Home / iGCSE Physics (0625) 2.2.1 Thermal expansion of solids, liquids and gases Paper 4 -Exam Style Questions

iGCSE Physics (0625) 2.2.1 Thermal expansion of solids, liquids and gases Paper 4 -Exam Style Questions- New Syllabus

Question

(a) (i) A rectangular glass tank contains $3.0\text{ m}^3$ of liquid. The mass of the liquid is $2800\text{ kg}$. Calculate the density of the liquid.
(ii) A metal block is placed in the tank and it sinks to the bottom. The top surface of the block is $1.1\text{ m}$ below the surface of the liquid. Fig. 4.1 shows the block on the bottom of the tank.
Calculate the force exerted by the liquid on the top of the block.
(b) Fig. 4.2 shows part of a steel railway track. There are small gaps between sections of the rail.
State why gaps are needed between the sections of rail. Explain your answer in terms of particles.
(c) Fig. 4.3 shows a saucepan of boiling water on the heating ring of an electric cooker.
The handle is made from plastic.
Explain why this material is suitable.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):

• Topic $1.4$ — Density (Part $\mathrm{(a)(i)}$)
• Topic $1.8$ — Pressure (Part $\mathrm{(a)(ii)}$)
• Topic $2.2.1$ — Thermal expansion of solids, liquids and gases (Part $\mathrm{(b)}$)
• Topic $2.3 .1$ — Conduction (Part $\mathrm{(c)}$)

▶️ Answer/Explanation
Part (a)(i)

Correct Answer: $930\text{ kg/m}^3$ (to 2 sig. figs) or $933\text{ kg/m}^3$

Detailed solution: Density is defined as mass per unit volume. Using the formula $\rho = \frac{m}{V}$, we substitute the given mass $m = 2800\text{ kg}$ and volume $V = 3.0\text{ m}^3$. Calculating $2800 / 3.0$ gives $933.33…$, which is rounded to $930\text{ kg/m}^3$ to match the significant figures of the input data.

Part (a)(ii)

Correct Answer: $1400\text{ N}$

Detailed solution: First, calculate the liquid pressure at depth $h = 1.1\text{ m}$ using $p = \rho gh = 933 \times 9.8 \times 1.1 \approx 10000\text{ Pa}$. Next, determine the area of the block’s top surface: $A = 0.45\text{ m} \times 0.30\text{ m} = 0.135\text{ m}^2$. Finally, find the force using $F = p \times A$, which gives $10000 \times 0.135 \approx 1350\text{ N}$ to $1400\text{ N}$ depending on rounding of intermediate values.

Part (b)

Correct Answer: To allow for thermal expansion and prevent buckling.

Detailed solution: When the temperature increases, the steel rails undergo thermal expansion. At a microscopic level, the particles gain kinetic energy and vibrate more vigorously, causing the average separation between particles to increase. The gaps provide space for this increase in length, preventing the tracks from bending or buckling under stress.

Part (c)

Correct Answer: Plastic is a poor conductor (good insulator) of heat.

Detailed solution: Plastic is a suitable material for the handle because it has a low thermal conductivity. This significantly reduces the rate at which thermal energy is transferred from the hot metal pan to the user’s hand. As a result, the handle remains at a safe temperature to touch even while the water inside the pan is boiling.

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