Secondary 4 Pure Physics Thermal Physics Quiz

Secondary 4 Pure Physics Quiz - Thermal Physics (Answer Key)

Total Marks: 50

Section A: Multiple Choice (Questions 1–5)

1. C [2 marks]

Teaching note: Internal energy is the sum of kinetic energy and potential energy of all molecules in a system. Temperature relates only to average kinetic energy. Heat refers to energy transfer, not stored energy.

Common mistake: Confusing heat (energy in transit) with internal energy (energy stored).

2. B [2 marks]

Working: $Q = mc\Delta\theta = 2.0 \times 900 \times (80-20) = 2.0 \times 900 \times 60 = 108\,000$ J

Teaching note: Use $\Delta\theta$ = final temperature − initial temperature. Always include units in calculation.

3. C [2 marks]

Teaching note: During melting, energy supplied breaks intermolecular bonds (increases potential energy) without changing average kinetic energy. Since temperature is proportional to average kinetic energy, temperature stays constant at the melting point.

Common mistake: Thinking temperature rise means kinetic energy increases — during phase change, this is not true.

4. D [2 marks]

Teaching note: A thermos flask minimizes all three heat transfer mechanisms: conduction (by vacuum/glass walls), convection (by preventing fluid movement), and radiation (by silvered surfaces that reflect infrared).

5. B [2 marks]

Teaching note: Specific latent heat of vaporisation is the energy required to change 1 kg of liquid to gas at constant temperature (boiling point). The value is for the phase change only, not for heating.

Section B: Short Answer and Structured Questions (Questions 6–15)

6. [2 marks]

Marking scheme: Any two distinct valid differences — 1 mark each.

7. [3 marks]

Metal is a good thermal conductor [1], while plastic is a poor thermal conductor/thermal insulator [1]. When the metal spoon is in hot soup, heat energy is quickly conducted from the soup through the metal to the handle, making it feel hot [1]. In the plastic spoon, heat transfer is slow, so the handle remains cooler.

Key concept: Thermal conductivity determines rate of heat transfer. Metals have free electrons that efficiently transfer kinetic energy.

8. [3 marks]

$Q = mc\Delta\theta$ [1]

$Q = 0.50 \times 900 \times (55-25)$ [1]

$Q = 0.50 \times 900 \times 30 = 13\,500$ J [1]

Teaching note: Always write the formula first, then substitute with units, then calculate.

9. [3 marks]

Method: Electrical heating method

Measurements needed:

• Mass of metal block, $m$ (using balance) [1]
• Initial and final temperatures, using thermometer (or temperature change $\Delta\theta$) [1]
• Current $I$ and potential difference $V$ of heater, and heating time $t$ [1]

Formula: Electrical energy supplied = thermal energy gained (assuming no heat loss) $VIt = mc\Delta\theta$

Therefore: $c = \frac{VIt}{m\Delta\theta}$

Common improvement: Insulate the block; use lid to minimize heat loss to surroundings.

10. [3 marks]

The horizontal plateau at 0°C represents melting (solid to liquid phase change) [1]. During melting, the energy supplied is used to break intermolecular bonds and increase the potential energy of molecules [1], not to increase their kinetic energy. Since temperature is proportional to average kinetic energy, the temperature remains constant [1].

Key concept: Latent heat — energy for phase change without temperature change.

11. [4 marks]

(a) Energy to melt ice: $Q = mL_f = 0.20 \times 334\,000 = 66\,800 \text{ J}$ [2]

(If using 334 kJ kg⁻¹: Q = 0.20 × 334 = 66.8 kJ = 66 800 J)

(b) Heat lost by warm water = heat gained by ice (to melt) + heat gained by melted ice (to warm up)

Let final temperature be $T$: $0.80 \times 4200 \times (40-T) = 66\,800 + 0.20 \times 4200 \times (T-0)$ [1]

$1344 \times (40-T) = 66\,800 + 840T$

$53\,760 - 1344T = 66\,800 + 840T$

$53\,760 - 66\,800 = 1344T + 840T$

$-13\,040 = 2184T$

This gives negative $T$, which is impossible. [1]

Interpretation: The water does not have enough energy to melt all the ice. Final temperature will be 0°C with some ice remaining unmelted.

Alternative correct answer: If student identifies that not all ice melts, award marks for correct reasoning.

12. [4 marks]

(a) Face A (matte black) [1]

(b) Matte black surfaces are good emitters of infrared radiation [1]. Shiny or light surfaces are poor emitters as they reflect radiation rather than absorb and re-emit it [1]. At the same temperature, a matte black surface emits thermal radiation at a higher rate than shiny or white surfaces [1], so the detector records a higher reading.

Key concept: Surface color and texture affect emissivity. Dark, rough surfaces ≈ black body (good emitter/absorber). Shiny/metallic surfaces = poor emitter, good reflector.

13. [4 marks]

(a) Energy supplied: $E = P \times t = 2200 \times (5 \times 60) = 2200 \times 300 = 660\,000 \text{ J}$ [2]

(Or 660 kJ)

(b) Energy required to heat water: $Q = mc\Delta\theta = 1.5 \times 4200 \times (100-20) = 1.5 \times 4200 \times 80 = 504\,000 \text{ J}$ [1]

$\text{Efficiency} = \frac{\text{useful energy output}}{\text{total energy input}} \times 100\% = \frac{504\,000}{660\,000} \times 100\% = 76.4\%$ [1]

Or accept 76% to 77% depending on rounding.

14. [4 marks]

(a) Evaporation occurs at the surface when faster-moving molecules escape [1]; it can happen at any temperature because there are always some molecules with enough energy [1]. Boiling occurs throughout the liquid at one specific temperature (boiling point) when the vapour pressure equals atmospheric pressure [1].

(Max 2 marks as allocated)

(b) During evaporation, the fastest-moving molecules escape from the surface [1]. This removes the molecules with highest kinetic energy, so the average kinetic energy of remaining molecules decreases [1], causing cooling.

15. [3 marks]

When the piston compresses the gas, the volume decreases so molecules are closer together [1]. The frequency of collisions with the walls increases [1]. Since temperature is constant, average molecular speed stays the same, but more frequent collisions means greater rate of momentum change on walls, hence higher pressure [1].

Key concept: Pressure = force/area; force arises from rate of change of momentum from molecular collisions.

Section C: Data Analysis and Extended Response (Questions 16–20)

16. [4 marks]

(a) Aluminium has a higher specific heat capacity (900 vs 390 J kg⁻¹ °C⁻¹) [1]. This means it can store more thermal energy for the same temperature rise, providing more even and sustained heating of food [1]. While copper conducts faster, aluminium's lower density and better heat storage make it practical for cookware.

(b) Wood has very low thermal conductivity (0.15 W m⁻¹ K⁻¹) [1], making it a good thermal insulator. It minimizes heat conduction from the hot pot to the hand, preventing burns and allowing safe handling [1].

17. [4 marks]

(a) Two factors to keep constant:

• Initial temperature of hot water [1]
• Volume/mass of water in each beaker [1]

(Other valid answers: room temperature, surface area exposed, type of container material if not the variable being tested, starting time)

(b) Sketch should show:

• Both curves starting at 80°C and decreasing exponentially to approach 25°C [1]
• Curve for Beaker B (bare) falling more steeply than Beaker A (wrapped) at all times [1]
• Both curves asymptotic to room temperature line

<image_placeholder> id: Q17-ans-fig1 type: graph linked_question: 17 description: Two temperature-time curves for cooling water, showing bare beaker cooling faster than insulated beaker labels: Temperature (°C) on y-axis; Time (min) on x-axis; room temperature 25°C dotted line; Curve A (insulated, slower cooling); Curve B (bare, faster cooling) values: Both start at 80°C, asymptote to 25°C; Curve B below Curve A at all t > 0 must_show: Initial common point; Curve B steeper initial slope; both approaching ambient temperature; clear labeling of which curve is which </image_placeholder>

18. [4 marks]

Energy released in condensation: $Q_1 = mL_v = 0.10 \times 2260\,000 = 226\,000 \text{ J}$ [1]

Energy released in cooling water: $Q_2 = mc\Delta\theta = 0.10 \times 4200 \times (100-20) = 0.10 \times 4200 \times 80 = 33\,600 \text{ J}$ [1]

Total energy released: $Q_{\text{total}} = Q_1 + Q_2 = 226\,000 + 33\,600 = 259\,600 \text{ J}$ [1]

Or 260 kJ or 2.60 × 10⁵ J (to 3 sig fig) [1]

Teaching note: Two separate calculations needed — condensation is a large energy contribution. Don't forget the temperature change of the resulting water.

19. [4 marks]

Energy transfers from Sun to Earth by radiation (electromagnetic waves/infrared radiation) [1]. This is the only mechanism possible because space is a vacuum [1]. Conduction requires a medium with particles to transfer kinetic energy through collisions [1]. Convection requires a fluid medium where heated particles can move and carry energy [1]. Neither can occur in the vacuum of space.

20. [5 marks]

(a) Water has a high specific heat capacity (4200 J kg⁻¹ °C⁻¹) [1]. This means it can absorb large amounts of thermal energy with only a small temperature rise, making it effective at removing heat from the engine without boiling [1].

(b) Temperature drop of coolant: $\Delta\theta = 90 - 80 = 10$°C [1]

Energy to be removed per cycle: $Q = mc\Delta\theta = 4.0 \times 4200 \times 10 = 168\,000 \text{ J}$ [1]

Rate of heat removal (power): $P = \frac{Q}{t} = \frac{168\,000}{30} = 5600 \text{ W} = 5.6 \text{ kW}$ [1]

END OF ANSWER KEY