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O Level Physics Thermal Physics Quiz
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Questions
O-Level Physics Quiz - Thermal Physics
Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 40
Duration: 45 minutes Total Marks: 40 Instructions: Answer ALL questions in the spaces provided. Show all working for calculation questions. State units in final answers. Use g = 10 N/kg where required.
Section A: Short Answer (10 marks)
Answer all questions in this section.
1. State the SI unit of temperature used in scientific measurements.
[1 mark]
2. A student observes smoke particles under a microscope and notices they move in a random, jerky motion. Name this phenomenon and state what causes it.
[2 marks]
3. Explain, in terms of particles, why a gas exerts pressure on the walls of its container.
[2 marks]
4. State two differences between boiling and evaporation.
[2 marks]
5. A metal spoon and a wooden spoon are both placed in a cup of hot water. After two minutes, the metal spoon feels much hotter than the wooden spoon. Explain this observation.
[3 marks]
Section B: Structured Questions (10 marks)
Answer all questions in this section.
6. A student investigates the cooling of hot water in two identical beakers. Beaker A is wrapped in shiny aluminium foil. Beaker B is wrapped in matt black paper. Both beakers start at 80 °C and are left in the same room.
(a) After 10 minutes, the water in Beaker B is cooler than the water in Beaker A. Explain why.
[2 marks]
(b) State one additional factor, other than surface colour, that affects the rate of thermal energy transfer by radiation.
[1 mark]
7. A 2.0 kg aluminium block is heated by an electric heater. The temperature of the block rises from 25 °C to 55 °C. The specific heat capacity of aluminium is 900 J/(kg °C).
(a) Calculate the thermal energy absorbed by the aluminium block.
[2 marks]
(b) The same heater is used to heat a 2.0 kg block of iron. The specific heat capacity of iron is 450 J/(kg °C). The heater supplies the same amount of energy as in part (a). Calculate the temperature rise of the iron block.
[2 marks]
8. A student heats a solid substance at a constant rate and records its temperature over time. The graph below describes the heating process.
[Note: A graph would show temperature on the y-axis and time on the x-axis, with a rising line, a flat plateau, another rising line, a second flat plateau, and a final rising line.]
(a) Explain why the temperature remains constant during the first plateau region, even though heating continues.
[2 marks]
(b) State what is happening to the particles of the substance during this plateau.
[1 mark]
9. A solar panel is used to heat water for a home. The panel has a matt black surface.
(a) Explain why a matt black surface is chosen for the solar panel.
[2 marks]
(b) On a sunny day, the panel transfers 8.4 × 10⁵ J of energy to 5.0 kg of water. The initial temperature of the water is 30 °C. The specific heat capacity of water is 4200 J/(kg °C). Calculate the final temperature of the water.
[3 marks]
(c) In practice, the final temperature of the water is lower than the calculated value. Suggest one reason for this difference.
[1 mark]
10. A vacuum flask keeps hot liquids hot and cold liquids cold. The diagram shows a cross-section of a vacuum flask.
[Note: A diagram would show a double-walled glass container with a vacuum between the walls, silvered surfaces, and a stopper.]
(a) Explain how the vacuum between the double walls reduces thermal energy transfer.
[1 mark]
(b) Explain how the silvered surfaces reduce thermal energy transfer.
[1 mark]
Section C: Data-Based and Application Questions (10 marks)
Answer all questions in this section.
11. A student investigates the specific latent heat of fusion of ice. She places 0.050 kg of ice at 0 °C into a beaker containing 0.200 kg of water at 40 °C. The ice melts completely, and the final temperature of the water is 18 °C. The specific heat capacity of water is 4200 J/(kg °C).
(a) Calculate the thermal energy lost by the warm water as it cools from 40 °C to 18 °C.
[2 marks]
(b) The melted ice (now water at 0 °C) warms up to the final temperature of 18 °C. Calculate the thermal energy gained by this melted ice as it warms up.
[2 marks]
(c) Using your answers to (a) and (b), calculate the specific latent heat of fusion of ice.
[3 marks]
12. A bimetallic strip is made by joining a strip of brass to a strip of iron. Brass expands more than iron when heated. The strip is straight at room temperature.
(a) Describe and explain what happens to the bimetallic strip when it is heated.
[2 marks]
(b) The bimetallic strip is used in a fire alarm. Explain how the strip can be used to complete an electric circuit and sound an alarm when the temperature rises above a certain level.
[2 marks]
(c) Suggest one other practical application of a bimetallic strip.
[1 mark]
13. A student places a thermometer in a beaker of water at room temperature. She then adds ice cubes and stirs gently until the temperature stops falling. She records the temperature every minute.
(a) Explain why the temperature of the water stops falling even though ice is still present.
[2 marks]
(b) State the expected final temperature reading on the thermometer, assuming no heat exchange with the surroundings.
[1 mark]
14. A metal block of mass 1.5 kg is heated to 100 °C and then placed into 0.80 kg of water at 20 °C in an insulated container. The final temperature of the water and block is 35 °C. The specific heat capacity of water is 4200 J/(kg °C).
(a) Calculate the thermal energy gained by the water.
[2 marks]
(b) Calculate the specific heat capacity of the metal block.
[2 marks]
15. A student investigates the cooling of water in a beaker. She wraps the beaker in different materials and records the temperature drop over 10 minutes. Her results are shown below.
| Material | Temperature drop (°C) |
|---|---|
| Cotton | 12 |
| Wool | 8 |
| Foil | 15 |
(a) Which material is the best insulator? Explain your choice.
[1 mark]
(b) Suggest one way the student could improve the reliability of her results.
[1 mark]
Section D: Extended Response (10 marks)
Answer all questions in this section.
16. A student investigates the factors affecting the rate of evaporation. She pours equal volumes of water into three identical dishes and places them in different conditions: Dish X in a warm room, Dish Y in a cool room, and Dish Z in a warm room with a fan blowing across it.
(a) State and explain which dish will have the highest rate of evaporation.
[2 marks]
(b) Explain, in terms of particles, how evaporation causes the remaining liquid to cool.
[2 marks]
17. A 500 W electric kettle contains 0.80 kg of water at 25 °C. The specific heat capacity of water is 4200 J/(kg °C). The kettle is switched on.
(a) Calculate the minimum time required to bring the water to its boiling point of 100 °C. Assume no energy losses.
[3 marks]
(b) In practice, the time taken is longer than the calculated value. Suggest two reasons why.
[2 marks]
18. A student places a sealed, empty plastic bottle in a freezer. After an hour, the bottle appears crushed.
(a) Explain this observation in terms of the behaviour of gas particles.
[2 marks]
(b) The student then removes the bottle from the freezer and leaves it at room temperature. Describe and explain what happens to the bottle over time.
[2 marks]
19. A copper rod and a glass rod of the same dimensions are held at one end. The other end of each rod is heated in a flame. The copper rod becomes too hot to hold much sooner than the glass rod.
(a) Explain this observation in terms of the mechanism of thermal conduction.
[2 marks]
(b) State one practical application where the difference in thermal conductivity between materials is important.
[1 mark]
20. A student investigates the specific latent heat of vaporisation of water. She uses an electric heater to boil water in a beaker. The heater supplies energy at a rate of 50 J/s. After the water reaches 100 °C, she collects the steam produced and finds that 10 g of water is boiled away in 5 minutes.
(a) Calculate the thermal energy supplied by the heater during the 5 minutes.
[1 mark]
(b) Calculate the specific latent heat of vaporisation of water from these results.
[2 marks]
(c) The accepted value for the specific latent heat of vaporisation of water is 2.26 × 10⁶ J/kg. Suggest one reason why the student's value is likely to be higher than the accepted value.
[1 mark]
END OF QUIZ
This quiz was generated by TuitionGoWhere AI. Content is syllabus-aligned and designed for O-Level Physics (6091) Thermal Physics practice.
Answers
O-Level Physics Quiz - Thermal Physics: Answer Key
Total Marks: 40
Section A: Short Answer (10 marks)
1. State the SI unit of temperature used in scientific measurements.
[1 mark]
Answer: Kelvin (K)
Marking notes: Accept "kelvin" (lowercase). Do not accept "degrees Kelvin" or "°K". The kelvin is the SI base unit of thermodynamic temperature.
2. A student observes smoke particles under a microscope and notices they move in a random, jerky motion. Name this phenomenon and state what causes it.
[2 marks]
Answer:
- Phenomenon: Brownian motion [1 mark]
- Cause: The smoke particles are being bombarded unevenly by fast-moving, invisible air molecules from all directions [1 mark]
Marking notes: Award 1 mark for "Brownian motion" or "Brownian movement". Award 1 mark for explanation involving collisions with air molecules/particles. Accept "random bombardment by air particles" or equivalent.
3. Explain, in terms of particles, why a gas exerts pressure on the walls of its container.
[2 marks]
Answer: Gas particles are in constant, random motion [1 mark]. They collide with the walls of the container, and each collision exerts a small force on the wall. The combined effect of many collisions per second over the surface area results in pressure [1 mark].
Marking notes: Award 1 mark for mentioning particle motion/collisions. Award 1 mark for linking collisions to force/pressure. Accept "particles bounce off walls" or "change in momentum of particles exerts force".
4. State two differences between boiling and evaporation.
[2 marks]
Answer: Any two from:
- Boiling occurs at a specific temperature (boiling point); evaporation occurs at any temperature below the boiling point [1 mark]
- Boiling occurs throughout the liquid; evaporation occurs only at the surface [1 mark]
- Boiling produces bubbles within the liquid; evaporation does not [1 mark]
- Boiling is a rapid process; evaporation is a slow process [1 mark]
Marking notes: Award 1 mark for each correct difference. Must clearly contrast boiling and evaporation.
5. A metal spoon and a wooden spoon are both placed in a cup of hot water. After two minutes, the metal spoon feels much hotter than the wooden spoon. Explain this observation.
[3 marks]
Answer: Metals are good conductors of thermal energy; wood is a poor conductor (good insulator) [1 mark]. In metals, free electrons can move through the material and transfer kinetic energy rapidly from the hot end to the cold end [1 mark]. Wood does not have free electrons, so thermal energy is transferred much more slowly through vibration of atoms only [1 mark].
Marking notes: Award 1 mark for identifying metal as good conductor and wood as poor conductor/insulator. Award 1 mark for explaining conduction in metals (free electrons). Award 1 mark for explaining why wood conducts poorly (no free electrons, relies on atomic vibrations only).
Section B: Structured Questions (10 marks)
6. (a) After 10 minutes, the water in Beaker B is cooler than the water in Beaker A. Explain why.
[2 marks]
Answer: Matt black surfaces are good emitters of thermal radiation [1 mark], while shiny surfaces are poor emitters of thermal radiation [1 mark]. Therefore, Beaker B (matt black) loses thermal energy by radiation at a faster rate than Beaker A (shiny), so its water cools more quickly.
Marking notes: Award 1 mark for stating that matt black is a good emitter. Award 1 mark for stating that shiny is a poor emitter OR for linking emission rate to cooling rate.
(b) State one additional factor, other than surface colour, that affects the rate of thermal energy transfer by radiation.
[1 mark]
Answer: Any one from:
- Surface temperature (higher temperature → greater rate of radiation)
- Surface area (larger area → greater rate of radiation)
- Nature/texture of the surface
Marking notes: Award 1 mark for any valid factor. Do not accept "colour" (already excluded in question).
7. (a) Calculate the thermal energy absorbed by the aluminium block.
[2 marks]
Answer:
- Temperature change Δθ = 55 - 25 = 30 °C [1 mark for correct substitution or temperature change]
- Energy = mcΔθ = 2.0 × 900 × 30 = 54 000 J (or 54 kJ) [1 mark for correct answer with units]
Marking notes: Award 1 mark for correct formula and substitution. Award 1 mark for correct numerical answer with correct unit. Accept 5.4 × 10⁴ J.
(b) Calculate the temperature rise of the iron block.
[2 marks]
Answer:
- Energy supplied = 54 000 J [1 mark for using answer from (a) or showing ECF]
- Δθ = E / (mc) = 54 000 / (2.0 × 450) = 54 000 / 900 = 60 °C [1 mark for correct answer with units]
Marking notes: Award 1 mark for correct formula and substitution. Award 1 mark for correct numerical answer. Allow error carried forward (ECF) from part (a). Accept "temperature rise = 60 °C".
8. (a) Explain why the temperature remains constant during the first plateau region, even though heating continues.
[2 marks]
Answer: The first plateau represents melting (change of state from solid to liquid) [1 mark]. The thermal energy supplied is used to break the bonds between particles (overcome the forces of attraction) rather than to increase the kinetic energy of the particles, so the temperature does not rise [1 mark].
Marking notes: Award 1 mark for identifying the plateau as melting/change of state. Award 1 mark for explaining that energy is used to break bonds/overcome attractive forces, not to increase kinetic energy.
(b) State what is happening to the particles of the substance during this plateau.
[1 mark]
Answer: The particles are gaining potential energy as they move further apart / The bonds between particles are being broken / The arrangement changes from regular (solid) to irregular (liquid) [1 mark].
Marking notes: Award 1 mark for any valid description of particle-level changes during melting. Accept "particles become less ordered" or "particles move out of fixed positions".
9. (a) Explain why a matt black surface is chosen for the solar panel.
[2 marks]
Answer: Matt black surfaces are good absorbers of thermal radiation [1 mark]. This allows the solar panel to absorb as much thermal energy as possible from the Sun's radiation, maximising the heating of the water [1 mark].
Marking notes: Award 1 mark for stating matt black is a good absorber. Award 1 mark for linking to maximising energy absorption from the Sun.
(b) Calculate the final temperature of the water.
[3 marks]
Answer:
- Energy transferred: E = 8.4 × 10⁵ J
- Mass: m = 5.0 kg
- Specific heat capacity: c = 4200 J/(kg °C)
- Temperature rise: Δθ = E / (mc) = 8.4 × 10⁵ / (5.0 × 4200) [1 mark for correct formula and substitution]
- Δθ = 840 000 / 21 000 = 40 °C [1 mark for correct temperature rise]
- Final temperature = 30 + 40 = 70 °C [1 mark for correct final temperature with units]
Marking notes: Award 1 mark for correct formula and substitution. Award 1 mark for correct temperature rise. Award 1 mark for correct final temperature. Accept working in stages.
(c) In practice, the final temperature of the water is lower than the calculated value. Suggest one reason for this difference.
[1 mark]
Answer: Any one from:
- Some thermal energy is lost to the surroundings (by conduction/convection/radiation)
- Not all the solar energy is absorbed by the water (some reflected)
- The solar panel itself absorbs some energy
- Heat losses from the pipes/container
Marking notes: Award 1 mark for any valid reason involving energy losses.
10. (a) Explain how the vacuum between the double walls reduces thermal energy transfer.
[1 mark]
Answer: A vacuum contains no particles, so conduction and convection cannot occur through it [1 mark].
Marking notes: Award 1 mark for stating that conduction and convection require a medium/particles, which are absent in a vacuum.
(b) Explain how the silvered surfaces reduce thermal energy transfer.
[1 mark]
Answer: Shiny/silvered surfaces are poor emitters and poor absorbers of thermal radiation [1 mark]. This reduces thermal energy transfer by radiation across the vacuum.
Marking notes: Award 1 mark for stating that shiny surfaces reduce radiation (either by being poor emitters or poor absorbers, or both).
Section C: Data-Based and Application Questions (10 marks)
11. (a) Calculate the thermal energy lost by the warm water as it cools from 40 °C to 18 °C.
[2 marks]
Answer:
- Temperature change Δθ = 40 - 18 = 22 °C [1 mark for correct temperature change]
- Energy lost = mcΔθ = 0.200 × 4200 × 22 = 18 480 J [1 mark for correct answer with units]
Marking notes: Award 1 mark for correct formula and substitution. Award 1 mark for correct numerical answer. Accept 1.848 × 10⁴ J.
(b) The melted ice (now water at 0 °C) warms up to the final temperature of 18 °C. Calculate the thermal energy gained by this melted ice as it warms up.
[2 marks]
Answer:
- Mass of melted ice = 0.050 kg
- Temperature change Δθ = 18 - 0 = 18 °C [1 mark for correct temperature change]
- Energy gained = mcΔθ = 0.050 × 4200 × 18 = 3780 J [1 mark for correct answer with units]
Marking notes: Award 1 mark for correct formula and substitution. Award 1 mark for correct numerical answer.
(c) Using your answers to (a) and (b), calculate the specific latent heat of fusion of ice.
[3 marks]
Answer:
- Energy lost by warm water = Energy gained by ice (melting + warming) [1 mark for energy conservation principle]
- Energy used to melt ice = Energy lost by water - Energy gained by melted ice warming up
- Energy used to melt ice = 18 480 - 3780 = 14 700 J [1 mark for correct subtraction]
- Specific latent heat of fusion Lf = E / m = 14 700 / 0.050 = 294 000 J/kg (or 2.94 × 10⁵ J/kg) [1 mark for correct answer with units]
Marking notes: Award 1 mark for applying conservation of energy. Award 1 mark for correct calculation of energy used for melting. Award 1 mark for correct specific latent heat with units. Allow ECF from parts (a) and (b).
12. (a) Describe and explain what happens to the bimetallic strip when it is heated.
[2 marks]
Answer: The strip bends [1 mark]. Brass expands more than iron when heated, so the brass side becomes longer than the iron side, causing the strip to curve with the brass on the outside of the curve [1 mark].
Marking notes: Award 1 mark for stating the strip bends/curves. Award 1 mark for explaining that brass expands more, causing differential expansion.
(b) The bimetallic strip is used in a fire alarm. Explain how the strip can be used to complete an electric circuit and sound an alarm when the temperature rises above a certain level.
[2 marks]
Answer: When the temperature rises, the bimetallic strip bends [1 mark]. This bending causes the strip to touch an electrical contact, completing the circuit and allowing current to flow to the alarm bell/siren [1 mark].
Marking notes: Award 1 mark for linking bending to temperature rise. Award 1 mark for explaining how bending completes the circuit.
(c) Suggest one other practical application of a bimetallic strip.
[1 mark]
Answer: Any one from:
- Thermostat (e.g., in an iron, oven, or air conditioner)
- Flashing indicator lights in cars
- Overload protection in circuit breakers
Marking notes: Award 1 mark for any valid application.
13. (a) Explain why the temperature of the water stops falling even though ice is still present.
[2 marks]
Answer: The water and ice reach thermal equilibrium at the melting point of ice (0 °C) [1 mark]. At this temperature, the rate of thermal energy transfer from the water to the ice equals the rate of thermal energy transfer from the ice to the water, so the temperature remains constant [1 mark].
Marking notes: Award 1 mark for identifying thermal equilibrium or 0 °C. Award 1 mark for explaining that energy transfer rates are balanced.
(b) State the expected final temperature reading on the thermometer, assuming no heat exchange with the surroundings.
[1 mark]
Answer: 0 °C [1 mark]
Marking notes: Award 1 mark for 0 °C. Accept "zero degrees Celsius" or "the melting point of ice".
14. (a) Calculate the thermal energy gained by the water.
[2 marks]
Answer:
- Temperature change Δθ = 35 - 20 = 15 °C [1 mark for correct temperature change]
- Energy gained = mcΔθ = 0.80 × 4200 × 15 = 50 400 J [1 mark for correct answer with units]
Marking notes: Award 1 mark for correct formula and substitution. Award 1 mark for correct numerical answer.
(b) Calculate the specific heat capacity of the metal block.
[2 marks]
Answer:
- Energy lost by metal block = Energy gained by water = 50 400 J [1 mark for energy conservation]
- Temperature change of metal block Δθ = 100 - 35 = 65 °C
- c = E / (mΔθ) = 50 400 / (1.5 × 65) = 50 400 / 97.5 ≈ 517 J/(kg °C) [1 mark for correct answer with units]
Marking notes: Award 1 mark for applying conservation of energy and correct formula. Award 1 mark for correct numerical answer with units. Accept values around 517 J/(kg °C).
15. (a) Which material is the best insulator? Explain your choice.
[1 mark]
Answer: Wool is the best insulator because it has the smallest temperature drop (8 °C), meaning it reduces thermal energy transfer the most [1 mark].
Marking notes: Award 1 mark for identifying wool and linking to the smallest temperature drop/least heat loss.
(b) Suggest one way the student could improve the reliability of her results.
[1 mark]
Answer: Any one from:
- Repeat the experiment and calculate an average temperature drop
- Ensure the starting temperature is the same for each trial
- Use the same volume/mass of water in each beaker
- Ensure the room temperature is constant
Marking notes: Award 1 mark for any valid method to improve reliability (consistency/repeatability).
Section D: Extended Response (10 marks)
16. (a) State and explain which dish will have the highest rate of evaporation.
[2 marks]
Answer: Dish Z (warm room with fan) will have the highest rate of evaporation [1 mark]. Evaporation rate increases with higher temperature (more particles have enough kinetic energy to escape) and with air movement (fan removes water vapour from above the surface, reducing humidity and allowing more particles to escape) [1 mark].
Marking notes: Award 1 mark for identifying Dish Z. Award 1 mark for explaining both factors (temperature and air movement/fan).
(b) Explain, in terms of particles, how evaporation causes the remaining liquid to cool.
[2 marks]
Answer: During evaporation, the most energetic particles (with the highest kinetic energy) escape from the liquid surface [1 mark]. This leaves behind particles with lower average kinetic energy, so the temperature of the remaining liquid decreases [1 mark].
Marking notes: Award 1 mark for stating that the most energetic particles escape. Award 1 mark for linking this to a decrease in average kinetic energy/temperature.
17. (a) Calculate the minimum time required to bring the water to its boiling point of 100 °C. Assume no energy losses.
[3 marks]
Answer:
- Temperature rise Δθ = 100 - 25 = 75 °C [1 mark for correct temperature change]
- Energy required = mcΔθ = 0.80 × 4200 × 75 = 252 000 J [1 mark for correct energy calculation]
- Power = Energy / time, so time = Energy / Power = 252 000 / 500 = 504 seconds [1 mark for correct time with units]
- Time = 504 s (or 8.4 minutes)
Marking notes: Award 1 mark for correct temperature change. Award 1 mark for correct energy calculation. Award 1 mark for correct time calculation with units.
(b) In practice, the time taken is longer than the calculated value. Suggest two reasons why.
[2 marks]
Answer: Any two from:
- Some thermal energy is lost to the surroundings (by conduction, convection, or radiation)
- Some energy is used to heat the kettle itself
- Some energy is lost as steam escapes before boiling point is reached
- The power rating may not be exact
Marking notes: Award 1 mark for each valid reason, up to 2 marks.
18. (a) Explain this observation in terms of the behaviour of gas particles.
[2 marks]
Answer: When the bottle is placed in the freezer, the temperature of the air inside decreases [1 mark]. The gas particles lose kinetic energy, move slower, and collide with the walls less frequently and with less force, reducing the pressure inside the bottle. The greater external atmospheric pressure crushes the bottle [1 mark].
Marking notes: Award 1 mark for linking temperature decrease to reduced particle kinetic energy/speed. Award 1 mark for explaining the pressure difference (internal pressure decreases, external pressure crushes bottle).
(b) The student then removes the bottle from the freezer and leaves it at room temperature. Describe and explain what happens to the bottle over time.
[2 marks]
Answer: The bottle expands back to its original shape [1 mark]. As the air inside warms up, the gas particles gain kinetic energy, move faster, and collide with the walls more frequently and with more force, increasing the internal pressure until it equals the external atmospheric pressure [1 mark].
Marking notes: Award 1 mark for stating the bottle expands/returns to original shape. Award 1 mark for explaining the pressure increase due to increased particle kinetic energy.
19. (a) Explain this observation in terms of the mechanism of thermal conduction.
[2 marks]
Answer: Copper is a good conductor of thermal energy because it has many free electrons that can move through the metal and transfer kinetic energy rapidly from the hot end to the cold end [1 mark]. Glass is a poor conductor (insulator) because it lacks free electrons; thermal energy is transferred only by vibrations of atoms, which is a much slower process [1 mark].
Marking notes: Award 1 mark for explaining conduction in copper (free electrons). Award 1 mark for explaining poor conduction in glass (no free electrons, atomic vibrations only).
(b) State one practical application where the difference in thermal conductivity between materials is important.
[1 mark]
Answer: Any one from:
- Cooking pans have metal bases (good conductor) and plastic/wooden handles (poor conductor/insulator)
- Insulation in houses (e.g., fibreglass, foam) uses poor conductors to reduce heat loss
- Heat sinks in electronics use metals (good conductors) to dissipate heat
Marking notes: Award 1 mark for any valid application that correctly identifies the use of good and/or poor conductors.
20. (a) Calculate the thermal energy supplied by the heater during the 5 minutes.
[1 mark]
Answer:
- Time = 5 minutes = 5 × 60 = 300 seconds
- Energy = Power × time = 50 × 300 = 15 000 J [1 mark for correct answer with units]
Marking notes: Award 1 mark for correct calculation and units.
(b) Calculate the specific latent heat of vaporisation of water from these results.
[2 marks]
Answer:
- Mass of water boiled = 10 g = 0.010 kg [1 mark for correct mass conversion]
- Specific latent heat Lv = E / m = 15 000 / 0.010 = 1 500 000 J/kg = 1.5 × 10⁶ J/kg [1 mark for correct answer with units]
Marking notes: Award 1 mark for correct mass conversion to kg. Award 1 mark for correct calculation and units.
(c) The accepted value for the specific latent heat of vaporisation of water is 2.26 × 10⁶ J/kg. Suggest one reason why the student's value is likely to be higher than the accepted value.
[1 mark]
Answer: Any one from:
- Some thermal energy is lost to the surroundings, so not all energy supplied goes into vaporising the water (this would actually make the calculated value higher if mass boiled is less than expected, but the question states the value is higher, so the reasoning should focus on energy losses meaning more energy is recorded as supplied than actually used for vaporisation, leading to an overestimate if mass is accurate) - Correction: If energy is lost, the calculated Lv = E/m would be higher if E is the total energy supplied but m is the actual mass vaporised, because some E was lost, so the true E used for vaporisation is less, making the true Lv lower. The student's value would be higher if they underestimate the mass vaporised (e.g., steam escapes before condensing) or if the heater supplies more energy than its rating.
- Better answer: Some water may have splashed out or steam escaped before being collected, leading to an underestimated mass of water vaporised. Since Lv = E/m, a smaller mass gives a larger calculated Lv.
- The heater may have supplied energy at a higher rate than 50 J/s (power rating inaccurate).
- Energy was absorbed from the surroundings, adding to the heater's energy.
Marking notes: Award 1 mark for any valid reason that would cause the calculated Lv to be higher than the accepted value. The most common reason is underestimation of the mass of water vaporised due to steam loss.
END OF ANSWER KEY
This answer key was generated by TuitionGoWhere AI. Content is syllabus-aligned and designed for O-Level Physics (6091) Thermal Physics practice.