A Level H1 Physics Thermal Physics Quiz

A-Level Physics H1 Quiz - Thermal Physics — Answer Key and Marking Scheme

Total Marks: 50

---

Section A: Short Answer and Conceptual Questions (1-5)

1. State the relationship between the Celsius temperature scale and the thermodynamic (Kelvin) temperature scale.

[2 marks]

Answer:

• T (in K) = θ (in °C) + 273.15 [B1]
• The size of one degree interval is the same on both scales / A change of 1 °C equals a change of 1 K [B1]

Accept 273 instead of 273.15.

---

2. Define the term specific heat capacity of a substance.

[2 marks]

Answer:

• The specific heat capacity of a substance is the amount of thermal energy required to raise the temperature of unit mass (1 kg) of the substance by 1 K (or 1 °C). [B2]

Award [B1] for "energy to raise temperature" without specifying unit mass or unit temperature change.

---

3. A student claims that when two objects at different temperatures are placed in thermal contact, the final temperature is always exactly halfway between the two initial temperatures. Explain why this claim is incorrect.

[2 marks]

Answer:

• The final temperature depends on the masses and specific heat capacities of the two objects, not just their initial temperatures. [B1]
• The claim would only be true if the two objects had equal mass and equal specific heat capacity (i.e., equal thermal capacity). [B1]

Accept: "Thermal energy gained by the colder object equals thermal energy lost by the hotter object. The temperature change depends on mass × specific heat capacity."

---

4. State the first law of thermodynamics in words.

[2 marks]

Answer:

• The increase in internal energy of a system is equal to the sum of the heat supplied to the system and the work done on the system. [B2]

Accept equivalent wording: ΔU = Q + W stated in words. Award [B1] if only the equation is given without explanation.

---

5. Explain why the temperature of a substance remains constant during a change of state, even though thermal energy continues to be supplied.

[2 marks]

Answer:

• During a change of state, the thermal energy supplied is used to overcome the intermolecular forces/bonds between particles, rather than to increase their kinetic energy. [B1]
• Since temperature is a measure of the average kinetic energy of the particles, and the kinetic energy does not increase during the change of state, the temperature remains constant. [B1]

---

Section B: Calculations and Data Analysis (6-10)

6. Distinguish between the terms heat and temperature.

[2 marks]

Answer:

• Heat (thermal energy) is the energy transferred from one body to another due to a temperature difference. It is measured in joules (J). [B1]
• Temperature is a measure of the average kinetic energy of the particles in a substance. It is measured in kelvin (K) or degrees Celsius (°C). It determines the direction of heat flow. [B1]

---

7. Explain what is meant by thermal equilibrium.

[2 marks]

Answer:

• Thermal equilibrium is the state in which two or more bodies in thermal contact have the same temperature. [B1]
• There is no net transfer of thermal energy between them. [B1]

---

8. A copper block of mass 0.500 kg is heated from 25.0 °C to 85.0 °C. The specific heat capacity of copper is 385 J kg⁻¹ K⁻¹.

(a) Calculate the thermal energy absorbed by the copper block.

[2 marks]

Working:

• Δθ = 85.0 − 25.0 = 60.0 °C (equivalent to 60.0 K) [M1]
• Q = mcΔθ = 0.500 × 385 × 60.0
• Q = 11 550 J ≈ 1.16 × 10⁴ J [A1]

Answer: 1.16 × 10⁴ J (or 11.6 kJ)

---

(b) The same amount of energy is supplied to an equal mass of water, initially at 25.0 °C. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹. Calculate the final temperature of the water.

[3 marks]

Working:

• Q = mcΔθ → Δθ = Q / (mc) [M1]
• Δθ = 11 550 / (0.500 × 4200) = 11 550 / 2100 [M1]
• Δθ = 5.50 °C
• Final temperature = 25.0 + 5.50 = 30.5 °C [A1]

Answer: 30.5 °C

---

9. An electric kettle rated at 2.20 kW contains 1.50 kg of water at 20.0 °C. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.

(a) Calculate the minimum time required to bring the water to its boiling point of 100 °C, assuming no heat losses.

[3 marks]

Working:

• Energy required: Q = mcΔθ = 1.50 × 4200 × (100 − 20.0) = 1.50 × 4200 × 80.0 [M1]
• Q = 504 000 J
• Power P = 2.20 kW = 2200 W
• Time t = Q / P = 504 000 / 2200 [M1]
• t = 229.09... s ≈ 229 s (or 3 min 49 s) [A1]

Answer: 229 s

---

(b) In practice, the time taken is longer than the calculated value. Suggest two reasons for this.

[2 marks]

Answer (any two, 1 mark each):

• Heat is lost to the surroundings (air, kettle body). [B1]
• Some energy is used to heat the kettle itself (the heating element and container). [B1]
• The kettle does not operate at 100% efficiency. [B1]
• Some water evaporates before reaching boiling point. [B1]

---

10. A student investigates the specific latent heat of vaporisation of water. An electric heater of power 50.0 W is used to heat water at its boiling point. In 300 s, the mass of water decreases by 22.0 g.

(a) Calculate the specific latent heat of vaporisation of water from these results.

[3 marks]

Working:

• Energy supplied: Q = Pt = 50.0 × 300 = 15 000 J [M1]
• Mass vaporised: m = 22.0 g = 0.0220 kg [M1]
• Lᵥ = Q / m = 15 000 / 0.0220 = 681 818 J kg⁻¹ ≈ 6.82 × 10⁵ J kg⁻¹ [A1]

Answer: 6.82 × 10⁵ J kg⁻¹

---

(b) The accepted value for the specific latent heat of vaporisation of water is 2.26 × 10⁶ J kg⁻¹. Calculate the percentage difference between the experimental result and the accepted value.

[2 marks]

Working:

• Difference = |6.82 × 10⁵ − 2.26 × 10⁶| = 1.578 × 10⁶ J kg⁻¹ [M1]
• Percentage difference = (1.578 × 10⁶ / 2.26 × 10⁶) × 100% = 69.8% ≈ 70% [A1]

Answer: 69.8% (or 70%)

---

(c) Suggest one reason for the difference between the experimental and accepted values.

[1 mark]

Answer (any one):

• Heat loss to the surroundings during the experiment. [B1]
• Not all the energy from the heater was transferred to the water (some heated the container). [B1]
• Some water may have been lost by splashing rather than vaporisation. [B1]

---

Section C: Structured and Data-Based Questions (11-15)

11. A metal cylinder of mass 0.800 kg at a temperature of 200 °C is placed into 0.400 kg of water at 20.0 °C in a well-insulated container. The final temperature of the water and cylinder is 38.0 °C. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.

(a) Calculate the thermal energy gained by the water.

[2 marks]

Working:

• Δθ_water = 38.0 − 20.0 = 18.0 °C (or 18.0 K) [M1]
• Q_water = mcΔθ = 0.400 × 4200 × 18.0 = 30 240 J ≈ 3.02 × 10⁴ J [A1]

Answer: 3.02 × 10⁴ J

---

(b) Hence, determine the specific heat capacity of the metal.

[4 marks]

Working:

• Energy lost by metal = Energy gained by water (assuming no heat loss to surroundings) [M1]
• Q_metal = 30 240 J
• Δθ_metal = 200 − 38.0 = 162 °C (or 162 K) [M1]
• Q_metal = m_metal × c_metal × Δθ_metal
• 30 240 = 0.800 × c_metal × 162 [M1]
• c_metal = 30 240 / (0.800 × 162) = 30 240 / 129.6 = 233.3... J kg⁻¹ K⁻¹ ≈ 233 J kg⁻¹ K⁻¹ [A1]

Answer: 233 J kg⁻¹ K⁻¹

---

12. Cooling curve investigation.

(a) On the grid below, plot a graph of temperature against time. Draw a smooth curve through the points.

[3 marks]

Marking:

• Axes correctly labelled with units (Temperature / °C on y-axis, Time / min on x-axis) [B1]
• All 9 points plotted accurately (± half a small square) [B1]
• Smooth curve drawn through points (not dot-to-dot straight lines) [B1]

---

(b) Use your graph to determine the rate of cooling at time t = 3 minutes. Show your working on the graph.

[3 marks]

Marking:

• Tangent drawn at t = 3 min on the curve [B1]
• Large triangle used to determine gradient [B1]
• Rate of cooling = −gradient (e.g., approx. −7 °C min⁻¹, accept 6.5 to 7.5 °C min⁻¹) with correct units [A1]

---

(c) Explain why the rate of cooling decreases as the liquid cools.

[2 marks]

Answer:

• The rate of cooling depends on the temperature difference between the liquid and the surroundings. [B1]
• As the liquid cools, the temperature difference decreases, so the rate of heat transfer to the surroundings decreases. [B1]

---

13. The kinetic theory of gases.

(a) State two assumptions of the kinetic theory of an ideal gas.

[2 marks]

Answer (any two, 1 mark each):

• The gas consists of a large number of identical molecules in random motion.
• The volume of the molecules is negligible compared to the volume of the container.
• There are no intermolecular forces between molecules (except during collisions).
• Collisions between molecules and with the walls are perfectly elastic.
• The duration of a collision is negligible compared to the time between collisions.

---

(b) Explain, in terms of molecular motion, why the pressure exerted by a gas on the walls of its container increases when the temperature of the gas is raised at constant volume.

[2 marks]

Answer:

• At a higher temperature, the average kinetic energy (and hence speed) of the molecules increases. [B1]
• The molecules collide with the walls more frequently and with greater momentum change per collision, resulting in a larger force (and hence pressure) on the walls. [B1]

---

14. A gas expands, doing 150 J of work on the surroundings, while 200 J of thermal energy is supplied.

(a) Calculate the change in internal energy of the gas.

[2 marks]

Working:

• ΔU = Q + W (where W is work done ON the gas) [M1]
• Work done BY gas = 150 J, so work done ON gas = −150 J
• ΔU = 200 + (−150) = 50 J [A1]

Answer: 50 J

---

(b) State and explain whether the temperature of the gas increases, decreases, or remains constant.

[2 marks]

Answer:

• The temperature increases. [B1]
• The internal energy of an ideal gas is proportional to its absolute temperature. Since the internal energy increases (ΔU > 0), the temperature must increase. [B1]

---

15. Describe an experiment to determine the specific heat capacity of a metal block.

[4 marks]

Answer:

• Diagram: Labelled diagram showing metal block with two holes for heater and thermometer, lagging, electrical circuit with ammeter, voltmeter, and power supply. [B1]
• Measurements: Mass of block (m), initial temperature (θ₁), final temperature (θ₂), current (I), potential difference (V), and time (t) for which heater is on. [B1]
• Calculation: Electrical energy supplied = VIt. Assuming no heat losses, thermal energy gained by block = mc(θ₂ − θ₁). [B1]
• Equate: VIt = mc(θ₂ − θ₁) → c = VIt / [m(θ₂ − θ₁)]. [B1]

Accept any valid experimental method. Award marks for clear description of procedure, measurements, and calculation.

---

Section D: Further Calculations and Applications (16-20)

16. Aluminium block heated by an electric heater.

(a) Calculate the total energy supplied by the heater.

[2 marks]

Working:

• Time = 5.0 min = 300 s [M1]
• Energy = Pt = 50 × 300 = 15 000 J [A1]

Answer: 15 000 J

---

(b) Calculate the specific heat capacity of aluminium.

[3 marks]

Working:

• Δθ = 64.0 − 22.0 = 42.0 °C (or 42.0 K) [M1]
• Q = mcΔθ → c = Q / (mΔθ) [M1]
• c = 15 000 / (0.250 × 42.0) = 15 000 / 10.5 = 1428.6... J kg⁻¹ K⁻¹ ≈ 1430 J kg⁻¹ K⁻¹ [A1]

Answer: 1430 J kg⁻¹ K⁻¹

---

17. Solar panel heating water.

(a) Calculate the total power received by the solar panel.

[2 marks]

Working:

• Power = Intensity × Area [M1]
• P = 800 × 2.0 = 1600 W [A1]

Answer: 1600 W

---

(b) If the panel transfers 60% of the received power to the water, calculate the mass of water heated per second.

[3 marks]

Working:

• Useful power = 0.60 × 1600 = 960 W [M1]
• Energy per second = 960 J
• Q = mcΔθ → m = Q / (cΔθ) [M1]
• Δθ = 45 − 15 = 30 °C (or 30 K)
• m = 960 / (4200 × 30) = 960 / 126 000 = 0.007619... kg ≈ 7.6 × 10⁻³ kg (or 7.6 g) [A1]

Answer: 7.6 × 10⁻³ kg s⁻¹ (or 7.6 g s⁻¹)

---

18. Ice melting in water.

(a) Calculate the energy required to melt all the ice.

[2 marks]

Working:

• Q = mL_f [M1]
• Q = 0.050 × 3.34 × 10⁵ = 16 700 J [A1]

Answer: 1.67 × 10⁴ J

---

(b) Calculate the final temperature of the water after all the ice has melted.

[4 marks]

Working:

• Energy lost by warm water cooling from 25.0 °C to final temperature θ: Q_lost = 0.300 × 4200 × (25.0 − θ) [M1]
• Energy gained by melted ice (now water at 0 °C) warming to θ: Q_gained = 0.050 × 4200 × (θ − 0) [M1]
• Energy balance: Energy lost by warm water = Energy to melt ice + Energy to warm melted ice 0.300 × 4200 × (25.0 − θ) = 16 700 + 0.050 × 4200 × θ [M1]
• 1260 × (25.0 − θ) = 16 700 + 210θ 31 500 − 1260θ = 16 700 + 210θ 31 500 − 16 700 = 1260θ + 210θ 14 800 = 1470θ θ = 14 800 / 1470 = 10.068... °C ≈ 10.1 °C [A1]

Answer: 10.1 °C

---

19. Gas heated at constant volume.

(a) Convert the initial and final temperatures to kelvin.

[1 mark]

Answer:

• T₁ = 27 + 273 = 300 K
• T₂ = 127 + 273 = 400 K [B1]

---

(b) Assuming the gas behaves ideally, calculate the final pressure of the gas.

[2 marks]

Working:

• At constant volume, p₁/T₁ = p₂/T₂ [M1]
• p₂ = p₁ × (T₂/T₁) = 1.00 × 10⁵ × (400/300) = 1.333... × 10⁵ Pa ≈ 1.33 × 10⁵ Pa [A1]

Answer: 1.33 × 10⁵ Pa

---

20. Pressure-temperature investigation.

(a) Plot a graph of pressure against temperature in °C. Draw the best-fit straight line.

[3 marks]

Marking:

• Axes correctly labelled with units (Pressure / kPa on y-axis, Temperature / °C on x-axis) [B1]
• All 5 points plotted accurately [B1]
• Best-fit straight line drawn through points [B1]

---

(b) Use your graph to determine the value of absolute zero in °C. Explain your method.

[3 marks]

Answer:

• Extend (extrapolate) the best-fit straight line backwards until it crosses the temperature axis (where pressure = 0). [B1]
• Read the temperature value at the intercept. [B1]
• The value should be approximately −273 °C (accept −270 °C to −280 °C depending on graph accuracy). [A1]

Method: The pressure of an ideal gas is proportional to absolute temperature. At absolute zero, the pressure would be zero. Extrapolating the graph to p = 0 gives the temperature of absolute zero on the Celsius scale.

---

END OF ANSWER KEY