OCR GCSE Physics: Matter — 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 available free from your board.

Related: Matter revision notes

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Questions

1. Describe how you would determine the density of (a) a regular block, (b) an irregular object, (c) a liquid. [6]

2. Explain why a change of state is a physical rather than a chemical change. [3]

3. A heating curve for a solid being heated at constant power shows two flat sections.

(a) State what each flat section represents. [2] (b) Explain why the temperature does not rise during these sections. [3] (c) Explain why the second flat section is longer than the first. [2]

4. 0.25 kg of a metal is heated by 2400 J and its temperature rises by 24 °C.

(a) Calculate the specific heat capacity. [3] (b) State one reason the experimental value would be higher than the true value. [2]

5. Explain, in terms of particles, why:

(a) the pressure of a gas increases when it is compressed at constant temperature [2] (b) the pressure increases when it is heated at constant volume [3]

6. Explain why a bicycle pump becomes warm when air is compressed inside it. [2]

7. Explain why the pressure in a liquid increases with depth. [2]

8. The specific latent heat of fusion of ice is 334 000 J/kg. Calculate the energy needed to melt 0.40 kg of ice at 0 °C. [2]

9. A gas has a pressure of 100 kPa in a container of volume 0.60 m³ at constant temperature. Calculate the new pressure if the volume is compressed to 0.20 m³. [3]

10. Calculate the total energy needed to turn 0.30 kg of ice at 0 °C into water at 15 °C. (Specific latent heat of fusion of ice = 334 000 J/kg; specific heat capacity of water = 4200 J/kg °C.) [4]

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Answers

1. (a) Measure the dimensions with a rule and calculate the volume; measure the mass on a balance; divide [1] [1]. (b) Measure the mass; find the volume by water displacement in a measuring cylinder or displacement can [1] [1]. (c) Measure the mass of an empty measuring cylinder, add a known volume of liquid, re-weigh and subtract [1] [1].

2. No new substance is formed [1]; the change is reversible [1]; only the arrangement and energy of the particles change, not the particles themselves [1].

3. (a) Melting and boiling [1] [1]. (b) The energy supplied is used to overcome the forces of attraction between particles [1], increasing their potential energy [1] rather than their kinetic energy, and temperature depends on kinetic energy [1]. (c) Boiling must completely separate the particles, whereas melting only loosens them [1], so more energy — and hence more time at constant power — is required [1].

4. (a) c = E ÷ (mΔθ) = 2400 ÷ (0.25 × 24) [1] [1] = 400 J kg⁻¹ °C⁻¹ [1]. (b) Some energy is lost to the surroundings rather than heating the metal [1], so more energy appears to be needed for the observed temperature rise, giving too high a value [1].

5. (a) The same number of particles occupies a smaller volume, so they hit the walls more frequently [1], and more frequent collisions mean a greater force per unit area [1]. (b) The particles gain kinetic energy and move faster [1], so they collide with the walls more frequently [1] and each collision exerts a greater force [1].

6. Work is done on the air as it is compressed [1], which increases its internal energy and therefore its temperature [1].

7. The weight of liquid above a given point increases with depth [1], increasing the force acting on a given area at that depth and so increasing the pressure [1].

8. E = mL = 0.40 × 334 000 [1] = 133 600 J [1].

9. p₁V₁ = p₂V₂, so p₂ = (p₁V₁) ÷ V₂ = (100 × 0.60) ÷ 0.20 [1] [1] = 300 kPa [1]. The volume falls to a third, so the pressure rises to three times its original value, since temperature is constant.

10. Two stages, two equations. Melting: E = mL = 0.30 × 334 000 [1] = 100 200 J. Heating: E = mcΔθ = 0.30 × 4200 × 15 [1] = 18 900 J [1]. Total = 100 200 + 18 900 = 119 100 J [1]. Melting the ice takes over five times as much energy as the heating that follows it — no temperature change happens during melting even though most of the energy is used there.

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Where marks are usually lost

• Forgetting to subtract the container mass when finding a liquid’s density.
• Explaining the flat sections without mentioning potential energy.
• Giving only one effect when a gas is heated at constant volume.
• Confusing pressure increasing with depth in a liquid with pressure decreasing with height in the atmosphere.
• Confusing specific heat capacity (E = mcΔθ, a temperature change) with specific latent heat (E = mL, a state change at constant temperature).
• Forgetting that p₁V₁ = p₂V₂ only holds at constant temperature — it does not apply if the gas is also heated or cooled.
• In a two-stage energy question, adding the latent heat and specific heat capacity terms in the wrong order, or forgetting one stage entirely — sketch the heating curve first to see how many sections the question covers.