OxfordAQA International AS and A-level Physics: First law of thermodynamics, heating and conduction (9630) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions are aligned to the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, sections 3.11.1 (energy transfer by heating and doing work) and 3.11.2 (energy transfer by conduction). Section 3.11 is International A-level only.

Read the study guide and revision notes first. See also the course page, the checklist and the free diagnostics.

Questions

1. (a) Define the internal energy of a body. [2] (b) A gas is compressed very quickly in a well-insulated cylinder. State what happens to its internal energy. [1]

2. (a) A gas receives 130 J of energy by heating and does 45 J of work on its surroundings. Calculate the change in its internal energy. [2] (b) A different gas loses 60 J by heating while the surroundings do 25 J of work on it. Calculate the change in its internal energy. [2]

3. A copper block of mass 0.45 kg at 92 °C is lowered into 0.30 kg of water at 18 °C in an insulated container. The specific heat capacity of copper is 385 J kg⁻¹ K⁻¹ and of water is 4180 J kg⁻¹ K⁻¹. (a) Calculate the energy the copper would lose in cooling to 25 °C. [2] (b) Calculate the final common temperature, assuming no energy is lost to the surroundings. [4]

4. A heater of power 150 W melts 0.120 kg of ice at 0 °C. The specific latent heat of fusion of ice is 3.34 × 10⁵ J kg⁻¹. Assume all the heater’s energy goes into the ice. (a) Calculate the energy needed to melt the ice. [2] (b) Calculate the time taken. [2] (c) A thermometer in the ice and water stays at 0 °C throughout. Explain this in terms of the energies of the molecules. [2]

5. A radiator receives 0.065 kg of steam at 100 °C. The steam condenses and the water then cools to 45 °C. The specific latent heat of vaporisation of water is 2.26 × 10⁶ J kg⁻¹. Calculate the total energy released. [4]

6. A water heater has a power of 7.5 kW. Of this, 92% reaches the water. Water flows through it at a steady 0.0900 kg s⁻¹. The specific heat capacity of water is 4180 J kg⁻¹ K⁻¹. (a) Calculate the temperature rise of the water. [3] (b) The density of water is 1000 kg m⁻³. Calculate the volume of water heated in one minute, in litres. [1]

7. A student determines the specific heat capacity of a metal block of mass 0.500 kg by an electrical method. The heater runs at 11.6 V and 3.40 A for 300 s. The block warms from 21.5 °C to 48.0 °C. (a) Calculate the electrical energy supplied. [2] (b) Calculate the specific heat capacity of the metal. [2] (c) Explain why the student’s value is likely to be larger than the true value, and suggest one improvement to the procedure. [2] (d) Each temperature reading has an uncertainty of ±0.5 K. Calculate the percentage uncertainty in the temperature rise. [2]

8. A uniform aluminium rod has length 0.60 m and diameter 12 mm. Its ends are held at 100 °C and 20 °C. The thermal conductivity of aluminium is 205 W m⁻¹ K⁻¹. (a) Calculate the rate of energy transfer along the rod in the steady state. [3] (b) An identical rod is made of a steel with thermal conductivity 50 W m⁻¹ K⁻¹. Calculate the rate of energy transfer for the same end temperatures. [1] (c) State one condition needed for the rate in part (a) to be correct for a real rod. [1]

9. A glass pane has area 1.4 m² and thickness 6.0 mm. The thermal conductivity of the glass is 1.0 W m⁻¹ K⁻¹. The inner face is at 21 °C and the outer face is at 3 °C. (a) Calculate the rate of energy transfer through the pane. [2] (b) Calculate the U-value of the pane. [1]

10. A small house has these parallel surfaces. The inside is at 20 °C and the outside at 2 °C.

Surface	Area / m²	U-value / W m⁻² K⁻¹
Walls	38	0.30
Roof	52	0.18
Windows	6.5	2.8
Door	2.0	1.6

(a) Calculate the total rate of energy loss through the surfaces. [4] (b) Calculate the energy lost in one day, in kilowatt-hours. [2] (c) The windows are replaced with ones of U-value 1.4 W m⁻² K⁻¹. Calculate the reduction in the total rate of energy loss. [2]

11. A cylindrical hot-water tank holds 120 kg of water at 60 °C in a room at 18 °C. Its surface area is 1.8 m² and the U-value of its lagging is 0.90 W m⁻² K⁻¹. The specific heat capacity of water is 4180 J kg⁻¹ K⁻¹. (a) Calculate the initial rate of energy loss from the tank. [3] (b) Estimate the time for the water temperature to fall by 1.0 K at this rate. [2] (c) Explain why the next 1.0 K fall takes longer than the first. [2]

Answers

1. (a) The sum of the kinetic energies [1] and potential energies of the particles, which are randomly distributed [1]. [2] (b) It increases, because work is done on the gas and almost no energy leaves by heating (ΔU = W) [1]. [1] Examiner insight: Name both energies and the random distribution. A definition with only kinetic energy is incomplete.

2. (a) Q = +130 J and W = −45 J, because the gas does the work [1]. ΔU = 130 − 45 = +85 J [1]. [2] (b) Q = −60 J and W = +25 J [1]. ΔU = −60 + 25 = −35 J [1]. [2] Examiner insight: Write the sign of Q and of W before adding. A bare answer with a wrong sign leaves little to credit.

3. (a) Δθ = 92 − 25 = 67 K [1]. Q = 0.45 × 385 × 67 = 1.2 × 10⁴ J (11 608 J) [1]. [2] (b) Energy lost by copper = energy gained by water [1]. 0.45 × 385 × (92 − θ) = 0.30 × 4180 × (θ − 18), that is 173.25 (92 − θ) = 1254 (θ − 18) [1]. 15 939 + 22 572 = (173.25 + 1254) θ, so θ = 38 511 / 1427.25 [1]. θ = 27 °C [1]. [4] Examiner insight: State the energy balance in words first. The final temperature must lie between 18 °C and 92 °C.

4. (a) Q = ml = 0.120 × 3.34 × 10⁵ [1] = 4.0 × 10⁴ J [1]. [2] (b) t = Q / P = 40 080 / 150 [1] = 2.7 × 10² s (267 s) [1]. [2] (c) The energy supplied increases the potential energy of the molecules as they move apart [1]. Their average kinetic energy does not change, so the temperature is constant [1]. [2] Examiner insight: Say which energy changes and which does not. Mentioning only “breaking bonds” leaves out the kinetic energy.

5. Condensation stage: Q = ml = 0.065 × 2.26 × 10⁶ [1] = 146 900 J [1]. Cooling stage: Q = mcΔθ = 0.065 × 4180 × 55 [1] = 14 944 J. Total = 146 900 + 14 944 = 161 844 J = 1.6 × 10⁵ J [1]. [4] Examiner insight: Split a mixed question into stages and add them. Using mcΔθ throughout gives a far smaller value.

6. (a) Useful power = 0.92 × 7500 = 6900 W [1]. Δθ = P / ((Δm/Δt) c) = 6900 / (0.0900 × 4180) [1] = 18 K (18.3 K) [1]. [3] (b) Mass in 60 s = 0.0900 × 60 = 5.4 kg, so volume = 5.4 / 1000 = 5.4 × 10⁻³ m³ = 5.4 litres [1]. [1] Examiner insight: Apply the efficiency to the power before dividing by the flow rate and specific heat capacity.

7. (a) P = VI = 11.6 × 3.40 = 39.44 W [1]. E = Pt = 39.44 × 300 = 1.18 × 10⁴ J (11 832 J) [1]. [2] (b) Δθ = 48.0 − 21.5 = 26.5 K [1]. c = E / (mΔθ) = 11 832 / (0.500 × 26.5) = 893 J kg⁻¹ K⁻¹ [1]. [2] (c) Some of the electrical energy is lost to the surroundings, so the block gains less energy than VIt and Δθ is smaller than it should be [1]. Improvement: lag the block or fit a lid [1]. [2] (d) Uncertainty in Δθ = 0.5 + 0.5 = 1.0 K [1]. Percentage = 1.0 / 26.5 × 100 = 3.8% [1]. [2] Examiner insight: Link the direction of the error to the reason: less energy reaches the block, so Δθ is too small and c too large.

8. (a) A = π (0.0060)² = 1.13 × 10⁻⁴ m² [1]. Rate = kAΔθ/L = 205 × 1.13 × 10⁻⁴ × 80 / 0.60 [1] = 3.1 W [1]. [3] (b) Rate scales with k: 3.09 × 50 / 205 = 0.75 W [1]. [1] (c) The rod must be in a steady state, with its sides lagged so that energy only passes along the length [1]. [1] Examiner insight: The diameter is given, so halve it before finding the area, and convert millimetres to metres.

9. (a) Δθ = 18 K and L = 0.0060 m [1]. Rate = 1.0 × 1.4 × 18 / 0.0060 = 4.2 × 10³ W [1]. [2] (b) U = k/L = 1.0 / 0.0060 = 1.7 × 10² W m⁻² K⁻¹ (167) [1]. [1] Examiner insight: Convert the thickness to metres first. Units of W m⁻² K⁻¹ are needed with the value.

10. (a) Rate = (ΣUA) Δθ with Δθ = 18 K [1]. UA values: walls 11.4, roof 9.36, windows 18.2, door 3.2 W K⁻¹ [1]. Sum = 42.16 W K⁻¹ [1]. Rate = 42.16 × 18 = 7.6 × 10² W (759 W) [1]. [4] (b) Energy = 0.759 kW × 24 h [1] = 18 kWh (18.2 kWh) [1]. [2] (c) Reduction = 6.5 × (2.8 − 1.4) × 18 [1] = 1.6 × 10² W (164 W) [1]. [2] Examiner insight: Show each surface’s UA separately before adding. The windows give the largest share of the total from the smallest area.

11. (a) Δθ = 60 − 18 = 42 K [1]. UA = 0.90 × 1.8 = 1.62 W K⁻¹ [1]. Rate = 1.62 × 42 = 68 W [1]. [3] (b) Energy for a 1.0 K fall = 120 × 4180 × 1.0 = 501 600 J [1]. t = 501 600 / 68 = 7.4 × 10³ s (about 2.0 hours) [1]. [2] (c) As the water cools the temperature difference with the room falls [1], so the rate of energy loss falls and each 1.0 K step takes longer [1]. [2] Examiner insight: In part (c), refer to the temperature difference, not just “less heat”.

Where marks are usually lost

• Using the wrong sign for W when a gas expands.
• Saying kinetic energy rises during a change of state.
• Leaving thicknesses in millimetres or masses in grams.
• Using mcΔθ for a stage where the temperature is constant, or ml where it changes.
• Using the diameter as the radius when finding the area of a rod.
• Adding the U-values themselves instead of adding the products UA.

Next steps

• Recheck the revision notes and the study guide.
• Visit the OxfordAQA A-level Physics course page and the checklist.
• Take the free 10-minute diagnostics.
• Next, book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams from May/June 2020, published by OxfordAQA. Sections 3.11.1 and 3.11.2.