 1.47 × 106 = 1.25 × 106 J
✓ 1
Award [2] for CNA. Award [1] max for 1.47 × 106 J (85 % omitted).
Part (c)(ii)
ΔT = 1.25 × 106 / (28.0 × 450) = 99 K
✓ 1
Allow ECF from (c)(i).
final temperature ≈ 35 + 99 = 134 °C, which is below 150 °C, so the discs would not reach 150 °C
✓ 1
MP2 needs a comparison and a conclusion consistent with the candidate's value.
Part (c)(iii)
energy is transferred from the discs to the air (by convection/radiation) throughout the long descent; OR air resistance does part of the work, so less than 85 % of the lost potential energy reaches the discs
✓ 1
Do not accept 'energy is lost' without saying how or to where.
Answers: (a)(i) 3.6 × 105 J · (a)(ii) 24.3 K · (a)(iii) 332 K · (c)(i) 1.25 × 106 J · (c)(ii) about 134 °C — no (the remaining parts are explanations — see the table above)
103B-2-19Density & the molecular modelPaper 2Easy10 marks
Gallium is a metal that melts at 29.8 °C. A student holds a 25.0 g bead of solid gallium, initially at 20.0 °C, in her closed hand. The skin of her hand is at 34.0 °C.
Specific heat capacity of solid gallium = 370 J kg−1 K−1. Specific latent heat of fusion of gallium = 8.0 × 104 J kg−1. At its melting point the density of solid gallium is 5.91 × 103 kg m−3 and the density of liquid gallium is 6.10 × 103 kg m−3.
(a)
State the melting point of gallium in kelvin.
(1)
(b)
Outline why there is a net transfer of thermal energy from the hand to the gallium, and how this energy is transferred by conduction.
(2)
(c)
(i)
Show that about 2.1 kJ must be transferred to the bead to melt it completely.
(1)
(ii)
The hand transfers energy to the bead at a mean rate of 2.5 W. Calculate the time taken for the bead to melt completely, in minutes.
(1)
(d)
(i)
Calculate the change in the volume of the bead when it melts.
(2)
(ii)
Most substances expand when they melt. Explain this, using the molecular model of solids and liquids.
(2)
(iii)
Suggest what your answer to (d)(i) shows about the arrangement of the atoms in solid gallium.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
29.8 + 273 = 303 K
✓ 1
Accept 302.95 K.
Part (b)
the hand is at a higher temperature than the gallium «and the net transfer is always from higher to lower temperature»
✓ 1
particles of the skin have a greater mean kinetic energy; in collisions at the surface of contact they pass kinetic energy to the gallium atoms, which pass it on to neighbouring atoms «free electrons also carry energy through the metal»
✓ 1
OWTTE. Do not accept 'heat particles' or 'cold flows into the hand'.
the volume decreases by 1.3 × 10−7 m3 «0.13 cm3, about 3 %»
✓ 1
A decrease must be stated or clear. Award [1 max] for the value without the direction.
Part (d)(ii)
in a solid the particles vibrate about fixed positions in a closely packed, ordered arrangement
✓ 1
in a liquid the particles can move past one another in a disordered arrangement, so their mean separation is slightly larger and the same mass occupies a larger volume «lower density»
✓ 1
OWTTE.
Part (d)(iii)
the atoms in solid gallium are arranged in an open «loosely packed» structure, so on average they are further apart than in the liquid
✓ 1
Accept 'the atoms are closer together in the liquid than in the solid'.
Answers: (a) 303 K · (c)(i) 2.09 × 103 J · (c)(ii) ≈ 14 min · (d)(i) −1.3 × 10−7 m3(the remaining parts are explanations — see the table above)
On a winter night a layer of ice forms on the surface of a pond. The upper surface of the ice is at −6.0 °C. The lower surface of the ice is in contact with the water of the pond and is at 0 °C. Assume that all of the energy conducted upwards through the ice is released by water freezing on the lower surface of the ice.
Thermal conductivity of ice = 2.2 W m−1 K−1. Density of ice = 920 kg m−3. Specific latent heat of fusion of ice = 3.34 × 105 J kg−1.
Axes for (c)
(a)
Energy is removed from the water that freezes on the lower surface of the ice. Explain, in terms of particles, why the temperature of this water does not fall below 0 °C while it freezes.
(2)
(b)
At one moment the ice is 4.0 cm thick.
(i)
Calculate the rate of thermal energy transfer through each square metre of the ice.
(2)
(ii)
Show that the mass of water that freezes on the lower surface of each square metre of ice in one hour is about 3.6 kg.
(1)
(iii)
Calculate the increase in the thickness of the ice in one hour.
(1)
(c)
On the axes, sketch a graph to show how the temperature varies with depth through the ice when it is 4.0 cm thick.
(2)
(d)
State and explain how the rate at which the ice thickens changes as the ice becomes thicker, if the temperature of its upper surface stays at −6.0 °C.
(2)
(e)
Suggest why a layer of snow lying on top of the ice slows the growth of the ice.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the energy removed reduces the intermolecular potential energy «as bonds form between the molecules»
✓ 1
the average «random» kinetic energy of the molecules does not change, so the temperature stays at 0 °C until all the water has frozen
✓ 1
OWTTE. Do not accept “the energy is used to freeze the water” alone.
Part (b)(i)
ΔQ/Δt = 2.2 × 1.0 × 6.0/0.040
✓ 1
= 330 W «per m²»
✓ 1
Award [1 max] for 3.3 W «thickness not converted to m».
Part (b)(ii)
m = 330 × 3600/3.34 × 105 = 3.56 kg
✓ 1
Must see full substitution OR answer to at least 3 s.f. Allow ECF from (b)(i).
Part (b)(iii)
Δx = 3.56/(920 × 1.0) = 3.9 × 10−3 m «3.9 mm»
✓ 1
Accept 3.9–4.0 mm. Allow ECF from (b)(ii).
Part (c)
straight line with a positive gradient
✓ 1
Do not accept a curve.
from −6.0 °C at depth 0 to 0 °C at 4.0 cm
✓ 1
Part (d)
the rate of thickening decreases
✓ 1
No mark for a bare statement without reasoning. MP1 only scores if MP2 scores.
the same temperature difference acts across a greater thickness, so the temperature gradient ΔT/Δx and the rate of conduction are smaller, so less water freezes each second
✓ 1
OWTTE.
Part (e)
snow «containing trapped air» is a poor conductor, so the upper surface of the ice is warmer than the air/closer to 0 °C, giving a smaller temperature gradient in the ice
✓ 1
OWTTE.
Answers: (b)(i) 330 W · (b)(ii) 3.56 kg · (b)(iii) 3.9 × 10−3 m (the remaining parts are explanations — see the table above)
105B-2-35Mechanisms of thermal transferPaper 2Easy11 marks
The figure shows a cross-section of a vacuum flask. The flask has a double glass wall with a vacuum between the two layers of glass. The glass surfaces facing the vacuum are silvered. The flask is closed by a hollow plastic stopper.
The flask is filled with 0.45 kg of tea at 88 °C. After 8.0 hours the temperature of the tea is 71 °C. Specific heat capacity of tea = 4180 J kg−1 K−1.
Diagram NOT accurately drawn
(a)
Annotate the figure to explain how the vacuum, the silvered surfaces and the stopper each reduce the rate of thermal energy transfer from the tea.
(3)
(b)
Determine the mean rate of thermal energy transfer from the tea during the 8.0 hours.
(3)
(c)
Explain why thermal energy can be transferred across a vacuum by radiation but not by conduction or convection.
(2)
(d)
State and explain how the rate of thermal energy transfer from the tea changes as the tea cools.
(2)
(e)
Suggest, with a reason, which part of the flask is responsible for most of the energy transferred from the tea.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
vacuum: contains «almost» no particles, so there is no conduction or convection across the gap
✓ 1
Annotations must be linked to the correct feature.
silvered surfaces: poor emitters and good reflectors of infrared radiation, so transfer by radiation is reduced
✓ 1
stopper: a poor conductor «containing trapped air» that reduces conduction AND stops warm air rising out of the flask «convection» or evaporation
✓ 1
Part (b)
Q = 0.45 × 4180 × (88 − 71)
✓ 1
Q = 3.2 × 104 J
✓ 1
rate = 3.2 × 104/(8.0 × 3600) = 1.1 W
✓ 1
Allow ECF from MP2. Award [3] for CNA.
Part (c)
conduction and convection need particles «that collide or that move as a fluid» to carry the energy
✓ 1
radiation is energy carried by electromagnetic waves, which need no medium
✓ 1
Part (d)
the rate decreases
✓ 1
No mark for a bare statement without reasoning.
because the temperature difference between the tea and the surroundings decreases
✓ 1
Part (e)
the stopper/neck of the flask: there is no vacuum there, so energy is conducted through the stopper and the glass where the two walls join
✓ 1
OWTTE.
Answers: (b) 1.1 W (the remaining parts are explanations — see the table above)
In a sauna, an electric heater keeps a pile of stones of total mass 18.0 kg at 260 °C. A bather pours a ladle of 0.150 kg of water at 20 °C onto the stones. All of the water is heated to 100 °C and then boils away. Assume that all the energy needed comes from the stones.
Specific heat capacity of the stones = 840 J kg−1 K−1; specific heat capacity of water = 4180 J kg−1 K−1; specific latent heat of vaporization of water = 2.26 × 106 J kg−1.
(a)
(i)
Calculate the energy needed to raise the temperature of the water to 100 °C.
(1)
(ii)
Show that the total energy needed to heat the water and boil it away is about 3.9 × 105 J.
(1)
(iii)
Calculate the decrease in the temperature of the stones.
(1)
(b)
The heater is now switched off. Ladles of water at 20 °C are poured onto the stones one after another. Determine the largest number of ladles that the stones can boil away completely.
(2)
(c)
State and explain how the average kinetic energy of the molecules in the liquid water at 100 °C compares with that of the molecules in the steam at 100 °C.
(2)
(d)
Some steam condenses on the skin of a bather, which is at 35 °C. Calculate the energy transferred to the skin when 2.0 g of steam at 100 °C condenses and the water formed cools to 35 °C.
(2)
(e)
In practice, the temperature of the stones falls by less than your answer to (a)(iii). Suggest one reason why.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Q = 0.150 × 4180 × 80 = 5.02 × 104 J
✓ 1
Part (a)(ii)
0.150 × 2.26 × 106 = 3.39 × 105 J, so total = 3.89 × 105 J
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (a)(iii)
ΔT = 3.89 × 105 / (18.0 × 840) = 25.7 K
✓ 1
Accept 26 K. Allow ECF.
Part (b)
the stones can cool by 160 K before they reach 100 °C
✓ 1
160 / 25.7 = 6.2, so 6 ladles
✓ 1
Do not accept 7. Allow ECF from (a)(iii).
Part (c)
the average kinetic energy is the same «in the liquid and in the steam»
✓ 1
because the average kinetic energy depends only on the «kelvin» temperature, which is the same; the energy supplied during boiling increases the intermolecular potential energy
✓ 1
MP2 only scores if MP1 scores. Accept reference to Ek = (3/2)kBT.
107B-2-41Mechanisms of thermal transferPaper 2Easy13 marks
On a sunny morning, solar radiation of mean intensity 450 W m−2 falls on a sandy beach and on the sea next to it for 4.0 hours. The albedo of the dry sand is 0.40 and the albedo of the sea is 0.06.
In a simple model, the energy absorbed by each square metre of beach heats only the top 0.15 m of sand, and the energy absorbed by each square metre of sea heats the top 3.0 m of water, which is mixed by the waves. Ignore all other energy transfers.
material
density / kg m−3
specific heat capacity / J kg−1 K−1
dry sand
1600
830
sea water
1030
3990
Diagram NOT accurately drawn
(a)
(i)
Calculate the intensity absorbed by the sand.
(1)
(ii)
Show that the temperature of the sand rises by about 20 K.
(1)
(b)
Calculate the rise in temperature of the sea water.
(2)
(c)
Explain why the rise in temperature of the sea is much smaller than that of the sand, even though the sea absorbs a greater intensity.
(3)
(d)
By midday the air above the beach is warmer than the air above the sea. Annotate the diagram with arrows to show the circulation of air that develops.
(2)
(e)
Explain how the difference in temperature between the beach and the sea produces this circulation.
(3)
(f)
State one limitation of the model used in (a) and (b).
Must see full substitution OR answer to at least 3 s.f. The mass heated, 240 kg per square metre, must be seen or implied.
Part (b)
energy absorbed per m² = 0.94 × 450 × 14 400 = 6.1 × 106 J and mass heated = 1030 × 3.0 = 3090 kg
✓ 1
ΔT = 6.1 × 106 / (3090 × 3990) = 0.49 K
✓ 1
Award [2] for CNA. Accept 0.5 K.
Part (c)
the energy is shared by a much greater mass per square metre «3090 kg compared with 240 kg» because the waves mix it to a greater depth
✓ 1
the specific heat capacity of sea water is about five times that of sand
✓ 1
so the energy needed per kelvin per square metre is about 62 times greater, which outweighs the absorbed intensity being only about 1.6 times greater «lower albedo of the sea»
✓ 1
Accept a comparison of 1.2 × 107 J K−1 with 2.0 × 105 J K−1. OWTTE
Part (d)
arrows showing air rising above the beach and sinking above the sea
✓ 1
arrows from the sea towards the land near the ground and from the land towards the sea at height, forming a closed loop
✓ 1
Award [1] max for a correct loop with no indication of direction.
Part (e)
air in contact with the hot sand is heated «by conduction», expands and its density decreases
✓ 1
the less dense air rises above the beach
✓ 1
cooler, denser air from above the sea flows in near the ground to replace it «and air sinks over the sea», forming a convection current
✓ 1
OWTTE
Part (f)
the sand and the sea also lose energy, e.g. by radiation, by conduction to deeper layers or to the air, or by evaporation; the depths heated are assumed / the intensity is not constant
✓ 1
Accept any one sensible limitation.
Answers: (a)(i) 270 W m−2 · (a)(ii) 19.5 K · (b) 0.49 K (the remaining parts are explanations — see the table above)
108B-2-53Latent heat & phase changePaper 2Easy11 marks
A cook heats 2.50 kg of water at 18.0 °C in an open pan on a gas burner, ready to cook pasta. In a simple model, the burner transfers energy to the water at a constant rate of 1.80 kW and no energy is lost from the water to the surroundings.
Specific heat capacity of water = 4180 J kg−1 K−1. Specific latent heat of vaporization of water = 2.26 × 106 J kg−1.
(a)
(i)
State, in kelvin, the rise in temperature of the water as it is heated to its boiling point of 100 °C.
(1)
(ii)
Show that the water takes about 8 minutes to reach 100 °C.
(1)
(b)
The water then boils for 12.0 minutes. Calculate the mass of water that boils away.
(2)
(c)
While the water is boiling, the cook turns the burner up so that energy is transferred to the water at 2.70 kW.
(i)
State and explain what happens to the temperature of the water.
(2)
(ii)
Calculate the new rate at which water boils away.
(1)
(d)
Calculate the average kinetic energy of the water molecules in the steam just above the surface, which is at 100 °C.
(1)
(e)
When a lid is put on the pan, energy must be supplied at a smaller rate to keep the water boiling. Outline two reasons for this.
(2)
(f)
In a town high in the mountains, water boils at 90 °C. Suggest why pasta takes longer to cook there, however strongly the water is heated.
(1)
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Notes
Part (a)(i)
82 K
✓ 1
Do not accept 355 K.
Part (a)(ii)
t = mcΔT/P = 2.50 × 4180 × 82/1800 = 476 s «= 7.9 min»
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)
Q = Pt = 1800 × 720 = 1.30 × 106 J
✓ 1
m = Q/L = 1.30 × 106/2.26 × 106 = 0.57 kg
✓ 1
Award [2] for CNA. Award [1] max for 9.6 × 10−3 kg «12.0 used as seconds».
Part (c)(i)
the temperature stays at 100 °C «the boiling point»
✓ 1
MP2 only scores if MP1 scores.
the extra energy increases the intermolecular potential energy of more molecules, separating them as they become vapour, while the average kinetic energy of the molecules «and so the temperature» does not change
✓ 1
Accept “the water boils away faster but does not get hotter” with a reference to potential energy. OWTTE.
Part (c)(ii)
2700/2.26 × 106 = 1.2 × 10−3 kg s−1 «72 g per minute»
✓ 1
Accept answers in any correct unit.
Part (d)
Ek = ³⁄₂ × 1.38 × 10−23 × 373 = 7.72 × 10−21 J
✓ 1
Award [0] if 100 is used for T.
Part (e)
steam condenses on the lid and the water drips back, so less water/vapour «and the energy it carries» escapes from the pan
✓ 1
Accept “less evaporation/steam lost”.
the lid stops convection currents in the air carrying energy away from the hot water surface «and reduces radiation from the surface»
✓ 1
Two distinct reasons needed for [2].
Part (f)
the water cannot be heated above its boiling point, so the pasta cooks at a lower temperature «90 °C instead of 100 °C»; extra heating only makes the water boil away faster
✓ 1
OWTTE.
Answers: (a)(i) 82 K · (a)(ii) 476 s · (b) 0.57 kg · (c)(ii) 1.2 × 10−3 kg s−1 · (d) 7.72 × 10−21 J (the remaining parts are explanations — see the table above)
109B-2-55Temperature & direction of transferPaper 2Easy12 marks
In sous-vide cooking, food sealed in a plastic bag is placed in a water bath kept at a constant temperature. An immersion circulator heats the water with a 1.20 kW heater and pumps it continuously around the container. The container holds 12.0 kg of water at 16.0 °C, and the circulator is set to 57.0 °C.
Specific heat capacity of water = 4180 J kg−1 K−1.
Axes for (b)(iv)
(a)
(i)
Calculate the minimum time needed to heat the water to 57.0 °C.
(2)
(ii)
Suggest one reason why the actual time is longer than your answer to (a)(i).
(1)
(b)
A steak at 5 °C, sealed in a bag, is placed in the bath at 57.0 °C.
(i)
State the direction of the resultant thermal energy transfer between the water and the steak, and give the reason for this direction.
(1)
(ii)
Explain why the steak cannot become hotter than 57.0 °C, however long it is left in the bath.
(1)
(iii)
Outline why pumping the water around the container shortens the time taken for the steak to reach 57.0 °C.
(2)
(iv)
Sketch, on the axes, a graph to show how the temperature at the centre of the steak varies with time after it is placed in the bath.
(2)
(c)
With no lid on the container, the heater must supply a mean power of 290 W to keep the bath at 57.0 °C. Water evaporates from the surface at a rate of 0.060 kg per hour. Specific latent heat of vaporization of water at 57 °C = 2.37 × 106 J kg−1.
(i)
Calculate the rate at which energy is lost from the bath by evaporation.
(1)
(ii)
A layer of floating hollow plastic balls is spread over the surface. It reduces the rate of evaporation by 80 %. Assume that the other energy losses are unchanged. Determine the new mean power that the heater must supply.
(2)
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Part (a)(i)
Q = 12.0 × 4180 × 41.0 = 2.06 × 106 J
✓ 1
t = Q/P = 1710 s «28.6 min»
✓ 1
Award [2] for CNA.
Part (a)(ii)
energy is also lost from the water to the surroundings «evaporation, conduction through the container, convection in the air» OR energy is also needed to warm the container and the circulator
✓ 1
Do not accept “energy is lost” without saying where.
Part (b)(i)
from the water to the steak, because the water is at a higher temperature
✓ 1
Both needed.
Part (b)(ii)
when the steak reaches 57.0 °C there is no temperature difference, so there is no resultant transfer of energy to it «it reaches thermal equilibrium with the water»
✓ 1
OWTTE.
Part (b)(iii)
water next to the bag is cooled as it gives energy to the steak
✓ 1
circulation continually replaces it with water at 57 °C «forced convection», keeping the temperature difference at the surface of the bag large, so the rate of transfer stays high
✓ 1
OWTTE.
Part (b)(iv)
curve starting at 5 °C at t = 0 and rising with a gradient that decreases
✓ 1
An initial slow start «delay» is acceptable.
curve approaches 57 °C «levels off» and never rises above it
✓ 1
Do not accept a straight line reaching 57 °C.
Part (c)(i)
0.060 × 2.37 × 106/3600 = 39.5 W
✓ 1
Accept 39–40 W.
Part (c)(ii)
reduction = 0.80 × 39.5 = 31.6 W
✓ 1
Allow ECF from (c)(i).
new mean power = 290 − 31.6 = 258 W
✓ 1
Award [2] for CNA. Award [1] max for 31.6 W or 7.9 W.
Answers: (a)(i) 1710 s · (c)(i) 39.5 W · (c)(ii) 258 W (the remaining parts are explanations — see the table above)
110B-2-62Specific heat capacityPaper 2Easy11 marks
A night storage heater contains a core of ceramic bricks of total mass 110 kg, surrounded by insulation inside a metal casing. Overnight, when electricity is cheap, heating elements in the core are switched on. During the day the core releases its energy to the room. Channels through the core allow air to flow through it when a damper at the top is opened, as shown in the figure.
Specific heat capacity of the bricks = 880 J kg−1 K−1.
Diagram NOT accurately drawn
(a)
The elements supply 2.55 kW for 7.0 hours, and the temperature of the core rises from 30 °C to 650 °C.
(i)
Calculate the electrical energy supplied.
(1)
(ii)
Calculate the energy stored in the bricks.
(1)
(iii)
Suggest what has happened to the difference between your answers to (a)(i) and (a)(ii).
(1)
(b)
During the following 15 hours, the core cools from 650 °C to 80 °C. Calculate the mean rate at which the core transfers energy during this time.
(2)
(c)
The insulation has a thermal conductivity of 0.022 W m−1 K−1, a thickness of 45 mm and an area of 1.40 m².
(i)
In the morning the inner surface of the insulation is at 600 °C and the outer surface is at 45 °C. Calculate the rate of thermal energy transfer through the insulation.
(2)
(ii)
Explain why this rate is smaller in the evening.
(1)
(d)
Explain how opening the damper increases the rate at which energy is transferred from the core to the room.
(2)
(e)
Calculate the peak wavelength of the radiation emitted by the bricks at 650 °C.
(1)
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Notes
Part (a)(i)
2550 × 7.0 × 3600 = 6.43 × 107 J
✓ 1
Accept 17.9 kW h.
Part (a)(ii)
110 × 880 × 620 = 6.00 × 107 J
✓ 1
Part (a)(iii)
it has been transferred to the surroundings/room «through the insulation and casing» during the night
✓ 1
Do not accept “wasted” or “lost” alone.
Part (b)
Q = 110 × 880 × 570 = 5.52 × 107 J
✓ 1
P = 5.52 × 107/(15 × 3600) = 1.0 × 103 W
✓ 1
Award [2] for CNA.
Part (c)(i)
ΔQ/Δt = 0.022 × 1.40 × 555/0.045
✓ 1
Thickness must be in metres.
= 380 W
✓ 1
Award [2] for CNA.
Part (c)(ii)
the core has cooled, so the temperature difference «gradient» across the insulation is smaller
✓ 1
OWTTE.
Part (d)
air in the channels is heated by the bricks, expands, becomes less dense and rises out through the top
✓ 1
cooler, denser air from the room is drawn in at the bottom to replace it, so a convection current carries energy from the core into the room
✓ 1
OWTTE.
Part (e)
λmax = 2.9 × 10−3/923 = 3.1 × 10−6 m
✓ 1
Kelvin temperature needed.
Answers: (a)(i) 6.43 × 107 J · (a)(ii) 6.00 × 107 J · (b) 1.0 × 103 W · (c)(i) 380 W · (e) 3.1 × 10−6 m (the remaining parts are explanations — see the table above)
111IBP2S-009-01-11Latent heat & phase changePaper 2Easy4 marks
A reusable hand warmer for winter climbers contains 0.090 kg of a supercooled liquid salt solution. When a metal disc in the pouch is clicked, the liquid crystallises. It stays at a constant 54 °C until it is completely solid, releasing thermal energy to the climber's hands. Specific latent heat of fusion of the salt = 2.6 × 105 J kg−1
(a)
Outline, in terms of the molecular model, what happens to the internal energy of the salt as it crystallises at constant temperature.
(2)
(b)
Calculate the thermal energy released as all the salt crystallises.
(1)
(c)
The crystallisation takes 15 minutes. Calculate the mean rate at which thermal energy is released.
(1)
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Part (a)
The mean kinetic energy of the particles is unchanged because the temperature is constant . The potential energy decreases as the particles bond into the lattice, so the internal energy decreases and energy is released to the surroundings .
112IBP2S-009-04-11Method of mixturesPaper 2Easy6 marks
In therapeutic cooling after cardiac arrest, cold saline is infused into a patient. A patient receives 2.0 kg of saline that enters the body at 4 °C and warms to the body temperature of 37 °C. Specific heat capacity of saline = 4.0×103 J kg−1 K−1.
(a)
State what is meant by specific heat capacity.
(1)
(b)
Calculate the energy transferred to the saline as it warms from 4 °C to 37 °C.
(2)
(c)
State the direction of the net thermal energy transfer between the patient's tissue and the saline, and give a reason.
(1)
(d)
The patient has a mass of 70 kg. The average specific heat capacity of the body is 3.5×103 J kg−1 K−1. Estimate the fall in the patient's mean body temperature. Assume the energy comes entirely from the body.
(2)
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Notes
Part (a)
The energy required to raise the temperature of 1 kg of a substance by 1 K .
✓ 1
Part (b)
Q = mcΔT = 2.0 × 4.0×103 × 33
✓ 1
Q = 2.6×105 J
✓ 1
Part (c)
From the tissue to the saline, because the tissue is at a higher temperature than the saline .
✓ 1
Part (d)
ΔT = Q/(mc) = 2.64×105/(70 × 3.5×103)
✓ 1
ΔT ≈ 1.1 K
✓ 1
ECF from (b).
—
Exam tip: The temperature change of 33 K equals 33 °C, so you do not need to convert to kelvin when only a change in temperature is involved.
113IBP2S-009-04-12Temperature & direction of transferPaper 2Easy5 marks
Squash players warm up the ball before a match by hitting it repeatedly. The ball has a mass of 24 g, and the specific heat capacity of its rubber is 1.7×103 J kg−1 K−1. During the warm-up its temperature rises by 15 K.
(a)
Calculate the energy needed to raise the temperature of the ball by 15 K.
(2)
(b)
Each impact with the racket or the wall transfers 0.80 J to the internal energy of the ball. Calculate the minimum number of impacts needed, assuming no energy is lost from the ball.
(1)
(c)
Outline why the actual number of impacts needed is greater than your answer to (b).
(2)
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Notes
Part (a)
Q = mcΔT = 0.024 × 1.7×103 × 15
✓ 1
Q = 612 J (accept 610 J or 6.1×102 J)
✓ 1
Part (b)
N = 612 / 0.80 = 765 impacts
✓ 1
ECF from (a): accept 763 if 610 J was used (762.5 rounded up to a whole number of impacts).
—
Part (c)
Once the ball is warmer than the surrounding air, it transfers energy to the surroundings (by conduction, convection and/or radiation) .
✓ 1
So the net gain in internal energy per impact is less than 0.80 J (this loss grows as the temperature difference increases), and more impacts are needed to reach the final temperature .
✓ 1
Common error: Convert grams to kilograms before using Q = mcΔT. Using 24 instead of 0.024 makes the answer 1000 times too large.
114IBP2S-009-00-11Temperature & molecular energyPaper 2Easy6 marks
The superconducting magnet in a hospital MRI scanner is cooled by liquid helium at a temperature of 4.2 K. Boltzmann constant kB = 1.38×10−23 J K−1.
(a)
State the temperature of the liquid helium in degrees Celsius.
(1)
(b)
State what is meant by absolute zero.
(1)
(c)
Outline the relationship between the kelvin temperature of a gas and the motion of its particles.
(2)
(d)
Calculate the average translational kinetic energy of a helium atom in the vapour above the liquid at 4.2 K.
(2)
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Notes
Part (a)
4.2 − 273 = −269 °C (accept −268.8 °C using 273, or −268.95 °C using 273.15)
✓ 1
Part (b)
The lowest possible temperature (0 K or −273 °C), at which a substance has its minimum internal energy / its particles have minimum kinetic energy / no further energy can be removed from it
✓ 1
Part (c)
The average random (translational) kinetic energy of the particles is directly proportional to the kelvin temperature
✓ 1
So the higher the absolute temperature, the faster the particles move on average (e.g. doubling T doubles the mean kinetic energy)
✓ 1
Part (d)
E = \(\frac{3}{2}\)kBT = 1.5 × 1.38×10−23 × 4.2
✓ 1
E = 8.7×10−23 J
✓ 1
Exam tip: The kinetic energy formula \(\frac{3}{2}k_BT\) needs T in kelvin. Here the temperature is already given in kelvin, so use it directly.
A computer processor (CPU) dissipates 65 W. All of this energy is conducted through a copper base plate into a cooling fan assembly. The base plate has a contact area of 4.0 cm2 and a thickness of 3.0 mm. The thermal conductivity of copper is 390 W m−1 K−1.
Diagram not to scale
(a)
Outline the mechanisms by which thermal energy is conducted through copper.
(2)
(b)
Calculate the temperature difference between the two faces of the copper plate.
(2)
(c)
A technician forgets to apply thermal paste, leaving a very thin layer of air between the CPU and the plate. Explain why the CPU temperature rises.
(2)
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Notes
Part (a)
Free/delocalised electrons gain kinetic energy at the hot side, move through the metal and transfer energy by collisions
✓ 1
Vibrating lattice ions pass energy to neighbouring ions (lattice vibrations)
✓ 1
Part (b)
ΔT = PΔx/(kA) = (65 × 3.0×10−3)/(390 × 4.0×10−4)
✓ 1
ΔT = 1.25 K ≈ 1.3 K
✓ 1
Part (c)
Air has a much lower thermal conductivity than the paste/copper
✓ 1
To conduct the same 65 W, a much larger temperature difference is needed across the gap, so the CPU must reach a higher temperature
✓ 1
Common error: Convert cm² to m² by multiplying by 10⁻⁴, not 10⁻².
116IBP2S-009-03-12Density & the molecular modelPaper 2Easy5 marks
In cryotherapy, liquid nitrogen is used to freeze and remove skin lesions. Liquid nitrogen boils at 77 K. Use kB = 1.38 × 10−23 J K−1.
(a)
Convert 77 K to degrees Celsius.
(1)
(b)
Calculate the average translational kinetic energy of a nitrogen molecule in the gas just above the liquid at 77 K.
(2)
(c)
While the liquid nitrogen boils, its temperature remains at 77 K even though energy is being supplied to it. Outline what happens to the energy supplied, in terms of the molecules.
(2)
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Notes
Part (a)
77 − 273 = −196 °C
✓ 1
Part (b)
Ek = (3/2)kBT = 1.5 × 1.38×10−23 × 77
✓ 1
Ek = 1.6×10−21 J
✓ 1
Part (c)
The average kinetic energy of the molecules stays constant, because the temperature is constant
✓ 1
The energy increases the intermolecular potential energy, as the molecules are separated against the intermolecular forces
✓ 1
Common error: Use the kelvin temperature directly in (3/2)kBT. Substituting −196 gives a negative kinetic energy, which is impossible.
117IBP2S-009-02-11Stefan–Boltzmann & WienPaper 2Easy5 marks
A non-contact medical thermometer detects radiation emitted from a patient's forehead. The skin surface temperature is 34 °C and the emissivity of skin is 0.98. Wien's constant b = 2.9×10−3 m K; σ = 5.67×10−8 W m−2 K−4.
(a)
Calculate the wavelength at which the emission from the skin is a maximum.
(2)
(b)
State the region of the electromagnetic spectrum in which this wavelength lies.
(1)
(c)
Calculate the power emitted per unit area of the skin.
(2)
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Notes
Part (a)
T = 34 + 273 = 307 K (accept 307.15 K) ; λmax = 2.9×10−3 / 307 = 9.4×10−6 m
✓ 2
Part (b)
Infrared, or a region consistent with the wavelength calculated in the previous part
✓ 1
Part (c)
P/A = eσT4 = 0.98 × 5.67×10−8 × 3074 ; = 494 W m−2 (accept 490 or 4.9×102 W m−2; accept answers using 307.15 K)
✓ 2
Common error: Using 34 instead of 307 K gives a peak wavelength of about 85 μm and a far too small power. Always convert the temperature to kelvin before using Wien's law or the Stefan–Boltzmann law.
118IBP2S-009-02-12Method of mixturesPaper 2Easy6 marks
In an engineering workshop, a steel gear of mass 1.2 kg at 850 °C is quenched by lowering it into a tank containing 15 kg of oil at 25 °C. Specific heat capacity of steel = 490 J kg−1 K−1; specific heat capacity of oil = 1900 J kg−1 K−1.
(a)
Calculate the final common temperature of the gear and the oil.
(3)
(b)
State one assumption made in your calculation in (a).
(1)
(c)
Explain whether the actual final temperature of the oil is likely to be higher or lower than your answer to (a).
(2)
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Notes
Part (a)
Energy lost by steel = energy gained by oil: 1.2 × 490 × (850 − T) = 15 × 1900 × (T − 25) ; 499800 − 588T = 28500T − 712500, so T = 1212300/29088 ; T = 41.7 °C (accept 42 °C)
✓ 3
Part (b)
Any one: no energy is transferred to the tank or the surroundings / no oil evaporates / the specific heat capacities are constant over the temperature range
✓ 1
Part (c)
Lower ; because some energy from the gear is transferred to the tank, to the air or to vaporising oil, so less energy is available to raise the oil's temperature
✓ 2
Exam tip: Set up 'energy lost = energy gained' with each temperature change written as a positive quantity: (850 − T) for the steel and (T − 25) for the oil.
119B-2-02Latent heat & phase changePaper 2Medium13 marks
Paraffin wax is used in hand warmers. A sample of solid wax of mass 0.150 kg, initially at 20 °C, is heated by an electric heater that supplies energy to it at a constant rate of 60.0 W. The graph shows how the temperature θ of the wax varies with time t. Assume that all the energy supplied by the heater is absorbed by the wax.
Graph drawn to scale
(a)
State the melting point of the wax in kelvin.
(1)
(b)
Show that the specific latent heat of fusion of the wax is about 2 × 105 J kg−1.
(1)
(c)
Calculate the specific heat capacity of the solid wax.
(2)
(d)
Deduce, using the gradients of the graph, whether the specific heat capacity of the liquid wax is greater or smaller than that of the solid wax.
(2)
(e)
Explain, with reference to the molecules, why the temperature of the wax is constant between t = 180 s and t = 720 s.
(2)
(f)
A second sample of the same wax, of mass 0.300 kg, is heated from 20 °C by the same heater. Sketch, on the graph, how the temperature of this sample varies with time for the first 900 s.
(3)
(g)
In a real experiment the wax loses some energy to the surroundings. Suggest how this affects the value of the specific latent heat found in (b).
Must see the melting time of 540 s read from the graph and full substitution OR answer to at least 3 s.f.
Part (c)
c = Pt/(mΔθ) = 60.0 × 180 / (0.150 × 36)
✓ 1
= 2.0 × 103 J kg−1 K−1
✓ 1
Accept 1.9–2.1 × 103 J kg−1 K−1. Award [2] for CNA.
Part (d)
gradient for the solid = 36/180 = 0.20 K s−1; for the liquid = 24/150 = 0.16 K s−1
✓ 1
Accept equivalent readings.
same P and m, so c = P/(m × gradient) is inversely proportional to the gradient: the liquid has the smaller gradient, so its specific heat capacity is greater (≈ 2.5 × 103 J kg−1 K−1)
✓ 1
ALTERNATIVE: calculating c for the liquid and comparing with (c) scores both marks. Do not accept a bare 'greater'.
Part (e)
the energy supplied increases the intermolecular potential energy: work is done separating the molecules against the attractive forces between them (breaking the solid structure)
✓ 1
the mean random kinetic energy of the molecules does not change, and temperature is a measure of the mean kinetic energy, so θ stays constant
✓ 1
OWTTE.
Part (f)
straight line from (0, 20 °C) with half the original gradient, reaching 56 °C at about 360 s
✓ 1
Accept 340–380 s.
then horizontal at the same temperature, 56 °C
✓ 1
the horizontal section continues to the end of the graph at 900 s (melting would last until 1440 s)
✓ 1
MP3: the line must not rise before 900 s.
Part (g)
during the melting some of the energy supplied is lost to the surroundings, so the energy actually used to melt the wax is less than 60.0 × 540 J / the melting takes longer than it would without losses
✓ 1
so the calculated value of L is too large (an overestimate)
✓ 1
MP2 only scores if MP1 scores.
Answers: (a) 329 K · (b) 2.16 × 105 J kg−1 · (c) 2.0 × 103 J kg−1 K−1 · (d) liquid: greater (the remaining parts are explanations — see the table above)
The outer wall of a mountain hut has an area of 15 m2. It is made of two layers in close contact: a timber board 20 mm thick on the inside and a layer of mineral wool 80 mm thick on the outside. The inner surface of the wall is at 20 °C and the outer surface is at −12 °C. The wall has reached a steady state.
Thermal conductivity of the timber = 0.15 W m−1 K−1. Thermal conductivity of the mineral wool = 0.040 W m−1 K−1.
Ignore the steel bolts shown in the diagram until (b).
Diagram NOT accurately drawn
(a)
(i)
Explain why, in the steady state, the rate of thermal energy transfer through the timber is equal to the rate of thermal energy transfer through the mineral wool.
(2)
(ii)
Show that the temperature at the boundary between the timber and the mineral wool is 18 °C.
(2)
(iii)
Calculate the rate of thermal energy transfer through the wall.
(2)
(b)
The two layers are held together by 40 steel bolts that pass right through the wall. Each bolt has a cross-sectional area of 1.0 × 10−4 m2, and its ends are at the temperatures of the two surfaces of the wall. Thermal conductivity of steel = 50 W m−1 K−1. Determine whether the bolts increase the rate of thermal energy transfer through the wall by more than 10 %.
(3)
(c)
Explain, in terms of particles, why steel conducts thermal energy far better than mineral wool, which consists of fine fibres with air trapped between them.
(2)
(d)
Suggest one change to the design of the wall, other than using thicker mineral wool, that would reduce the rate of thermal energy transfer through it.
(1)
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Part (a)(i)
in the steady state the temperature at every point of the wall, including the boundary, is constant, so the internal energy of the layers does not change
✓ 1
so, by conservation of energy, the energy transferred per second into the boundary through the timber must all leave it through the mineral wool
✓ 1
OWTTE. Do not accept 'the layers are in series' without an energy argument.
Accept 220–230 W. Award [2] for CNA. Award [1] max for 240 W or 3.6 × 103 W (the whole temperature difference of 32 K used across one layer).
Part (b)
rate through one bolt = 50 × 1.0 × 10−4 × 32 / 0.100 = 1.6 W
✓ 1
Length of bolt = 0.100 m (both layers) must be used.
rate through 40 bolts = 64 W
✓ 1
Allow ECF from MP1.
64/225 = 28 % «increase», which is more than 10 %, so yes
✓ 1
Allow ECF from (a)(iii). MP3 needs a comparison and a consistent conclusion.
Part (c)
steel contains free / delocalised electrons that move quickly through the metal and carry kinetic energy from hotter to colder regions «as well as the vibrations passed between closely packed ions»
✓ 1
in mineral wool energy must pass through thin fibres touching at few points and through air, whose molecules are far apart, so collisions transfer kinetic energy only slowly
✓ 1
OWTTE.
Part (d)
replace the steel bolts by fixings of low thermal conductivity (e.g. wood/plastic) OR use bolts that do not pass right through the wall OR use bolts of smaller cross-sectional area / fewer bolts
✓ 1
Accept a thicker timber board or a material of lower thermal conductivity. Do not accept 'thicker wool'.
Answers: (a)(ii) 18 °C · (a)(iii) 225 W · (b) 64 W, a 28 % increase — yes (the remaining parts are explanations — see the table above)
121B-2-13Density & the molecular modelPaper 2Medium12 marks
Liquid nitrogen boils at −196 °C at atmospheric pressure. At this temperature the density of liquid nitrogen is 808 kg m−3. At room temperature (20 °C) and atmospheric pressure, the density of nitrogen gas is 1.16 kg m−3.
A technician pours 1.00 × 10−3 m3 of liquid nitrogen into an open vacuum flask in a laboratory at 20 °C. Over the following hours all of the liquid boils away.
(a)
State the boiling point of nitrogen in kelvin.
(1)
(b)
Calculate the average kinetic energy of the molecules of nitrogen gas at its boiling point.
(1)
(c)
Show that, when the gas has warmed to room temperature, its volume is about 700 times the volume of the liquid from which it formed.
(1)
(d)
Estimate the ratio of the mean separation of the molecules in the gas at room temperature to the mean separation of the molecules in the liquid.
(2)
(e)
Describe the differences between liquid nitrogen and nitrogen gas in terms of the arrangement, the forces and the motion of their molecules.
(3)
(f)
Explain how the internal energy of the nitrogen changes as the liquid boils at its boiling point and as the gas then warms to room temperature.
(3)
(g)
The flask is not being heated. Explain why the nitrogen nevertheless boils.
Must see the ratio of densities (or mass 0.808 kg and volume 0.697 m3) OR answer to at least 3 s.f.
Part (d)
the volume occupied by each molecule is about 700 times larger, and the separation is proportional to the cube root of the volume per molecule
✓ 1
ratio ≈ 7001/3 = 8.9 ≈ 9
✓ 1
Accept 8.8–9. Award [1] max for 700 or 26 (√700).
Part (e)
arrangement: in the liquid the molecules are close together (almost touching) in a disordered arrangement; in the gas they are far apart
✓ 1
forces: intermolecular forces are significant in the liquid but negligible in the gas (except during collisions)
✓ 1
motion: in the liquid the molecules vibrate and slide past one another; in the gas they move freely and randomly at high speed in straight lines between collisions
✓ 1
Part (f)
internal energy = total intermolecular potential energy + total random kinetic energy of the molecules
✓ 1
May be implied by the answers for the two stages.
boiling: the potential energy increases because the molecules are separated against the attractive forces, while the kinetic energy stays the same (constant temperature)
✓ 1
warming: the random kinetic energy increases because the temperature rises, while the potential energy stays (almost) the same — so the internal energy increases in both stages
✓ 1
Part (g)
the laboratory/surroundings are at a higher temperature than the nitrogen, so there is a net thermal energy transfer into the nitrogen, which is used to boil it
✓ 1
Answers: (a) 77 K · (b) 1.59 × 10−21 J · (c) 697 · (d) 8.9 (the remaining parts are explanations — see the table above)
122B-2-20Continuous-flow heatingPaper 2Medium13 marks
In a nuclear power station the fission of uranium-235 in the reactor releases energy at a rate of 3.2 GW. The electrical power output of the station is 1.1 GW. The rest of the energy is removed in the condenser by cooling water that is taken from a river and returned to it at a higher temperature.
Energy released per fission = 200 MeV. Mass of a uranium-235 nucleus = 235 u. Specific heat capacity of water = 4180 J kg−1 K−1.
(a)
(i)
Calculate the overall efficiency of the power station.
(1)
(ii)
Determine the number of fissions that occur in the reactor each second.
(2)
(iii)
Estimate the mass of uranium-235 that undergoes fission in one day.
(2)
(b)
Regulations allow the station to take no more than 5.0 × 104 kg of water per second from the river, and the water may be returned no more than 8.0 K warmer than when it was taken.
(i)
Determine whether the station can operate at full power without breaking either regulation.
(4)
(ii)
Outline how the control rods can be used to reduce the power of the reactor.
(2)
(c)
The warm water returned to the river tends to remain near the surface instead of mixing with the cooler water below it. Explain why.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
η = 1.1/3.2 = 0.34 OR 34 %
✓ 1
Accept 0.344 or 34.4 %.
Part (a)(ii)
energy per fission = 200 × 1.60 × 10−13 = 3.2 × 10−11 J
✓ 1
number per second = 3.2 × 109/3.2 × 10−11 = 1.0 × 1020 s−1
✓ 1
Award [2] for CNA.
Part (a)(iii)
mass of one nucleus = 235 × 1.66 × 10−27 = 3.90 × 10−25 kg
✓ 1
mass per day = 1.0 × 1020 × 86 400 × 3.90 × 10−25 ≈ 3.4 kg
✓ 1
Accept 3.3–3.4 kg. Allow ECF from (a)(ii).
Part (b)(i)
power to be removed = 3.2 − 1.1 = 2.1 GW «2.1 × 109 W»
✓ 1
greatest power that can be removed = 5.0 × 104 × 4180 × 8.0
✓ 1
= 1.67 × 109 W «1.7 GW»
✓ 1
this is less than 2.1 × 109 W, so the station cannot run at full power
✓ 1
ALTERNATIVE 1: flow needed for an 8.0 K rise = 2.1 × 109/(4180 × 8.0) = 6.3 × 104 kg s−1 ✓✓, more than 5.0 × 104 kg s−1 ✓. ALTERNATIVE 2: rise at the greatest flow = 2.1 × 109/(5.0 × 104 × 4180) = 10 K ✓✓, more than 8.0 K ✓. MP4 requires a comparison of calculated values.
Part (b)(ii)
control rods absorb neutrons «without undergoing fission»
✓ 1
inserting them further means fewer neutrons go on to cause fissions, so the fission rate «and the power» decreases
✓ 1
OWTTE.
Part (c)
warm water has expanded, so its density is lower than that of the cooler river water
✓ 1
the less dense water is buoyed up «resultant upward force» and stays above the denser water, so convection does not carry it downwards
✓ 1
OWTTE.
Answers: (a)(i) 34 % · (a)(ii) 1.0 × 1020 s−1 · (a)(iii) 3.4 kg · (b)(i) 2.1 × 109 W needed, 1.67 × 109 W possible: no (the remaining parts are explanations — see the table above)
123B-2-22Latent heat & phase changePaper 2Medium13 marks
A cyclist rides at a constant speed of 8.0 m s−1 against a total resistive force of 30 N. The cyclist's muscles convert chemical energy from food into useful mechanical work with an efficiency of 20 %; the rest of the energy released becomes thermal energy in the body.
Mass of cyclist = 68 kg. Mean specific heat capacity of the human body = 3500 J kg−1 K−1. Specific latent heat of vaporization of sweat at skin temperature = 2.4 × 106 J kg−1.
(a)
(i)
Calculate the useful mechanical power developed by the cyclist.
(1)
(ii)
Show that thermal energy is produced in the cyclist's body at a rate of about 960 W.
(2)
(b)
During a 2.0 hour ride, 75 % of the thermal energy produced is removed by the evaporation of sweat.
The cyclist drinks 1.5 kg of water during the ride. Determine whether this replaces the water lost by evaporation.
(4)
(c)
(i)
Estimate the time it would take for the cyclist's body temperature to rise by 2.0 K if none of the thermal energy produced could leave the body.
(2)
(ii)
Explain, in terms of molecules, why the evaporation of sweat cools the skin.
(3)
(iii)
Suggest why cycling on a hot, humid day is more likely to cause overheating than cycling on a hot, dry day.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
P = Fv = 30 × 8.0 = 240 W
✓ 1
Part (a)(ii)
input power = 240/0.20 = 1200 W
✓ 1
thermal power = 1200 − 240 = 960 W
✓ 1
Award [0] for 0.80 × 240 = 192 W. Must see both steps.
Part (b)
power removed by evaporation = 0.75 × 960 = 720 W
✓ 1
Allow ECF from (a)(ii).
rate of evaporation = 720/2.4 × 106 = 3.0 × 10−4 kg s−1
✓ 1
mass evaporated in 2.0 h = 3.0 × 10−4 × 7200 = 2.2 kg
✓ 1
ALTERNATIVE: energy removed = 720 × 7200 = 5.2 × 106 J ✓, mass = 5.2 × 106/2.4 × 106 = 2.2 kg ✓.
2.2 kg is more than 1.5 kg, so the water drunk does not replace the water lost
✓ 1
MP4 requires a comparison with a calculated value.
Part (c)(i)
t = mcΔT/P = 68 × 3500 × 2.0/960
✓ 1
t ≈ 500 s «about 8 minutes»
✓ 1
Accept 470–500 s «use of 1 kW gives 476 s».
Part (c)(ii)
only the fastest/most energetic molecules have enough energy to overcome the intermolecular forces and escape from the liquid surface
✓ 1
so the average kinetic energy of the molecules remaining in the liquid decreases
✓ 1
temperature is a measure of the average kinetic energy, so the sweat cools and energy is transferred to it from the skin
✓ 1
Part (c)(iii)
in humid air the rate of evaporation is lower «molecules return to the liquid almost as fast as they leave», so less energy is removed each second «sweat drips off without evaporating»
✓ 1
OWTTE.
Answers: (a)(i) 240 W · (b) 2.2 kg lost: no · (c)(i) ≈ 500 s (the remaining parts are explanations — see the table above)
124B-2-23Stefan–Boltzmann & WienPaper 2Medium12 marks
A potter monitors the temperature inside a kiln by analysing the radiation that emerges from a small peephole in the kiln wall. The peephole has an area of 4.0 cm² and behaves as a black body at the temperature of the kiln interior.
The graph shows the spectrum of the radiation from the peephole at one moment during a firing.
Graph drawn to scale
(a)
(i)
State what is meant by a black body.
(1)
(ii)
Using the graph, determine the temperature of the kiln interior, in °C.
(2)
(iii)
The kiln later cools. Sketch, on the graph, the spectrum of the radiation from the peephole when the temperature of the kiln interior is lower.
(3)
(b)
Calculate the power radiated through the peephole at the temperature found in (a)(ii).
(2)
(c)
(i)
Calculate the energy of a photon at the peak wavelength of the spectrum.
(2)
(ii)
The interior of the kiln appears bright orange to the potter. Explain this observation, given that the peak of the spectrum is in the infrared.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
a body that absorbs all the electromagnetic radiation that falls on it «and so emits the maximum possible radiation at each wavelength for its temperature»
✓ 1
OWTTE.
Part (a)(ii)
λmax = 2.0 μm read from the graph AND T = 2.9 × 10−3/2.0 × 10−6 = 1450 K
✓ 1
Accept λmax from 1.9 to 2.1 μm.
T = 1450 − 273 ≈ 1180 °C
✓ 1
Accept 1110–1250 °C. Award [1 max] for an answer left in kelvin.
Part (a)(iii)
peak at a longer wavelength than 2.0 μm
✓ 1
peak lower than the original peak
✓ 1
curve below the original at every wavelength, with a similar shape
✓ 1
Curves must not cross.
Part (b)
P = σAT4 = 5.67 × 10−8 × 4.0 × 10−4 × 14504
✓ 1
Allow ECF from (a)(ii).
P = 1.0 × 102 W
✓ 1
Award [1 max] for 44 W «temperature in °C used». Award [1 max] if the area is not converted to m².
Part (c)(i)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/2.0 × 10−6
✓ 1
E = 9.9 × 10−20 J «0.62 eV»
✓ 1
Allow ECF from the value of λmax read in (a)(ii).
Part (c)(ii)
the spectrum is continuous and extends into the visible region, so some visible light is emitted
✓ 1
within the visible region the intensity is greatest at the red «long-wavelength» end and very small at the blue end, so the light looks orange-red
✓ 1
OWTTE.
Answers: (a)(ii) 1450 K ≈ 1180 °C · (b) 1.0 × 102 W · (c)(i) 9.9 × 10−20 J (the remaining parts are explanations — see the table above)
125B-2-37Stefan–Boltzmann & WienPaper 2Medium13 marks
At an airport, a thermal imaging camera screens travellers for fever. The camera detects infrared radiation of wavelengths between 8 μm and 14 μm emitted by the skin of a traveller’s forehead. It calculates the temperature of the skin by assuming that the skin radiates like a surface of emissivity 0.98.
The forehead skin of a healthy traveller is at 34.0 °C. The graph shows how the intensity of the radiation emitted by a black body at 34.0 °C varies with wavelength λ.
Graph drawn to scale
(a)
(i)
Show that the wavelength at which the emission from the skin of the healthy traveller is most intense is about 9.4 μm.
(1)
(ii)
Outline why the camera is designed to detect wavelengths between 8 μm and 14 μm.
(1)
(b)
A traveller with a fever has forehead skin at 37.5 °C. Calculate the percentage increase in the power radiated per unit area by the skin, compared with the skin of the healthy traveller.
(2)
(c)
Sketch, on the graph, the spectrum of the radiation emitted by a black body at 37.5 °C.
(2)
(d)
Many travellers wear glasses. In the image produced by the camera, the lenses of the glasses appear much colder than the skin around them. Glass does not transmit infrared radiation of wavelengths between 8 μm and 14 μm. Explain this observation.
(3)
(e)
The forehead of another traveller, who has a fever, is covered with a cosmetic cream of emissivity 0.92. The skin is at 37.5 °C. The camera flags a traveller when the temperature it calculates is 37.0 °C or more. Ignore radiation reflected by the skin. Determine whether the camera flags this traveller.
(3)
(f)
Suggest one reason, other than the emissivity of the skin, why the camera may fail to identify a traveller who has a fever.
Must see full substitution OR answer to at least 3 s.f. The temperature must be in kelvin.
Part (a)(ii)
this band contains the peak of the emission spectrum of skin, so skin emits most strongly / the camera receives most power at these wavelengths
✓ 1
OWTTE
Part (b)
ratio of powers per unit area = (310.5/307)4 = 1.046
✓ 1
Accept temperatures converted with 273.15.
increase = 4.6 %
✓ 1
Award [2] for CNA. Accept 4.5–4.7 %. Award [0] for 48 % (Celsius temperatures used).
Part (c)
curve above the given curve at every wavelength «peak about 6 % higher»
✓ 1
The two curves must not cross.
peak at a slightly shorter wavelength «about 9.3 μm», with a similar shape
✓ 1
Do not accept a large shift of the peak (more than about 1 μm).
Part (d)
the radiation emitted by the skin behind the lenses is absorbed by the glass / does not reach the camera
✓ 1
the camera detects instead the radiation emitted by the outer surface of the lenses
✓ 1
the outer surface of the lens is at a lower temperature than the skin «it is separated from the skin by air and cooled by the surrounding air», so it emits less power per unit area
✓ 1
OWTTE
Part (e)
power per unit area emitted = 0.92σ(310.5)4 «= 485 W m−2»
✓ 1
camera: 0.98σT4 = 0.92σ(310.5)4, so T = 310.5 × (0.92/0.98)1/4 = 305.6 K
✓ 1
= 32.6 °C, which is below 37.0 °C, so the traveller is not flagged
✓ 1
Allow ECF. MP3 needs a comparison and a conclusion consistent with the candidate’s value.
Part (f)
the skin temperature is not the same as the internal body temperature, e.g. a traveller who has just come in from cold outside air, or whose forehead is sweating, has cooler skin
✓ 1
Accept hair or a hat covering the forehead; the traveller too far from the camera. OWTTE
Answers: (a)(i) 9.45 × 10−6 m · (b) 4.6 % · (e) about 32.6 °C — not flagged (the remaining parts are explanations — see the table above)
A high-power light-emitting diode (LED) in a car headlamp is mounted on an aluminium heat sink. A thin thermal pad fills the gap between the base of the LED and the heat sink, as shown.
In normal use the current in the LED is 1.50 A and the potential difference across it is 3.20 V. Of the electrical power supplied to the LED, 35 % is emitted as light and the rest is transferred as thermal energy through the pad to the heat sink.
The pad has an area of 6.0 × 10−5 m², a thickness of 0.25 mm and a thermal conductivity of 1.2 W m−1 K−1. The surface of the heat sink in contact with the pad is at 60 °C. The base of the LED must not exceed 85 °C.
Diagram NOT accurately drawn
(a)
Show that thermal energy is transferred through the pad at a rate of about 3.1 W.
(1)
(b)
Calculate the temperature of the base of the LED.
(2)
(c)
Without the pad, the LED and the heat sink would touch only at a few points, leaving a layer of air about 0.020 mm thick between them. Assume that all the thermal energy is conducted through a layer of air of this thickness over the same area. Thermal conductivity of air = 0.026 W m−1 K−1. Determine whether the LED would then exceed its maximum safe temperature.
(3)
(d)
When the current in the LED is 0.75 A, the potential difference across it is 3.00 V. Deduce whether the LED is an ohmic conductor.
(2)
(e)
The equation ΔQ/Δt = kAΔT/Δx has the same form as the equation for the current in a conductor of resistivity ρ, length L and cross-sectional area A. Outline this similarity.
(2)
(f)
The headlamp is designed so that the fins of the heat sink are vertical. Explain how this helps the heat sink to transfer thermal energy to the surrounding air.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
P = 1.50 × 3.20 = 4.80 W and 0.65 × 4.80 = 3.12 W
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Award [2] for CNA. Accept 70–71 °C. Award [1] max for 10.8 °C.
Part (c)
ΔT = 3.12 × 2.0 × 10−5 / (0.026 × 6.0 × 10−5)
✓ 1
Allow ECF from (a).
= 40 K, so the base would be at about 100 °C
✓ 1
this is above 85 °C, so the LED would exceed its maximum safe temperature
✓ 1
MP3 needs a comparison and a consistent conclusion.
Part (d)
R = V/I = 2.13 Ω at 1.50 A and 4.00 Ω at 0.75 A
✓ 1
Accept a comparison of V/I ratios or a statement that halving I does not halve V.
the resistance is not constant / I is not proportional to V, so the LED is non-ohmic
✓ 1
MP2 only scores if MP1 scores.
Part (e)
rate of thermal energy transfer ΔQ/Δt corresponds to current I «rate of flow of charge» and temperature difference ΔT corresponds to potential difference V
✓ 1
I = V/R = VA/(ρL), so k corresponds to 1/ρ and the thickness Δx to the length L «a thicker layer is like a longer wire»
✓ 1
OWTTE
Part (f)
air next to the fins is heated «by conduction», expands and its density decreases
✓ 1
the warm air rises «convection» and is replaced by cooler air drawn in from below
✓ 1
vertical fins let the air flow freely upwards along their surfaces, so the flow is continuous / horizontal fins would trap the warm air
✓ 1
OWTTE
Answers: (a) 3.12 W · (b) 71 °C · (c) about 100 °C — yes · (d) 2.13 Ω and 4.00 Ω — non-ohmic (the remaining parts are explanations — see the table above)
127B-2-54Specific heat capacityPaper 2Medium13 marks
At a solar power tower, thousands of mirrors reflect sunlight onto a receiver at the top of a tower. Molten salt is pumped through the receiver, where it is heated from 290 °C to 565 °C, and is then stored in an insulated hot tank. When electricity is needed, the hot salt passes through a steam generator, where it transfers energy to water, and returns to a cold tank at 290 °C.
Specific heat capacity of the molten salt = 1.50 × 103 J kg−1 K−1. Density of the molten salt = 1.90 × 103 kg m−3.
(a)
(i)
Show that each kilogram of salt stores about 4 × 105 J when it is heated in the receiver.
(1)
(ii)
The plant can deliver an electrical power of 110 MW for 10.0 hours using only stored salt. The efficiency of conversion of the stored thermal energy to electrical energy is 41 %. Determine the mass of hot salt that must be stored.
(3)
(iii)
Calculate the volume of the hot tank needed.
(1)
(b)
The outer surface of the receiver has an area of 1.10 × 103 m² and is at a mean temperature of 600 °C. Assume that it radiates as a black body.
(i)
Calculate the power radiated by the receiver.
(2)
(ii)
Calculate the wavelength at which the radiation from the receiver is most intense, and state the region of the electromagnetic spectrum in which it lies.
(2)
(iii)
The receiver absorbs 5.6 × 108 W of concentrated solar radiation. Calculate the percentage of this power that is lost by radiation from the receiver.
(1)
(iv)
Heating the salt to a much higher temperature would store more energy in each kilogram. Suggest, with reference to (b)(i), one disadvantage of operating the receiver at a much higher temperature.
(1)
(c)
The water enters the steam generator at 240 °C. Explain why the salt cannot leave the steam generator at a temperature below 240 °C.
Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
thermal power needed = 110/0.41 = 268 MW
✓ 1
Award [0] for MP1 if 0.41 × 110 is used.
energy to be stored = 268 × 106 × 10.0 × 3600 = 9.66 × 1012 J
✓ 1
m = 9.66 × 1012/4.125 × 105 = 2.3 × 107 kg
✓ 1
Allow ECF. Award [3] for CNA. Award [2] max for 9.6 × 106 kg «efficiency omitted».
Part (a)(iii)
V = m/ρ = 2.34 × 107/1.90 × 103 = 1.2 × 104 m³
✓ 1
Allow ECF from (a)(ii).
Part (b)(i)
P = σAT⁴ = 5.67 × 10−8 × 1.10 × 103 × 873⁴
✓ 1
Kelvin temperature must be used.
= 3.6 × 107 W
✓ 1
Award [2] for CNA. Award [1] max for 8.1 × 106 W «600 used for T».
Part (b)(ii)
λmax = 2.9 × 10−3/873 = 3.3 × 10−6 m
✓ 1
Allow 3.3 μm.
infrared
✓ 1
Allow ECF for a region consistent with their wavelength.
Part (b)(iii)
3.62 × 107/5.6 × 108 × 100 = 6.5 %
✓ 1
Allow ECF from (b)(i). Accept 6–7 %.
Part (b)(iv)
the power radiated is proportional to T⁴, so radiation losses from the receiver would rise very steeply «e.g. a 20 % increase in T roughly doubles them» and a smaller fraction of the absorbed power would be stored
✓ 1
OWTTE. Accept other sensible physics, e.g. greater losses from the hot tank.
Part (c)
thermal energy is transferred only from a body at a higher temperature to one at a lower temperature «the resultant transfer is set by the temperature difference»
✓ 1
salt colder than 240 °C would not transfer energy to water at 240 °C «energy would flow from the water to the salt», so the salt cannot be cooled below the temperature of the incoming water
✓ 1
OWTTE.
Answers: (a)(i) 4.125 × 105 J · (a)(ii) 2.3 × 107 kg · (a)(iii) 1.2 × 104 m³ · (b)(i) 3.6 × 107 W · (b)(ii) 3.3 × 10−6 m · (b)(iii) 6.5 % (the remaining parts are explanations — see the table above)
128B-2-56Latent heat & phase changePaper 2Medium11 marks
Some vaccines must be kept below −60 °C during transport. They are packed in a box with polyurethane foam walls 50 mm thick, together with blocks of solid carbon dioxide (dry ice). At atmospheric pressure dry ice changes directly from solid to gas (sublimes) at −78 °C. The figure shows a cross-section of the box.
Total area of the walls and lid, measured at the middle of their thickness = 0.54 m². Thermal conductivity of the foam = 0.025 W m−1 K−1. Specific latent heat of sublimation of carbon dioxide = 5.7 × 105 J kg−1.
Diagram NOT accurately drawn
(a)
Outline how the internal energy of the carbon dioxide changes as it sublimes, and why its temperature stays at −78 °C.
(2)
(b)
During a journey the outside of the box is at 30 °C. Assume that the inner surface of the foam is at −78 °C and that all the energy entering the box sublimes dry ice.
(i)
Show that thermal energy is conducted into the box at a rate of about 29 W.
(1)
(ii)
The box contains 6.0 kg of dry ice. Determine whether this is enough for a journey of 48 hours.
(3)
(iii)
Suggest one change to the design of the box, other than adding more dry ice, that would make the dry ice last longer.
(1)
(c)
(i)
The carbon dioxide gas leaves through a vent. At 20 °C the density of carbon dioxide gas is 1.84 kg m−3. Calculate the volume of gas, measured at 20 °C, that leaves the box each hour.
(1)
(ii)
Suggest why the box must not be completely sealed.
(1)
(d)
The dry ice is placed above the vials, as shown in the figure, rather than below them. Explain, with reference to convection, why this keeps all the vials cold.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the internal energy increases because the intermolecular potential energy increases «molecules are separated against the attractive forces between them»
✓ 1
the average random kinetic energy of the molecules stays the same, so the temperature does not change «until all the solid has sublimed»
✓ 1
OWTTE.
Part (b)(i)
ΔQ/Δt = 0.025 × 0.54 × 108/0.050 = 29.2 W
✓ 1
Must see full substitution, including ΔT = 108 K, OR answer to at least 3 s.f.
Part (b)(ii)
rate of sublimation = 29.2/5.7 × 105 = 5.1 × 10−5 kg s−1
✓ 1
mass needed for 48 h = 5.12 × 10−5 × 48 × 3600 = 8.8 kg
✓ 1
8.8 kg is more than 6.0 kg, so the dry ice is not enough
✓ 1
MP3 requires a comparison of calculated values. ALTERNATIVE: time for 6.0 kg = 6.0 × 5.7 × 105/29.2 = 1.2 × 105 s ✓ = 33 h ✓, less than 48 h so not enough ✓. Allow ECF from (b)(i).
Part (b)(iii)
thicker foam walls OR a material of lower thermal conductivity «e.g. vacuum panels» OR a box with smaller surface area «more compact shape» OR a reflective/white outer surface
✓ 1
Any one.
Part (c)(i)
mass per hour = 5.12 × 10−5 × 3600 = 0.18 kg AND V = 0.184/1.84 = 0.10 m³
✓ 1
Allow ECF from (b)(ii).
Part (c)(ii)
the gas produced would build up and increase the pressure inside the box, which could burst it
✓ 1
OWTTE.
Part (d)
gas/air next to the dry ice is cooled, becomes denser and sinks, flowing down over the vials
✓ 1
warmer, less dense gas rises to the dry ice to be cooled, so a convection current circulates through the whole box; with the dry ice at the bottom the cold dense gas would stay there and the upper vials would warm
✓ 1
OWTTE.
Answers: (b)(i) 29.2 W · (b)(ii) 8.8 kg needed · (c)(i) 0.10 m³ (the remaining parts are explanations — see the table above)
A scuba diver wears a wetsuit made of neoprene, a rubber foam that contains many small closed bubbles of gas. At the surface the suit is 6.0 mm thick, has a thermal conductivity of 0.054 W m−1 K−1 and covers an area of 1.9 m². The inner surface of the suit is at 30 °C and the outer surface is at the temperature of the sea, 14 °C.
Thermal conductivity of water = 0.60 W m−1 K−1. Thermal conductivity of air = 0.026 W m−1 K−1.
(a)
A person in water at 14 °C loses energy much faster than a person in air at 14 °C. Explain, in terms of particles, why water conducts energy so much better than air.
(2)
(b)
(i)
Show that energy is conducted through the suit at a rate of about 270 W.
(1)
(ii)
The diver’s body produces thermal energy at a rate of 210 W. The diver has a mass of 78 kg and the mean specific heat capacity of the body is 3500 J kg−1 K−1. Calculate the time taken for the mean temperature of the diver’s body to fall by 1.0 K.
(2)
(c)
At a depth of 20 m the pressure of the water is about three times the pressure at the surface.
(i)
Outline why the suit is thinner at this depth than at the surface.
(1)
(ii)
At this depth the suit is 3.0 mm thick and its thermal conductivity is 0.070 W m−1 K−1. While swimming, the diver’s body produces thermal energy at 350 W. Assume that the surface temperatures of the suit are unchanged. Determine whether the diver’s body temperature can stay constant.
(2)
(d)
A thin layer of water is trapped between the skin and the suit. Explain why a tight-fitting suit keeps the diver warmer than a loose-fitting one.
(2)
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Part (a)
the molecules in water are much closer together «about 10 times closer» than those in air
✓ 1
so energy is passed from more energetic to less energetic molecules by collisions/interactions much more frequently, transferring kinetic energy away from the skin faster
✓ 1
OWTTE. Do not accept “water is colder”.
Part (b)(i)
ΔQ/Δt = 0.054 × 1.9 × 16/6.0 × 10−3 = 274 W
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
net rate of loss = 274 − 210 = 63.6 W
✓ 1
Allow ECF from (b)(i).
t = mcΔT/P = 78 × 3500 × 1.0/63.6 = 4.3 × 103 s «≈ 72 min»
✓ 1
Award [2] for CNA. Accept 4.2 × 103–4.6 × 103 s. Award [1] max for 1.0 × 103 s «210 W not subtracted» or 1.3 × 103 s «210 W used as the rate of loss».
Part (c)(i)
the greater pressure compresses the gas bubbles «their volume falls as the pressure rises at constant temperature», so the foam becomes thinner
✓ 1
OWTTE.
Part (c)(ii)
new rate = 0.070 × 1.9 × 16/3.0 × 10−3 = 710 W
✓ 1
710 W is greater than 350 W, so the body loses energy faster than it produces it and its temperature falls: it cannot stay constant
✓ 1
MP2 needs a comparison and a consistent conclusion. Allow ECF.
Part (d)
in a tight suit the thin layer of trapped water cannot flow away, so it is warmed by the skin «to close to skin temperature» and stays in place
✓ 1
in a loose suit cold sea water flows in and out «convection/flushing», continually replacing the warmed water, so the temperature difference at the skin stays large and more energy is lost
✓ 1
OWTTE.
Answers: (b)(i) 274 W · (b)(ii) 4.3 × 103 s · (c)(ii) 710 W (the remaining parts are explanations — see the table above)
130B-2-61Density & the molecular modelPaper 2Medium13 marks
The envelope of a hot-air balloon has a volume of 2800 m³ and is open at the bottom, so the air inside is at the same pressure as the air outside. On a day when the outside air is at 15 °C and has a density of 1.225 kg m−3, a burner heats the air in the envelope to 100 °C.
At constant pressure, the density of air is inversely proportional to its kelvin temperature.
Diagram NOT accurately drawn
(a)
(i)
Show that the density of the air in the envelope is about 0.95 kg m−3.
(1)
(ii)
Explain, in terms of molecules, why heating the air in the open envelope reduces its density.
(2)
(b)
(i)
The balloon floats at a constant height. Draw, on the dot in the figure, the forces acting on the whole balloon, including the hot air inside it. Ignore any motion of the surrounding air.
(2)
(ii)
Determine the maximum total mass of the envelope, basket, burner, fuel and passengers.
(2)
(c)
To keep the air at 100 °C, energy must be supplied to it at a rate of 0.45 MW to replace the energy transferred to the surroundings. The fuel, propane, has an energy density of 4.6 × 107 J kg−1, and 85 % of the energy released by burning it is transferred to the air in the envelope.
(i)
Calculate the mass of propane burned in one hour.
(2)
(ii)
Outline two mechanisms by which energy is transferred from the hot air, through the envelope, to the surroundings.
(2)
(d)
On a hot day the outside air is at 35 °C. Determine the maximum total mass in (b)(ii) when the air in the envelope is again at 100 °C.
(2)
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Part (a)(i)
ρ = 1.225 × 288/373 = 0.946 kg m−3
✓ 1
Must see kelvin temperatures in the ratio OR answer to at least 3 s.f.
Part (a)(ii)
the molecules gain kinetic energy «move faster» and the air expands «at the same pressure» as the molecules move further apart
✓ 1
some air leaves through the opening, so there are fewer molecules «less mass» in the same volume
✓ 1
OWTTE.
Part (b)(i)
downward arrow labelled weight/gravitational force AND upward arrow labelled buoyancy/upthrust
✓ 1
Both arrows must start on the dot.
the two arrows of equal length
✓ 1
Award [1] max if any extra force is drawn.
Part (b)(ii)
buoyancy = weight: ρoutsideVg = (M + ρinsideV)g
✓ 1
M = (1.225 − 0.946) × 2800 = 780 kg
✓ 1
Award [2] for CNA. Award [1] max for 3.4 × 103 kg «mass of hot air not subtracted».
Part (c)(i)
energy needed per hour = 0.45 × 106 × 3600 = 1.62 × 109 J AND energy from the fuel = 1.62 × 109/0.85
✓ 1
mass = 1.91 × 109/4.6 × 107 = 41 kg
✓ 1
Award [2] for CNA. Award [1] max for 35 kg «efficiency omitted».
Part (c)(ii)
a first mechanism named and outlined, e.g. conduction: molecules of the hot air collide with the inner surface and the energy is passed through the fabric by particle interactions
✓ 1
Names alone score [1] max.
a second, different mechanism outlined, e.g. convection: outside air warmed by the envelope becomes less dense and rises, carrying energy away; OR radiation: the warm envelope emits infrared radiation
✓ 1
Part (d)
outside density = 1.225 × 288/308 = 1.145 kg m−3
✓ 1
M = (1.145 − 0.946) × 2800 = 560 kg «less than on the cool day»
✓ 1
Allow ECF from (a)(i). Award [2] for CNA.
Answers: (a)(i) 0.946 kg m−3 · (b)(ii) 780 kg · (c)(i) 41 kg · (d) 560 kg (the remaining parts are explanations — see the table above)
131IBP2S-009-01-12Method of mixturesPaper 2Medium6 marks
Hospital autoclaves sterilise surgical instruments using steam at 100 °C. A technician's hand is accidentally exposed to 2.0 g of this steam, which condenses on the skin. The resulting water cools to skin temperature, 37 °C. Specific latent heat of vaporisation of water = 2.26 × 106 J kg−1 Specific heat capacity of water = 4180 J kg−1 K−1
(a)
Explain why steam at 100 °C causes a more severe burn than the same mass of water at 100 °C.
(2)
(b)
Calculate the total thermal energy transferred to the skin by the 2.0 g of steam.
(3)
(c)
Determine the ratio \(\frac{\text{energy transferred by 2.0 g of steam}}{\text{energy transferred by 2.0 g of water cooling from 100 °C to 37 °C}}\).
(1)
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Part (a)
As steam condenses on the skin it releases its latent heat of vaporisation at 100 °C, in addition to the energy released when the condensed water then cools . The latent heat is much larger than the energy released by water cooling from 100 °C to 37 °C, so far more energy is transferred to the skin .
5047 / 527 = 9.6 . Allow error carried forward from the previous part.
✓ 1
Common error: Remember the cooling stage after condensation. Leaving out mcΔT for the condensed water loses a mark in the total-energy calculation and changes the ratio.
132IBP2S-009-01-14Mechanisms of thermal transferPaper 2Medium6 marks
A crewed space station must get rid of 2.5 kW of waste thermal energy from its electronics. It does this with a flat radiator panel that emits from one face only. The panel's surface is at 290 K and has an emissivity of 0.85. σ = 5.67 × 10−8 W m−2 K−4; Wien constant = 2.9 × 10−3 m K
(a)
State why the radiator must lose energy by radiation rather than by convection.
(1)
(b)
Calculate the peak wavelength of the radiation emitted by the panel.
(1)
(c)
Calculate the area of panel needed, ignoring any radiation the panel absorbs.
(2)
(d)
When the station moves into sunlight, the panel absorbs solar radiation. Outline how this affects the area of panel needed to keep it at 290 K.
(2)
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Part (a)
Space is a vacuum, so there is no fluid to carry energy away by convection .
133IBP2S-009-01-15Internal energy & the molecular modelPaper 2Medium7 marks
Liquid nitrogen is used in dermatology to freeze skin lesions. It is stored in a flask at its boiling point of 77 K. Specific latent heat of vaporisation of nitrogen = 1.99 × 105 J kg−1 Molar mass of nitrogen = 0.028 kg mol−1 NA = 6.02 × 1023 mol−1; kB = 1.38 × 10−23 J K−1
(a)
Distinguish between the temperature and the internal energy of the liquid nitrogen.
(2)
(b)
Determine the energy needed to vaporise one molecule of nitrogen.
(2)
(c)
Calculate the average translational kinetic energy of a nitrogen molecule at 77 K.
(1)
(d)
The average translational kinetic energy of a molecule at 77 K is much smaller than the energy needed to vaporise one molecule. Explain why the liquid nitrogen still evaporates continuously from its flask.
(2)
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Part (a)
Temperature is a measure of the average random kinetic energy of the molecules . Internal energy is the total of the random kinetic and intermolecular potential energies of all the molecules .
✓ 2
Part (b)
Mass of one molecule = 0.028 / 6.02 × 1023 = 4.65 × 10−26 kg .
The molecules have a distribution of kinetic energies, and a small fraction of molecules near the surface have enough energy to escape . Energy absorbed from the surroundings replaces the energy carried away by the escaping molecules, so more molecules keep reaching the escape energy .
✓ 2
Common error: Temperature and internal energy are not the same quantity. A large mass of cold liquid can have more internal energy than a small mass of hot liquid.
134IBP2S-009-04-13Specific heat capacityPaper 2Medium7 marks
A 0.40 kg block of a new alloy for a satellite component is heated by a 24.0 W electric heater. The block is well insulated. The table shows its temperature θ at time t.
135IBP2S-009-04-14Temperature & direction of transferPaper 2Medium8 marks
A rack of servers in a data centre dissipates 12 kW. The rack is cooled by water flowing through cold plates attached to the processors. The water enters at 18 °C and leaves at 30 °C. Specific heat capacity of water = 4200 J kg−1 K−1.
(a)
Explain, in terms of particles, why thermal energy transfers from the processors to the water.
(2)
(b)
Calculate the mass flow rate of water required.
(2)
(c)
The pump fails, and 25 kg of water is trapped in the loop at 30 °C. The servers continue to dissipate 12 kW into this water. Assume that all of this energy goes into the water. Estimate the time taken for the water to reach 60 °C.
(2)
(d)
A coolant with a smaller specific heat capacity is used instead of water, with the same inlet and outlet temperatures. State and explain the effect on the mass flow rate required.
(2)
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Notes
Part (a)
The processors are at a higher temperature, so their particles have a greater mean random kinetic energy .
✓ 1
Collisions at the boundary transfer energy, on average, from the faster to the slower particles, giving a net transfer to the water .
✓ 1
Part (b)
Mass flow rate = P/(cΔT) = 12000/(4200 × 12)
✓ 1
= 0.24 kg s−1
✓ 1
Part (c)
t = mcΔT/P = 25 × 4200 × 30/12000
✓ 1
t ≈ 260 s (accept any value from 2.6×102 s to 263 s)
✓ 1
Part (d)
The mass flow rate must increase .
✓ 1
Each kilogram absorbs less energy for the same 12 K rise, and mass flow rate = P/(cΔT) .
✓ 1
Exam tip: For flowing fluids, divide Q = mcΔT by time to get P = (Δm/Δt)cΔT. The rate of mass flow then takes the place of the mass.
136IBP2S-009-04-15Method of mixturesPaper 2Medium6 marks
A blacksmith quenches a 0.80 kg steel tool at 850 °C in a tank containing 6.0 kg of oil at 25 °C. Assume no oil evaporates. Specific heat capacity of steel = 490 J kg−1 K−1 Specific heat capacity of oil = 1.9×103 J kg−1 K−1
(a)
State the condition at which the net transfer of thermal energy between the tool and the oil stops.
(1)
(b)
Determine the final temperature, assuming no energy is lost to the surroundings.
(3)
(c)
The measured final temperature is 49 °C. Calculate the energy transferred to the surroundings.
(2)
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Part (a)
The tool and the oil reach the same temperature (thermal equilibrium) .
✓ 1
Part (b)
Energy lost by steel = energy gained by oil:
—
0.80 × 490 × (850 − T) = 6.0 × 1900 × (T − 25)
✓ 1
333200 − 392T = 11400T − 285000
✓ 1
T ≈ 52 °C
✓ 1
Part (c)
Energy lost by steel − energy gained by oil = 392 × (850 − 49) − 11400 × (49 − 25)
✓ 1
= 3.14×105 − 2.74×105 ≈ 4.0×104 J
✓ 1
Common error: Write each temperature change as a positive quantity, for example (850 − T) for the steel and (T − 25) for the oil. Mixing up the signs gives a final temperature outside the starting range, which is impossible.
137IBP2S-009-00-13Luminosity & apparent brightnessPaper 2Medium6 marks
A solar-powered probe is sent to study an asteroid orbiting 2.8 AU from the Sun. Luminosity of the Sun = 3.8×1026 W; 1 AU = 1.5×1011 m.
(a)
Define apparent brightness.
(1)
(b)
Calculate the apparent brightness of the Sun at the asteroid.
(2)
(c)
The probe needs 500 W of electrical power. Its solar panels convert 28% of the incident solar power into electrical power. Determine the minimum area of the panels.
(2)
(d)
State one assumption made in your answer to (c).
(1)
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Part (a)
The power received per unit area, measured perpendicular to the direction of the radiation, at the observer
✓ 1
Part (b)
d = 2.8 × 1.5×1011 = 4.2×1011 m; b = L/(4πd2) = 3.8×1026/(4π(4.2×1011)2)
✓ 1
b = 171 W m−2 (accept 170 W m−2)
✓ 1
Part (c)
Allow error carried forward from (b). Electrical power per m2 = 0.28 × 171 = 48 W m−2
✓ 1
A = 500/48 = 10.4 m2 (accept 10 m2)
✓ 1
Part (d)
Any one: panels are perpendicular to the Sun's rays / no absorption or scattering of radiation between the Sun and the probe / the Sun radiates equally in all directions / the efficiency remains 28%
✓ 1
Exam tip: Convert AU to metres before squaring; an error in d is magnified by the inverse-square law.
An outdoor-clothing company tests a synthetic insulating fabric for mountaineering jackets. Samples of area 0.050 m2 but different thicknesses are placed between a heater and a cold plate. The temperature difference across each sample is kept at 20 K and the steady power P conducted through it is measured.
thickness x / mm: 2.0 | 4.0 | 6.0 | 8.0 power P / W: 15.0 | 7.4 | 5.1 | 3.7
Diagram not to scale. The thickness x of the sample is varied.
(a)
Show that the data are consistent with P being inversely proportional to x.
(2)
(b)
Determine the thermal conductivity of the fabric.
(3)
(c)
The fabric contains many tiny pockets of trapped air. Explain why this makes it a good insulator.
(2)
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Part (a)
Calculates P × x for (at least three) samples: 30, 29.6, 30.6, 29.6 W mm
✓ 1
Products are (approximately) constant, within about 2%, so P ∝ 1/x
✓ 1
Part (b)
Mean value of P × x ≈ 30 W mm = 0.030 W m
✓ 1
k = Px/(AΔT) = 0.030/(0.050 × 20)
✓ 1
k = 0.030 W m−1 K−1
✓ 1
Part (c)
Air (a gas) is a poor conductor because its particles are far apart so collisions transferring energy are infrequent
✓ 1
Trapping the air in small pockets prevents convection currents from forming
✓ 1
Exam tip: Using a mean of all data pairs (or the gradient of P against 1/x) gives a more reliable value than a single row of the table.
139IBP2S-009-03-13Density & the molecular modelPaper 2Medium6 marks
An engineer is designing a domestic central-heating system that contains 120 kg of water. The table gives the density of water at several temperatures.
Temperature / °C : Density / kg m−3 0 : 999.8 4 : 1000.0 10 : 999.7 20 : 998.2 40 : 992.2 60 : 983.2 80 : 971.8
(a)
State the temperature, from the data, at which water has its maximum density.
(1)
(b)
Outline, using the molecular model, why the density of water decreases between 20 °C and 80 °C.
(2)
(c)
The water is filled at 20 °C and heated to 80 °C. Determine the increase in volume of the water, which must be absorbed by an expansion vessel.
(3)
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Part (a)
4 °C
✓ 1
Part (b)
At higher temperature the molecules have greater average kinetic energy and move or vibrate more vigorously
✓ 1
The mean separation of the molecules increases, so the same mass occupies a larger volume and the density falls
140IBP2S-009-03-14Internal energy & the molecular modelPaper 2Medium9 marks
The gap in a double-glazed window is filled with argon, which reduces energy transfer through the window. The gap measures 1.2 m × 0.80 m × 0.016 m. The argon is at 20 °C and has a density of 1.66 kg m−3. The molar mass of argon is 0.040 kg mol−1. Use NA = 6.02 × 1023 mol−1 and kB = 1.38 × 10−23 J K−1.
(a)
Calculate the mass of argon in the gap.
(2)
(b)
Show that the number of argon atoms in the gap is about 3.8 × 1023.
(2)
(c)
Calculate the average translational kinetic energy of an argon atom in the gap.
(2)
(d)
Treat the argon as an ideal gas. Determine the internal energy of the argon in the gap.
(3)
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Part (a)
V = 1.2 × 0.80 × 0.016 = 0.01536 m3
✓ 1
m = ρV = 1.66 × 0.01536 = 0.0255 kg
✓ 1
Part (b)
n = m/M = 0.0255/0.040 = 0.637 mol
✓ 1
N = nNA = 0.637 × 6.02×1023 = 3.84×1023
✓ 1
Part (c)
T = 20 + 273 = 293 K
✓ 1
Ek = 1.5 × 1.38×10−23 × 293 = 6.07×10−21 J
✓ 1
Part (d)
In an ideal gas there are no intermolecular forces, so the potential energy is zero and the internal energy is the total random kinetic energy
✓ 1
U = N × Ek = 3.84×1023 × 6.07×10−21 — ECF from (b) and (c)
✓ 1
U = 2.3×103 J
✓ 1
Exam tip: In a 'show that' part, give the answer to one more significant figure than the value stated. Carry the unrounded value into later parts.
141IBP2S-009-02-14Mechanisms of thermal transferPaper 2Medium6 marks
A communications satellite must reject 2.4 kW of waste thermal energy from its electronics. It uses a flat radiator panel that radiates from one face only. The panel's surface temperature is 290 K and its emissivity is 0.85. Radiation absorbed from space can be neglected. σ = 5.67×10−8 W m−2 K−4.
(a)
Outline why the satellite must rely on radiation to remove the waste energy.
(2)
(b)
Calculate the area of the radiator panel required.
(2)
(c)
The panel's temperature rises to 300 K while its area stays the same. Determine the percentage increase in the power radiated.
(2)
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Notes
Part (a)
Space is a near vacuum, so there is no medium for conduction or convection to the surroundings ; electromagnetic radiation needs no medium to travel
✓ 2
Part (b)
A = P/(eσT4) = 2400/(0.85 × 5.67×10−8 × 2904) ; A = 7.0 m2
✓ 2
Part (c)
Power ratio = (300/290)4 = 1.145 ; increase = 14.5% (accept 15%)
✓ 2
Exam tip: In the final part the area and emissivity cancel, so work with the temperature ratio rather than recalculating the whole power.
142IBP2S-009-02-15Method of mixturesPaper 2Medium7 marks
A student estimates the specific latent heat of fusion of ice. She adds ice, which she assumes to be at 0 °C, to warm water in a polystyrene cup. She stirs until all the ice has melted. Her results are shown below.
Mass of empty cup
0.012 kg
Mass of cup + warm water
0.262 kg
Mass of cup + water after ice has melted
0.302 kg
Initial temperature of warm water
22.0 °C
Final temperature of mixture
8.0 °C
Specific heat capacity of water = 4200 J kg−1 K−1. The heat capacity of the cup is negligible.
(a)
Calculate the energy transferred from the warm water as it cools.
(2)
(b)
Determine the specific latent heat of fusion of ice from these data, assuming the ice was at 0 °C.
(3)
(c)
The ice was actually taken straight from a freezer at −18 °C. Explain the effect this has on the value of L obtained in (b).
(2)
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Part (a)
Mass of warm water = 0.262 − 0.012 = 0.250 kg ; Q = 0.250 × 4200 × 14.0 = 1.47×104 J
✓ 2
Part (b)
Mass of ice = 0.302 − 0.262 = 0.040 kg ; energy to warm the melted ice from 0 °C to 8.0 °C = 0.040 × 4200 × 8.0 = 1344 J ; L = (14700 − 1344)/0.040 = 3.3×105 J kg−1 . Allow ECF from (a).
✓ 3
Part (c)
Some of the energy from the water was used to warm the ice from −18 °C to 0 °C, but the calculation attributes all of that energy to melting ; so the calculated value of L is larger than the true value
✓ 2
Common error: Once the ice has melted, its water must also warm to the final temperature. Leaving out this term overestimates L.
143IBP2S-009-02-16Stefan–Boltzmann & WienPaper 2Medium6 marks
A 60 W incandescent lamp in a vintage-style light fitting has a tungsten filament at 2800 K. The emissivity of tungsten at this temperature is 0.35. Assume all the electrical power supplied is radiated by the filament. Wien's constant b = 2.9×10−3 m K; σ = 5.67×10−8 W m−2 K−4.
(a)
Calculate the peak wavelength of the radiation emitted by the filament.
(2)
(b)
Explain, with reference to your answer to (a), why incandescent lamps are inefficient sources of visible light.
(2)
(c)
Calculate the surface area of the filament.
(2)
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Part (a)
λmax = 2.9×10−3/2800 ; λmax = 1.0×10−6 m
✓ 2
Part (b)
The peak lies in the infrared, beyond the visible range of about 400–700 nm ; so most of the radiated energy is infrared rather than visible light . ECF from (a).
✓ 2
Part (c)
A = P/(eσT4) = 60/(0.35 × 5.67×10−8 × 28004) ; A = 4.9×10−5 m2
✓ 2
Exam tip: Quote the visible range when explaining an answer from Wien's law. Saying only that the peak is 'in the infrared' may not earn both marks.
144B-2-04Luminosity & apparent brightnessPaper 2Hard13 marks
A space probe moves around the Sun in a circular orbit of radius 0.40 AU. It carries a flat sunshade that is always perpendicular to the Sun's rays.
Luminosity of the Sun = 3.85 × 1026 W. 1 AU = 1.50 × 1011 m. The Earth orbits the Sun at 1.00 AU with a period of 365 days.
Axes for (b)(iii)
(a)
(i)
Show that the intensity of the Sun's radiation at a distance of 1.00 AU from the Sun is about 1.4 kW m−2.
(1)
(ii)
Calculate the intensity of the Sun's radiation at the probe.
(1)
(b)
The sunshade absorbs all the radiation incident on its front face. It emits as a black body from both its front face and its back face.
(i)
Show that the equilibrium temperature of the sunshade is about 520 K.
(2)
(ii)
Deduce the equilibrium temperature that the same sunshade would have at a distance of 1.00 AU from the Sun.
(1)
(iii)
Sketch, on the axes, a graph to show how the equilibrium temperature T of the sunshade varies with its distance r from the Sun, for r from 0.20 AU to 1.00 AU.
(2)
(c)
The front face of the sunshade is given a coating that reflects 70 % of the incident solar radiation. Both faces still emit as black bodies at infrared wavelengths. Determine the new equilibrium temperature of the sunshade.
(3)
(d)
(i)
Determine, using Kepler's third law, the orbital period of the probe in days.
(2)
(ii)
Explain why the orbital period of the probe does not depend on its mass.
(1)
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Part (a)(i)
b = L/4πd2 = 3.85 × 1026 / (4π × (1.50 × 1011)2) = 1.36 × 103 W m−2
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Accept 380–390 K. Award [3] for CNA. Award [2] max for 479 K (0.70 used as the absorbed fraction).
Part (d)(i)
T2 ∝ r3 so T = 365 × 0.403/2
✓ 1
T = 92 days
✓ 1
Accept 92–93 days. Award [2] for CNA.
Part (d)(ii)
the gravitational force on the probe and the centripetal force it needs (mv2/r) are both proportional to its mass, so the mass cancels
✓ 1
OWTTE.
Answers: (a)(i) 1.36 × 103 W m−2 · (a)(ii) 8.5 × 103 W m−2 · (b)(i) 523 K · (b)(ii) 331 K · (c) 387 K · (d)(i) 92 days (the remaining parts are explanations — see the table above)
145B-2-14Mechanisms of thermal transferPaper 2Hard14 marks
A steel panel radiator is filled with hot water at 70 °C. It transfers thermal energy to a room at a total rate of 1.6 kW. The outer surface of the radiator has a total area of 1.8 m2 and an emissivity of 0.90. The steel wall of the radiator is 1.2 mm thick and the thermal conductivity of steel is 50 W m−1 K−1. The air and the walls of the room are at 20 °C.
Diagram NOT accurately drawn
(a)
(i)
Show that the temperature difference across the steel wall of the radiator is less than 0.1 K.
(1)
(ii)
Hence state the temperature of the outer surface of the radiator.
(1)
(b)
(i)
Calculate the power emitted as radiation by the outer surface of the radiator.
(2)
(ii)
The net rate of energy transfer by radiation from the radiator is given by eσA(Tr4 − Tw4), where Tr is the temperature of the radiator and Tw the temperature of the walls of the room. Explain why this is smaller than your answer to (b)(i).
(1)
(iii)
Determine whether more than half of the energy transferred from the radiator to the room is transferred by radiation.
(4)
(c)
(i)
Annotate the diagram to show the convection current that the radiator sets up in the room.
(2)
(ii)
Explain how the radiator sets up this convection current.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
ΔT = (ΔQ/Δt)Δx/(kA) = 1600 × 1.2 × 10−3 / (50 × 1.8) = 0.021 K, which is less than 0.1 K
✓ 1
Must see full substitution OR the value 0.021 K.
Part (a)(ii)
70 °C / 343 K (the same as the water, to within 0.1 K)
✓ 1
Part (b)(i)
P = eσAT4 = 0.90 × 5.67 × 10−8 × 1.8 × 3434
✓ 1
Kelvin temperature must be used.
= 1.27 × 103 W
✓ 1
Accept 1.27–1.28 × 103 W. Award [2] for CNA.
Part (b)(ii)
the radiator also absorbs radiation emitted by the walls of the room (at 293 K), which partly offsets what it emits
less than half, so no: the remaining ≈ 1.0 kW (about two-thirds) is transferred by convection (the name 'radiator' is misleading)
✓ 1
MP4 needs a conclusion consistent with the candidate's value.
Part (c)(i)
arrows rising from the radiator to the ceiling, then across the ceiling towards the far wall
✓ 1
arrows down the far side of the room and back along the floor to the radiator, forming a closed loop
✓ 1
Ignore small loops near the window.
Part (c)(ii)
air next to the radiator is heated and expands
✓ 1
so its density decreases and it rises (buoyancy)
✓ 1
cooler, denser air sinks elsewhere and flows in to replace it, so a continuous circulation is set up
✓ 1
OWTTE.
Answers: (a)(i) 0.021 K · (a)(ii) 70 °C · (b)(i) 1.27 × 103 W · (b)(iii) 5.9 × 102 W, 37 % — no (the remaining parts are explanations — see the table above)
146B-2-38Luminosity & apparent brightnessPaper 2Hard15 marks
R is a red dwarf star with a luminosity of 0.0040 L☉ and a surface temperature of 3100 K, where L☉ is the luminosity of the Sun. Treat R and the Sun as black bodies. The apparent brightness of R measured at the Earth is 3.4 × 10−12 W m−2.
L☉ = 3.83 × 1026 W; radius of the Sun R☉ = 6.96 × 108 m; surface temperature of the Sun = 5800 K; 1 ly = 9.46 × 1015 m; 1 AU = 1.50 × 1011 m.
The Hertzsprung–Russell (HR) diagram shows the main sequence and the position of the Sun.
HR diagram drawn to scale (logarithmic axes)
(a)
(i)
State what is meant by the luminosity of a star.
(1)
(ii)
Calculate the wavelength at which the emission spectrum of R has its peak.
(1)
(b)
Show that the radius of R is about 0.22 R☉.
(2)
(c)
Draw a cross on the HR diagram to show the position of R. Label the cross R.
(1)
(d)
Calculate, in light years, the distance of R from the Earth.
(2)
(e)
A planet P orbits R at the distance at which the intensity of the radiation from R is 1360 W m−2, the same as the solar constant at the Earth.
(i)
Calculate the radius of the orbit of P, in AU.
(2)
(ii)
A student claims that, because R is much cooler than the Sun, P must have a lower equilibrium temperature than the Earth would have with the same albedo and no atmosphere. The albedo of P and of the Earth is 0.30; assume that both radiate as black bodies. Evaluate this claim.
(3)
(f)
Outline why plants on P would receive much less visible light than plants on the Earth, even though P receives the same total intensity.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
the total power emitted «as electromagnetic radiation» by the star
✓ 1
Do not accept ‘energy’ or ‘brightness’ without reference to power/rate.
Part (a)(ii)
λmax = 2.9 × 10−3 / 3100 = 9.35 × 10−7 m
✓ 1
Accept 9.4 × 10−7 m or 935 nm.
Part (b)
L = 0.0040 × 3.83 × 1026 = 1.53 × 1024 W
✓ 1
radius of R = √(L/4πσT4) = 1.53 × 108 m = 0.219 R☉
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (c)
cross at T = 3100 K and L = 4 × 10−3L☉, labelled R «on the lower main sequence»
✓ 1
Accept a position within half a grid interval of the correct point on both axes.
Part (d)
d = √(L/4πb) = √(1.53 × 1024 / (4π × 3.4 × 10−12)) = 1.9 × 1017 m
✓ 1
= 20 ly
✓ 1
Award [2] for CNA. Accept 19–21 ly. Allow ECF from (b) for L.
Part (e)(i)
d = √(L/(4π × 1360)) = 9.5 × 109 m
✓ 1
= 0.063 AU
✓ 1
Award [2] for CNA. ALTERNATIVE: d = √0.0040 AU because the intensity at 1 AU from the Sun is 1360 W m−2; award [2].
Part (e)(ii)
the equilibrium temperature depends on the intensity absorbed, (1 − α)S/4, not on the temperature of the star
✓ 1
for both: T = (0.70 × 1360 / 4σ)1/4 = 255 K
✓ 1
Accept a statement that the temperatures are equal because intensity and albedo are the same, with a calculated value.
so the claim is incorrect: P has the same equilibrium temperature as the Earth would have
✓ 1
MP3 only scores if MP1 or MP2 scores.
Part (f)
the peak of the spectrum of R is in the infrared «935 nm», whereas that of the Sun «500 nm» is in the visible
✓ 1
so a much smaller fraction of the power emitted by R is at visible wavelengths «400–700 nm»
✓ 1
the total intensity at P is the same as at the Earth, so the intensity of visible light at P is much smaller «less energy available for photosynthesis»
✓ 1
OWTTE
Answers: (a)(ii) 9.35 × 10−7 m · (b) 0.219 R☉ · (d) 20 ly · (e)(i) 0.063 AU · (e)(ii) 255 K for both (the remaining parts are explanations — see the table above)
147B-2-57Stefan–Boltzmann & WienPaper 2Hard13 marks
A filament lamp is rated 230 V, 60 W. In normal use its tungsten filament is at a temperature of 2700 K. In a simple model, the filament is a uniform cylinder of diameter d and length l that radiates as a black body from its curved surface, and all the electrical power supplied is radiated.
Resistivity of tungsten at 2700 K = 7.9 × 10−7 Ω m.
(a)
(i)
Calculate the wavelength at which the radiation from the filament is most intense.
(1)
(ii)
Hence explain why a filament lamp is an inefficient source of visible light.
(2)
(b)
Show that the surface area of the filament is about 2.0 × 10−5 m².
(1)
(c)
(i)
Show that the resistance of the filament at its working temperature is about 880 Ω.
(1)
(ii)
Determine d and l.
(3)
(d)
The filament of a halogen lamp works at 3200 K.
(i)
Calculate (power radiated per unit area of the halogen filament) / (power radiated per unit area of the filament at 2700 K).
(1)
(ii)
Suggest why the halogen lamp is a more efficient source of visible light.
(2)
(e)
When the lamp is switched on, the filament is at room temperature and its resistance is about one-fifteenth of its working value. Explain why filament lamps most often fail at the moment they are switched on.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
λmax = 2.9 × 10−3/2700 = 1.1 × 10−6 m
✓ 1
Accept 1.07 μm.
Part (a)(ii)
λmax is in the infrared «longer than about 700 nm»
✓ 1
so most of the power is radiated at infrared wavelengths and only a small fraction is visible light
✓ 1
MP2 only scores if MP1 scores. OWTTE.
Part (b)
A = P/σT⁴ = 60/(5.67 × 10−8 × 2700⁴) = 1.99 × 10−5 m²
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (c)(i)
R = V²/P = 230²/60 = 882 Ω
✓ 1
Must see the substitution.
Part (c)(ii)
A = πdl AND R = ρl/(πd²/4) = 4ρl/πd²
✓ 1
Both equations needed.
eliminating l: d³ = 4ρA/(π²R) so d = 1.9 × 10−5 m
✓ 1
Allow ECF from (b) and (c)(i).
l = A/πd = 0.33 m
✓ 1
Allow ECF from MP2. Award [3] for CNA for both values.
Part (d)(i)
(3200/2700)⁴ = 1.97
✓ 1
Accept 2.0.
Part (d)(ii)
λmax is shorter «9.1 × 10−7 m», closer to the visible region
✓ 1
so a greater fraction of the radiated power is emitted at visible wavelengths
✓ 1
OWTTE.
Part (e)
the initial current is about 15 times the working current, so the initial power is far greater «≈ 15 × 60 W»
✓ 1
the filament heats up extremely rapidly, so the thinnest part overheats/is under great thermal stress and breaks
✓ 1
OWTTE.
Answers: (a)(i) 1.1 × 10−6 m · (b) 1.99 × 10−5 m² · (c)(i) 882 Ω · (c)(ii) d = 1.9 × 10−5 m, l = 0.33 m · (d)(i) 1.97 (the remaining parts are explanations — see the table above)
148B-2-59Latent heat & phase changePaper 2Hard12 marks
A snow cannon sprays fine droplets of water into cold, dry air. Water enters the cannon at 2.0 °C. As the droplets fly through the air, part of the water evaporates; the rest freezes, and the ice crystals formed cool to −4.0 °C before they land as snow.
Specific heat capacity of water = 4.18 × 103 J kg−1 K−1; specific heat capacity of ice = 2.10 × 103 J kg−1 K−1; specific latent heat of fusion of ice = 3.34 × 105 J kg−1; specific latent heat of vaporization of water near 0 °C = 2.50 × 106 J kg−1.
Axes for (c)
(a)
Describe the changes in the arrangement and the motion of the water molecules when a droplet freezes.
(2)
(b)
(i)
Show that about 3.5 × 105 J must be removed from each kilogram of water that becomes snow at −4.0 °C.
(2)
(ii)
In a simple model, all of the energy removed from the water that becomes snow is used to evaporate the rest of the water. Determine the fraction of the water entering the cannon that evaporates.
(3)
(c)
Energy is removed from a droplet at a constant rate from the moment it leaves the cannon until it lands. Sketch, on the axes, a graph to show how the temperature of the droplet varies with time.
(3)
(d)
Suggest why a snow cannon produces snow more effectively in dry air than in humid air at the same temperature.
(1)
(e)
Water is supplied to the cannon at 12.0 kg s−1. Calculate the mass of snow produced in one hour.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
arrangement: from a disordered arrangement «close together» to a regular, ordered arrangement «lattice»
✓ 1
motion: from moving randomly past one another to vibrating about fixed positions
149B-2-60Luminosity & apparent brightnessPaper 2Hard12 marks
Star K has a luminosity of 0.30 L☉ and a surface temperature of 4900 K. Its apparent brightness at the Earth is 2.0 × 10−10 W m−2. Treat K as a black body.
L☉ = 3.83 × 1026 W; 1 ly = 9.46 × 1015 m.
Axes for (e)
(a)
State what is meant by the apparent brightness of a star.
(1)
(b)
Calculate, in light years, the distance of K from the Earth.
(2)
(c)
Show that the radius of K is about 5.3 × 108 m.
(2)
(d)
A planet orbits K. Each time the planet passes between K and the Earth, the apparent brightness of K falls by 1.2 %. The planet emits a negligible amount of visible light.
(i)
Explain why the fractional decrease in the apparent brightness is approximately (radius of planet / radius of K)².
(2)
(ii)
Determine the radius of the planet.
(1)
(e)
Each fall in brightness lasts 2.8 hours. Sketch, on the axes, a graph to show how b/b0 varies with time t over 12 hours that include one of these events, where b0 is the normal apparent brightness of K. Ignore the short times during which the planet is only partly in front of K.
(2)
(f)
The side of the planet facing K is at about 1300 K. Suggest, with a calculation, why astronomers try to detect the radiation emitted by the planet itself at infrared rather than visible wavelengths.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the power received per unit area «perpendicular to the direction of the light» at the observer
✓ 1
Do not accept “how bright the star appears” or “energy per unit area”.
Part (b)
L = 0.30 × 3.83 × 1026 = 1.15 × 1026 W AND d = √(L/4πb) = 2.1 × 1017 m
✓ 1
= 23 ly
✓ 1
Award [2] for CNA. Allow ECF for L.
Part (c)
L = 4πR²σT⁴ so R = √(L/4πσT⁴)
✓ 1
R = √(1.15 × 1026/(4π × 5.67 × 10−8 × 4900⁴)) = 5.29 × 108 m
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (d)(i)
the planet blocks the light from an area πr² of the star’s disc, of total area πR²
✓ 1
each part of the disc sends the same power per unit area to the Earth «uniform disc», so the fraction of the light blocked is πr²/πR² = (r/R)²
✓ 1
OWTTE.
Part (d)(ii)
r = 5.3 × 108 × √0.012 = 5.8 × 107 m
✓ 1
Allow ECF from (c). Award [0] for 6.4 × 106 m «no square root».
Part (e)
horizontal line at b/b0 = 1.000 before and after a dip, with a flat-bottomed dip to 0.988
✓ 1
dip lasting 2.8 h, wholly within the 12 h shown
✓ 1
Award [1] max for a dip of correct depth but wrong duration.
Part (f)
λmax of the planet = 2.9 × 10−3/1300 = 2.2 × 10−6 m «infrared», while λmax of K = 5.9 × 10−7 m «visible»
✓ 1
the planet emits almost all its radiation in the infrared, where the star emits relatively less, so the planet’s emission is a much larger fraction of the total there «easier to distinguish from the starlight»
✓ 1
OWTTE.
Answers: (b) 23 ly · (c) 5.29 × 108 m · (d)(ii) 5.8 × 107 m (the remaining parts are explanations — see the table above)
An outdoor ice-skating venue uses a natural frozen lake. On a still winter night, the ice layer is 5.0 cm thick. The upper surface of the ice is at −10 °C, and the water under the ice is at 0 °C. Thermal conductivity of ice = 2.2 W m−1 K−1 Density of ice = 917 kg m−3 Specific latent heat of fusion of ice = 3.34 × 105 J kg−1 Liquid water has its maximum density at about 4 °C. In parts (b) to (d), assume the ice thickness stays at 5.0 cm throughout the hour.
Diagram not to scale
(a)
Explain why a lake freezes from the top downwards.
(2)
(b)
Calculate the rate of thermal energy conduction through 1.0 m2 of the ice.
(2)
(c)
Show that about 4.7 kg of water freezes beneath each square metre of ice in one hour.
(2)
(d)
Determine the increase in thickness of the ice in one hour, in mm.
(2)
(e)
Suggest why, in reality, the ice thickens more slowly as the night goes on, even if the surface temperature stays at −10 °C.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Water is densest at about 4 °C. Surface water that cools below 4 °C becomes less dense, so it stays at the top instead of sinking by convection .
✓ 1
Ice is less dense than water and floats, so freezing begins at the surface and the ice layer grows downwards .
✓ 1
Part (b)
ΔQ/Δt = kAΔT/Δx = 2.2 × 1.0 × 10 / 0.050 .
✓ 1
= 440 W .
✓ 1
Part (c)
Energy conducted in one hour = 440 × 3600 = 1.58 × 106 J .
✓ 1
m = 1.58 × 106 / 3.34 × 105 = 4.74 kg ≈ 4.7 kg .
✓ 1
ECF from (b).
—
Part (d)
Volume of ice per square metre = 4.74 / 917 = 5.17 × 10−3 m3, so the thickness increase = 5.17 × 10−3 m3 / 1.0 m2 = 5.17 × 10−3 m .
✓ 1
= 5.17 × 10−3 × 1000 ≈ 5.2 mm .
✓ 1
ECF from (c).
—
Part (e)
The ice gets thicker, so Δx increases and the conduction rate falls for the same temperature difference. Less energy is removed from the water each second, so less water freezes per second .
✓ 1
Exam tip: In freezing problems, the energy conducted away equals the latent heat released by the water that freezes. Link the two with mL = (ΔQ/Δt) × t.
151IBP2S-009-04-16Latent heat & phase changePaper 2Hard9 marks
A new battery coolant is tested in a continuous-flow apparatus. The liquid flows steadily through a tube past an electric heater. The heater power is adjusted so that the rise in temperature between inlet and outlet is 5.0 K. The inlet temperature is the same in both runs, and the room temperature stays constant throughout.
Run 1: mass flow rate 2.0×10−3 kg s−1, heater power 28.0 W Run 2: mass flow rate 3.5×10−3 kg s−1, heater power 46.0 W
Diagram not to scale
(a)
Explain why the rate of energy loss to the surroundings, h, can be taken as the same in both runs.
(2)
(b)
Show that the specific heat capacity of the coolant is about 2.4×103 J kg−1 K−1.
(3)
(c)
Determine h.
(2)
(d)
A student uses only run 1 and ignores energy losses. Calculate the percentage difference between the student's value of the specific heat capacity and the value in (b), expressed as a percentage of the value in (b).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The inlet and outlet temperatures are the same in both runs, so the apparatus is at the same temperature in both runs .
✓ 1
The room temperature is constant, so the temperature difference between the apparatus and the surroundings is unchanged, and the rate of energy loss is the same .
✓ 1
Part (b)
P1 = ṁ1cΔT + h and P2 = ṁ2cΔT + h
✓ 1
Subtracting eliminates h: P2 − P1 = (ṁ2 − ṁ1)cΔT
✓ 1
c = 18.0/(1.5×10−3 × 5.0) = 2.40×103 J kg−1 K−1
✓ 1
Part (c)
h = P1 − ṁ1cΔT = 28.0 − 2.0×10−3 × 2400 × 5.0 (or the same calculation using run 2)
✓ 1
h = 4.0 W
✓ 1
Allow ECF from (b).
—
Part (d)
c' = 28.0/(2.0×10−3 × 5.0) = 2.8×103 J kg−1 K−1
✓ 1
Percentage difference = (2800 − 2400)/2400 × 100 = 16.7 % (accept 17 %); the student's value is an overestimate
✓ 1
Allow ECF from (b).
—
Exam tip: In a 'show that' part, write out the full working and give the answer to one more significant figure than the value quoted, e.g. 2.40×10³ rather than 2.4×10³.
152IBP2S-009-00-15Luminosity & apparent brightnessPaper 2Hard9 marks
A red supergiant star is observed with a space telescope. Its emission spectrum peaks at a wavelength of 850 nm. Its apparent brightness at Earth is 1.1×10−7 W m−2 and its distance from Earth is 6.6×1018 m. Wien constant = 2.9×10−3 m K; σ = 5.67×10−8 W m−2 K−4.
(a)
Calculate the surface temperature of the star.
(2)
(b)
Calculate the luminosity of the star.
(2)
(c)
Determine the radius of the star.
(3)
(d)
The surface temperature of this star is lower than that of the Sun, yet its luminosity is far greater. Explain how this is possible.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
T = 2.9×10−3/850×10−9
✓ 1
T = 3.4×103 K
✓ 1
Part (b)
L = 4πd2b = 4π(6.6×1018)2 × 1.1×10−7
✓ 1
L = 6.0×1031 W
✓ 1
Part (c)
ECF from (a) and (b). Uses L = 4πR2σT4 rearranged: R = √(L/(4πσT4))
✓ 1
Correct substitution: R = √(6.0×1031/(4π × 5.67×10−8 × 34104))
✓ 1
R = 7.9×1011 m
✓ 1
Part (d)
Luminosity depends on R2 as well as T4
✓ 1
The star's radius is so much larger (about 1000 times the Sun's) that the huge surface area outweighs the lower power emitted per unit area
✓ 1
Common error: Convert nm to m (×10⁻⁹) before using Wien's law, and remember the square root when finding R.
Vaccines are transported in a box packed with dry ice (solid carbon dioxide). The inside of the box stays at −78 °C while the outside air is at 25 °C. Assume the inner and outer surfaces of the walls are at these temperatures. The walls have a total area of 1.2 m2 and consist of a 2.0 mm aluminium outer skin (k = 205 W m−1 K−1) bonded to a 40 mm layer of polyurethane foam (k = 0.025 W m−1 K−1). Specific latent heat of sublimation of dry ice = 5.7×105 J kg−1.
Diagram not to scale
(a)
State the temperature difference across the wall in kelvin.
(1)
(b)
Explain why the temperature at the boundary between the aluminium and the foam is very close to 25 °C.
(2)
(c)
Calculate the rate of thermal energy transfer into the box. Ignore the aluminium layer.
(2)
(d)
Determine the mass of dry ice that sublimes in one day.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
25 − (−78) = 103 K
✓ 1
Part (b)
The same power passes through both layers in series (steady state)
✓ 1
Aluminium has a far higher k and is much thinner, so its thermal resistance (Δx/kA) is tiny compared with the foam's. Only a very small temperature difference is needed across it, so almost all of the 103 K drop is across the foam
✓ 1
Part (c)
ECF from (a). P = kAΔT/Δx = 0.025 × 1.2 × 103/0.040
✓ 1
P = 77 W
✓ 1
Part (d)
ECF from (c). Energy in one day Q = 77 × 86 400 = 6.7×106 J
✓ 1
m = Q/L = 6.7×106/5.7×105 = 11.7 ≈ 12 kg
✓ 1
Exam tip: In a time-based question, convert the time to seconds (1 day = 86 400 s) before multiplying by a power in watts.
154IBP2S-009-03-16Density & the molecular modelPaper 2Hard9 marks
A hydrogen-powered aircraft stores liquid hydrogen with a density of 71 kg m−3. At 273 K and atmospheric pressure, hydrogen gas has a density of 0.090 kg m−3. The molar mass of hydrogen (H2) is 2.0 g mol−1. Use NA = 6.02 × 1023 mol−1 and kB = 1.38 × 10−23 J K−1.
(a)
Show that the mass of one hydrogen molecule is about 3.3 × 10−27 kg.
(1)
(b)
Estimate the mean spacing between molecules in liquid hydrogen.
(3)
(c)
Determine the ratio \(\frac{\text{mean molecular spacing in the gas at 273 K}}{\text{mean molecular spacing in the liquid}}\).
(2)
(d)
Determine a typical (root mean square) speed of a hydrogen molecule in the gas at 273 K.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
m = 2.0×10−3 / 6.02×1023 = 3.32×10−27 kg
✓ 1
Part (b)
Number of molecules per unit volume n = ρ/m = 71/3.32×10−27 = 2.14×1028 m−3
155IBP2S-009-02-13Luminosity & apparent brightnessPaper 2Hard10 marks
An exoplanet orbits close to a red dwarf star. Data for the system are given below.
Peak wavelength of star's spectrum
9.0×10−7 m
Luminosity of star
3.8×1024 W
Orbital radius of planet
3.0×109 m
Wien's constant b = 2.9×10−3 m K; σ = 5.67×10−8 W m−2 K−4.
(a)
Determine the surface temperature of the star.
(2)
(b)
Show that the radius of the star is about 2.2×108 m.
(2)
(c)
Calculate the intensity of the star's radiation at the planet.
(2)
(d)
The planet has radius r, no atmosphere, rotates rapidly and behaves as a black body. Determine the equilibrium surface temperature of the planet.
(3)
(e)
Suggest one reason why the actual surface temperature of the planet may differ from your answer to (d).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
T = b/λmax = 2.9×10−3/9.0×10−7 ; T = 3.2×103 K (3222 K)
✓ 2
Part (b)
Rearranges L = σ4πR2T4 to R = √(L/(4πσT4)) ; substitution: R = √(3.8×1024/(4π × 5.67×10−8 × 32224)) = 2.22×108 m, answer given to at least 3 s.f. . Accept 2.25×108 m if the rounded value 3.2×103 K is used. ECF from (a).
✓ 2
Part (c)
I = L/(4πd2) = 3.8×1024/(4π × (3.0×109)2) ; I = 3.4×104 W m−2
✓ 2
Part (d)
Power absorbed = I × πr2 (cross-sectional area) ; power emitted = σ × 4πr2 × T4 (whole surface radiates, since rapid rotation keeps the temperature uniform), and equating gives T4 = I/(4σ) ; T = (3.36×104/(4 × 5.67×10−8))1/4 = 620 K . ECF from (c).
✓ 3
Part (e)
Any one: the planet reflects some of the incident radiation (non-zero albedo), so it is not a perfect absorber / its emissivity is less than 1 / internal heating, such as tidal or geological heating / the orbit is not perfectly circular, so the intensity received varies
✓ 1
Exam tip: The planet intercepts radiation over its cross-section πr² but radiates from its whole surface 4πr². This gives the factor of 4.
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