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Secondary 3 Physics Thermal Physics Quiz
Free Sec 3 Physics Thermal Physics quiz, LongCat Exam version, with questions, answers, and O Level-style practice for Singapore students.
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Secondary 3 Physics Quiz - Thermal Physics
Answer Key
Section A: Multiple Choice Questions
1. B — J °C⁻¹ [1]
Heat capacity is the energy required to raise the temperature of a body by 1 °C. Its unit is joules per degree Celsius (J °C⁻¹). Option A is the unit for specific heat capacity.
2. A — 60 K [1]
A change in temperature in °C is numerically equal to the change in K. ΔT = 85 − 25 = 60 °C = 60 K.
3. B — overcome the intermolecular forces between molecules [1]
During boiling, energy is used to break intermolecular bonds (change of state), not to increase kinetic energy/temperature.
4. C — Radiation [1]
Radiation transfers energy via electromagnetic waves and does not require a medium. Conduction and convection require matter.
5. B — X has a higher temperature than Y [1]
Heat flows from a region of higher temperature to a region of lower temperature, regardless of mass, heat capacity, or volume.
6. B — raise the temperature of 1 kg of the substance by 1 °C [1]
Specific heat capacity is defined per unit mass (1 kg) per unit temperature change (1 °C or 1 K).
7. A — absorber and poor emitter [1]
A shiny silver surface is a poor absorber and a poor emitter of thermal radiation. It is a good reflector.
8. C — Solid [1]
In solids, particles are closely packed and transfer energy efficiently through vibrations. Conduction is most efficient in solids.
9. B — 10 °C [1]
Q = mcΔT → ΔT = Q / (mc) = 18 000 / (2 × 900) = 10 °C.
10. D — loses thermal energy but its temperature remains constant [1]
During condensation, the gas releases latent heat. The temperature remains constant during the change of state.
Section B: Short Answer and Structured Questions
11.
(a) Specific heat capacity is the amount of thermal energy required to raise the temperature of 1 kg of a substance by 1 °C (or 1 K). [2]
Marking: 1 mark for "1 kg", 1 mark for "by 1 °C / 1 K".
(b) Specific latent heat of fusion is the amount of thermal energy required to change 1 kg of a substance from solid to liquid at its melting point, without a change in temperature. [2]
Marking: 1 mark for "1 kg", 1 mark for "solid to liquid at constant temperature / without temperature change".
12. During melting, the thermal energy supplied is used to overcome the intermolecular forces holding the molecules in their fixed positions in the solid lattice. [1]
The energy increases the potential energy of the molecules, not their kinetic energy. [1]
Since temperature is a measure of the average kinetic energy of the molecules, and kinetic energy does not increase, the temperature remains constant. [1]
[3]
13.
| Boiling | Evaporation | |
|---|---|---|
| Difference 1 | Occurs at a fixed temperature (boiling point) | Occurs at any temperature |
| Difference 2 | Occurs throughout the bulk of the liquid | Occurs only at the surface of the liquid |
[2]
Marking: 1 mark per correct difference. Accept other valid differences, e.g., boiling requires a continuous heat source / evaporation is a slower process / boiling produces bubbles.
14.
(a) Q = mcΔT [1]
Q = 0.5 × 4 200 × (100 − 20)
Q = 0.5 × 4 200 × 80
Q = 168 000 J [1]
(b) P = E / t → t = E / P [1]
t = 168 000 / 500
t = 336 s [1]
15.
(a) Thermal energy travels from the heated end (near the Bunsen burner) towards the cooler end / from left to right along the rod. [1]
(b) The wax blobs near the heated end are closest to the heat source. [1] Thermal energy is transferred along the rod by conduction, so the nearest blobs receive thermal energy first and reach the melting point before the others. [1]
(c) Conduction [1]
16. The metal handle is in direct contact with the saucepan body, which is heated by the stove. [1] Thermal energy is transferred through the metal by conduction — the particles in the hot part of the saucepan vibrate more vigorously and pass on their kinetic energy to neighbouring particles along the handle. [1] Since metals are good conductors of thermal energy, the handle becomes hot. [1]
[2] — Award marks for: mention of conduction + explanation of particle vibration/energy transfer.
Section C: Longer Response and Application Questions
17.
(a) Thermal energy lost by copper:
Q = mcΔT [1]
Q = 0.8 × 385 × (150 − 30)
Q = 0.8 × 385 × 120
Q = 36 960 J (or 3.70 × 10⁴ J) [1]
(b) Thermal energy gained by water:
Q = mcΔT [1]
Q = 1.2 × 4 200 × (30 − 25)
Q = 1.2 × 4 200 × 5
Q = 25 200 J [1]
(c) Some thermal energy is lost to the surroundings (e.g., absorbed by the container, lost to the air). [1]
Accept: heat absorbed by the container / heat lost to the environment.
18.
Feature A (Plastic cap): Plastic is a poor conductor of heat (thermal insulator), so it reduces thermal energy loss by conduction through the top of the flask. [1]
Feature B (Vacuum layer): The vacuum (empty space between the walls) prevents thermal energy transfer by convection and conduction, as both require a medium and there are no particles in a vacuum. [1]
Feature C (Silvered walls): The silvered surfaces are poor emitters and good reflectors of thermal radiation, so they reflect thermal radiation back into the liquid and reduce energy loss by radiation. [1]
19.
(a) Q = m × l_f [1]
Q = 1.5 × 3.34 × 10⁵
Q = 5.01 × 10⁵ J (or 501 000 J) [1]
(b) During melting, the thermal energy supplied is used to break the intermolecular bonds that hold the water molecules in the solid ice lattice. [1] This energy increases the potential energy of the molecules, not their kinetic energy. Since temperature depends on the average kinetic energy of the molecules, the temperature remains constant during the change of state. [1]
[2]
20.
(a) The matt black can cools faster. [1] A matt black surface is a better emitter of thermal radiation compared to a polished silver surface, so it loses thermal energy to the surroundings at a faster rate. [1]
(b) Graph:
- Both curves start at 80 °C and decrease towards 20 °C (room temperature).
- The matt black can curve drops more steeply (cools faster).
- The polished silver can curve drops more gradually (cools slower).
- Both curves level off asymptotically towards 20 °C.
- Both curves must be clearly labelled. [2]
Marking: 1 mark for correct shape (decreasing, asymptotic to 20 °C), 1 mark for correct relative steepness and labelling.
End of Answer Key