Secondary 3 Physics Thermal Physics Quiz

Secondary 3 Physics Quiz - Thermal Physics (Answer Key)

Total Marks: 40

Section A: Multiple Choice Questions (10 marks)

1. [1 mark] Answer: B

Explanation: Internal energy is the sum of kinetic and potential energies of all molecules in a substance. During melting at constant temperature, heat energy (latent heat) is supplied to overcome intermolecular forces, increasing the potential energy component of internal energy while kinetic energy (temperature) remains constant.
Common mistake: Option A is incorrect because internal energy includes potential energy. Option C is incorrect because internal energy also depends on mass and state. Option D confuses internal energy (a property of the system) with heat (energy in transit).

2. [1 mark] Answer: B

Explanation: Water has a much higher specific heat capacity ($4200 \text{ J/(kg·°C)}$) than copper ($385 \text{ J/(kg·°C)}$). The water also has a larger mass (500 g vs 200 g). The heat capacity ($mc$) of water is $0.5 \times 4200 = 2100 \text{ J/°C}$, while for copper it is $0.2 \times 385 = 77 \text{ J/°C}$. The final temperature will be much closer to the initial water temperature (20°C) because water can absorb much more heat per degree temperature change.

3. [1 mark] Answer: B

Explanation: $Q = mc\Delta\theta$. For equal $m$ and $Q$, $\Delta\theta \propto \frac{1}{c}$. Since $c_X = 900$ and $c_Y = 450$, $c_X = 2c_Y$. Therefore $\Delta\theta_Y = 2\Delta\theta_X$. Substance Y has half the specific heat capacity, so its temperature rises twice as much for the same heat input.

4. [1 mark] Answer: B

Explanation: During boiling, the supplied heat energy (latent heat of vaporisation) is used to overcome intermolecular forces holding molecules together in the liquid phase and to do work against atmospheric pressure as the volume expands significantly during the phase change to gas. The kinetic energy of molecules (and thus temperature) does not increase during the phase change.

5. [1 mark] Answer: C

Explanation: Convection is heat transfer through fluid motion. Warm air rising above a radiator is a classic example of natural convection. Option A is radiation (through vacuum). Option B is conduction (through solid metal). Option D describes a vacuum which prevents conduction and convection; heat transfer in a thermos flask vacuum is by radiation only.

6. [1 mark] Answer: A

Explanation: Electrical energy supplied = $VIt$. This energy melts mass $m$ of ice: $VIt = mL$. Therefore $L = \frac{VIt}{m}$.

7. [1 mark] Answer: B

Explanation: The plateau on a cooling curve represents the phase change from liquid to solid (freezing). The temperature remains constant at the freezing point during this phase change. The graph shows a plateau at 40°C, so the freezing point is 40°C.

8. [1 mark] Answer: A

Explanation: Matt black surfaces are good emitters of infrared radiation, while shiny white surfaces are poor emitters. Both cans lose heat by radiation (and convection/conduction). The matt black can radiates heat faster, so its water cools more quickly.

9. [1 mark] Answer: A

Explanation: A bimetallic strip bends because the two metals have different coefficients of thermal expansion. The metal with the higher expansion coefficient (brass) expands more and becomes the outer curve when heated. Brass expands more than steel for the same temperature rise.

10. [1 mark] Answer: B

Explanation: A thermocouple has a very small sensing junction (low mass, low heat capacity), so it reaches thermal equilibrium with the measured object very quickly. Liquid-in-glass thermometers have a larger bulb with more thermal mass, resulting in slower response times.

Section B: Structured Questions (18 marks)

11. [3 marks]

(a) [1 mark] Specific heat capacity of a substance is the amount of heat energy required to raise the temperature of 1 kg of the substance by 1°C (or 1 K).
Accept: "per unit mass per degree Celsius/Kelvin"

(b) [2 marks]
$Q = mc\Delta\theta$
$Q = 0.5 \times 900 \times (75 - 25)$
$Q = 0.5 \times 900 \times 50$
$Q = 22\,500 \text{ J}$ (or $22.5 \text{ kJ}$)
1 mark for correct substitution, 1 mark for correct answer with unit

12. [4 marks]

(a) [2 marks]
$Q = mc\Delta\theta$
$Q = 1.5 \times 4200 \times (100 - 20)$
$Q = 1.5 \times 4200 \times 80$
$Q = 504\,000 \text{ J}$ (or $504 \text{ kJ}$)
1 mark for correct substitution, 1 mark for correct answer with unit

(b) [1 mark]
$P = \frac{Q}{t} \Rightarrow t = \frac{Q}{P}$
$t = \frac{504\,000}{2000} = 252 \text{ s}$ (or $4 \text{ min } 12 \text{ s}$)
1 mark for correct answer with unit

(c) [1 mark] Heat losses to the surroundings / heat absorbed by the kettle itself / some energy used to heat the kettle body / not all electrical energy converted to heat in water.
Accept any reasonable heat loss explanation.

13. [4 marks]

(a) [1 mark]
$E = VIt = 12 \times 3.0 \times 300 = 10\,800 \text{ J}$

(b) [2 marks]
$E = mL \Rightarrow L = \frac{E}{m}$
$m = 18 \text{ g} = 0.018 \text{ kg}$
$L = \frac{10\,800}{0.018} = 600\,000 \text{ J/kg} = 6.0 \times 10^5 \text{ J/kg}$
1 mark for correct mass conversion and substitution, 1 mark for correct answer with unit

(c) [1 mark] Heat losses to surroundings (not all electrical energy goes into vaporising water) / some steam escapes without being condensed / condensation incomplete / heat absorbed by apparatus / measurement errors in mass/time/voltage/current.
Accept any valid experimental error/heat loss reason.

14. [3 marks]

(a) [2 marks] During melting, the supplied heat energy (latent heat of fusion) is used to overcome the intermolecular forces holding the molecules in fixed positions in the solid lattice. The energy increases the potential energy of the molecules as they gain freedom to move past each other, but the average kinetic energy of the molecules (which determines temperature) does not change. Hence temperature remains constant.

(b) [1 mark] In a solid, molecules are closely packed in a fixed, ordered arrangement (lattice) and vibrate about fixed positions. In a liquid, molecules are still closely packed but have no long-range order; they can move past each other and slide over one another.

15. [4 marks]

(a) [2 marks]
Heat gained by water = $m_w c_w \Delta\theta_w$
$= 0.200 \times 4200 \times (35 - 25)$
$= 0.200 \times 4200 \times 10$
$= 8400 \text{ J}$
1 mark for correct substitution, 1 mark for correct answer with unit

(b) [2 marks]
Heat lost by metal = Heat gained by water (no heat losses)
$m_m c_m \Delta\theta_m = 8400$
$0.100 \times c_m \times (150 - 35) = 8400$
$0.100 \times c_m \times 115 = 8400$
$c_m = \frac{8400}{0.100 \times 115} = \frac{8400}{11.5} = 730 \text{ J/(kg·°C)}$ (or $730.4 \text{ J/(kg·°C)}$)
1 mark for correct principle and substitution, 1 mark for correct answer with unit

Section C: Longer Structured Questions (12 marks)

16. [6 marks]

(a) [2 marks]
Graph sketch requirements:

• Two curves starting at (0, 80°C)
• Curve A (uncovered) steeper, reaching ~35°C at 20 min
• Curve B (covered) less steep, reaching ~50°C at 20 min
• Both curves labelled clearly
• Room temperature (25°C) indicated as horizontal asymptote 1 mark for correct shape and labelling of both curves, 1 mark for correct relative steepness and asymptotic approach to room temperature

(b) [2 marks] The lid reduces heat loss by convection (traps air above water, preventing warm air from rising and being replaced by cooler air) and evaporation (prevents water vapour from escaping, which would carry away latent heat). The uncovered beaker loses heat by convection, evaporation, and radiation from the water surface, while the covered beaker mainly loses heat by conduction through the lid and radiation from the outer lid surface.

(c) [2 marks]
$Q = mc\Delta\theta = 0.200 \times 4200 \times (80 - 72) = 0.200 \times 4200 \times 8 = 6720 \text{ J}$
Time = 2 minutes = 120 s
Rate of heat loss = $\frac{Q}{t} = \frac{6720}{120} = 56 \text{ W}$
1 mark for correct heat energy calculation, 1 mark for correct rate with unit

17. [6 marks]

(a) [1 mark] Black surfaces are good absorbers of radiation. Painting the panel black maximises absorption of solar radiation (visible and infrared), converting it to thermal energy to heat the water.

(b) [2 marks] The glass cover creates a greenhouse effect: it allows short-wavelength solar radiation to pass through and be absorbed by the black panel, but traps long-wavelength infrared radiation re-emitted by the hot panel (glass is opaque to IR). It also reduces convection losses by trapping a layer of air above the panel.

(c) [3 marks]
Incident solar power = $800 \times 2.5 = 2000 \text{ W}$
Power absorbed by water = $0.60 \times 2000 = 1200 \text{ W}$
Energy per second = $1200 \text{ J/s}$
Mass flow rate = $0.05 \text{ kg/s}$
$Q = mc\Delta\theta \Rightarrow \Delta\theta = \frac{Q}{mc} = \frac{1200}{0.05 \times 4200} = \frac{1200}{210} = 5.71 \text{ °C}$
1 mark for incident power, 1 mark for absorbed power, 1 mark for correct $\Delta\theta$ with unit

18. [5 marks]

(a) [2 marks]
$Q = mc\Delta\theta = 0.8 \times 1200 \times (60 - 20) = 0.8 \times 1200 \times 40 = 38\,400 \text{ J}$
$P = 2000 \text{ W}$
$t = \frac{Q}{P} = \frac{38\,400}{2000} = 19.2 \text{ s}$
1 mark for correct heat energy, 1 mark for correct time with unit

(b) [2 marks]
$Q = mL = 0.8 \times 1.5 \times 10^5 = 120\,000 \text{ J}$
$t = \frac{Q}{P} = \frac{120\,000}{2000} = 60 \text{ s}$
1 mark for correct heat energy, 1 mark for correct time with unit

(c) [1 mark]
*
Graph sketch requirements:

• Axes labelled: Temperature (°C) vertical, Time (s) horizontal
• Line from (0, 20) to (19.2, 60) with positive slope
• Horizontal line from (19.2, 60) to (79.2, 60)
• Key points labelled: 20°C, 60°C, 19.2 s, 79.2 s 1 mark for correct shape with labels

19. [5 marks]

(a) [1 mark] Argon has a lower thermal conductivity ($0.016 \text{ W/(m·K)}$) than air ($0.026 \text{ W/(m·K)}$), so it reduces conductive heat transfer across the gap more effectively, improving the window's insulation.

(b) [4 marks]
Thermal resistance of one glass pane: $R_g = \frac{d}{kA} = \frac{0.004}{1.0 \times 2.0} = 0.002 \text{ K/W}$
Two glass panes: $2 \times 0.002 = 0.004 \text{ K/W}$
Thermal resistance of argon gap: $R_{Ar} = \frac{0.015}{0.016 \times 2.0} = \frac{0.015}{0.032} = 0.46875 \text{ K/W}$
Total resistance: $R_{total} = 0.004 + 0.46875 = 0.47275 \text{ K/W}$
Heat conduction rate: $\frac{\Delta T}{R_{total}} = \frac{20}{0.47275} = 42.3 \text{ W}$
1 mark for glass resistance, 1 mark for argon gap resistance, 1 mark for total resistance, 1 mark for final answer with unit

20. [5 marks]

(a) [2 marks]
Electrical energy supplied = $Pt = 50 \times (10 \times 60) = 50 \times 600 = 30\,000 \text{ J}$
$Q = mc\Delta\theta \Rightarrow c = \frac{Q}{m\Delta\theta} = \frac{30\,000}{1.0 \times (65 - 25)} = \frac{30\,000}{40} = 750 \text{ J/(kg·°C)}$
1 mark for correct energy, 1 mark for correct specific heat capacity with unit

(b) [1 mark]
Percentage difference = $\frac{|750 - 450|}{450} \times 100\% = \frac{300}{450} \times 100\% = 66.7\%$
1 mark for correct calculation with unit

(c) [2 marks]

1. Insulate the metal block (e.g., wrap with polystyrene/lagging) to reduce heat losses to surroundings.
2. Use a thermometer with smaller heat capacity / better thermal contact (e.g., digital probe in a drilled hole with oil) to measure block temperature more accurately.
   1 mark each for any two valid improvements addressing heat loss or measurement accuracy

Marking Notes:

• Allow ecf (error carried forward) where appropriate in multi-part calculations.
• Units must be included in final answers for full marks.
• Significant figures: typically 2-3 SF acceptable unless specified.
• For graph questions, shape, labels, and key values must be clear.
• Molecular explanations should reference kinetic/potential energy and intermolecular forces.