# Thermal Expansion and Specific Heat Capacity

> Physics · CIE IGCSE 0625
> Source: https://www.owlsprep.com/study/cie-0625-u2-thermal-expansion-and-specific-heat/

This guide covers thermal expansion of solids, liquids, gases, and specific heat capacity calculations for CIE IGCSE Physics 0625 Core and Extended tiers, including practical methods and exam-style examples.

**Prerequisites:** [Kinetic particle theory for states of matter](https://www.owlsprep.com/study/cie-0625-u2-kinetic-particle-theory/); [Basic energy transfer concepts](https://www.owlsprep.com/study/cie-0625-u1-energy-transfers/)

## Learning objectives

- Explain thermal expansion of solids, liquids and gases using kinetic particle theory
- State that a rise in the temperature of an object increases its internal energy (Core)
- Recall and apply the specific heat capacity equation Q = mcΔT for calculations (Extended)
- Describe experiments to measure the specific heat capacity of a solid and a liquid (Extended)
- Solve multi-step specific heat capacity energy transfer problems (Extended only)
- Identify common applications and hazards of thermal expansion for exams

## 1. Thermal Expansion (Core)

**Thermal Expansion** — The increase in volume or length of a substance when heated, caused by particles gaining kinetic energy and moving further apart, with no change of state.

*Example:* A 1 m long steel rod expands by 1 mm when heated by 100 °C.

The extent of expansion for the same temperature rise follows the order: **solids < liquids < gases**. This is because interparticle forces are strongest in solids, holding particles close together, and negligible in gases, so particles can move much further apart when heated.

> **mnemonic**
>
> S < L < G: Solids expand Least, Gases expand Greatest for equal heat input.

**Worked example:** Explain why glass tumblers often crack when very hot water is poured into them, using particle theory.

1. 1. When hot water is poured into the cold glass, the inner surface of the glass heats up rapidly, and its particles gain kinetic energy and expand.
2. 2. Glass is a poor conductor of heat, so the outer surface of the glass remains cool and does not expand at the same rate.
3. 3. The uneven expansion creates stress in the glass, causing it to crack.

> **Exam tip:** Core questions frequently ask for applications/hazards of thermal expansion: common examples include railway track gaps, bimetallic strips in thermostats, and overhead power line slack in summer.

## 2. Internal Energy and Temperature (Core)

Everything is made of particles that are always moving, so every object stores energy in the movement and arrangement of its particles. This stored energy is called the object's internal energy.

**Internal energy** — The total energy stored in an object in the movement (kinetic energy) and arrangement (potential energy) of its particles.

> **info**
>
> Core outcome (2.2.2): a rise in the temperature of an object increases its internal energy. Heating an object makes its particles move faster, so the object's internal energy increases.

> **note**
>
> Extended only: a rise in the temperature of an object corresponds to an increase in the average kinetic energy of all of the particles in the object.

**Worked example:** A metal block is heated by a flame so its temperature rises. State and explain what happens to the internal energy of the block.

1. Step 1: State the change:
2. The internal energy of the block increases.
3. Step 2: Explain why:
4. Raising the temperature makes the particles of the block move faster, so the total energy stored in the block - its internal energy - increases.

> **Exam tip:** For Core (Paper 1/3) you only need the link between temperature and internal energy - the specific heat capacity equation, calculations and measuring experiments are all Extended (Supplement) content.

## 3. Specific Heat Capacity and Its Formula (Extended)

**Specific Heat Capacity** — The amount of thermal energy required to raise the temperature of 1 kg of a substance by 1 °C, measured in J/(kg °C).

*Notation:* $c$

*Example:* Water has a specific heat capacity of 4200 J/(kg °C).

$$Q = m c \Delta T$$

Where $Q$ = thermal energy transferred (J), $m$ = mass of substance (kg), $c$ = specific heat capacity, and $\Delta T$ = change in temperature ($T_{final} - T_{initial}$). Always convert mass from grams to kilograms before substituting values into the formula.

**Worked example:** Calculate the energy required to heat 3 kg of water from 15 °C to 65 °C. Use $c_{water} = 4200$ J/(kg °C).

1. Step 1: Identify known values: $m = 3$ kg, $c = 4200$ J/(kg °C), $\Delta T = 65 - 15 = 50$ °C
2. Step 2: Substitute into the formula $Q = mc\Delta T$:
3. $$Q = 3 \times 4200 \times 50$$
4. Step 3: Calculate the result: $Q = 630,000$ J = 630 kJ

> **Exam tip:** Always show full substitution steps for SHC calculations to earn all available marks, even if you can solve the problem mentally.

## 4. Measuring Specific Heat Capacity (Extended)

You are expected to describe the method to measure the specific heat capacity of a solid (e.g. aluminium block) or liquid (e.g. water) in the exam:

1. 1. Measure the mass of the substance using a balance, and record its initial temperature with a thermometer.
2. 2. Wrap the substance in insulating material (e.g. foam) to reduce thermal energy loss to the surroundings.
3. 3. Heat the substance using an electric heater of known power for a measured time, then record the final maximum temperature.
4. 4. Calculate energy supplied by the heater: $Q = power (W) \times time (s)$, then rearrange $c = Q/(m\Delta T)$ to find specific heat capacity.

**Worked example:** A 1 kg copper block is heated by a 100 W heater for 90 seconds. Its temperature rises from 22 °C to 47 °C. Calculate the specific heat capacity of copper.

1. Step 1: Calculate energy supplied: $Q = P \times t = 100 \times 90 = 9000$ J
2. Step 2: Calculate $\Delta T = 47 - 22 = 25$ °C, $m = 1$ kg
3. Step 3: Rearrange the formula to $c = Q/(m\Delta T)$:
4. $$c = 9000 / (1 \times 25)$$
5. Step 4: Calculate $c = 360$ J/(kg °C)

> **warning**
>
> The calculated SHC from practicals will always be slightly higher than the true value, because some energy supplied by the heater is lost to the surroundings instead of heating the test substance.

> **Exam tip:** You may be asked to identify sources of error in SHC practicals: the most common error is thermal energy loss to the surroundings.

## 5. Extended: Multi-Substance Energy Transfers

For Extended tier, you will solve problems where thermal energy is transferred from a hotter substance to a cooler one, with the assumption that no energy is lost to the surroundings. This means: **energy lost by hot substance = energy gained by cold substance**.

**Worked example:** A 0.3 kg block of brass heated to 180 °C is dropped into 0.4 kg of water at 18 °C. Assuming no energy loss, calculate the final temperature of the mixture. Use $c_{brass} = 380$ J/(kg °C) and $c_{water} = 4200$ J/(kg °C).

1. Step 1: Let $T$ = final temperature of the mixture. Set energy lost by brass equal to energy gained by water:
2. $$m_{brass} c_{brass} (180 - T) = m_{water} c_{water} (T - 18)$$
3. Step 2: Substitute known values:
4. $$0.3 \times 380 \times (180 - T) = 0.4 \times 4200 \times (T - 18)$$
5. $$114(180 - T) = 1680(T - 18)$$
6. Step 3: Expand and rearrange to solve for T:
7. $$20520 - 114T = 1680T - 30240$$
8. $$50760 = 1794T$$
9. Step 4: Calculate final temperature: $T \approx 28.3$ °C

> **Exam tip:** For energy transfer problems, always ensure temperature change terms are positive: use $T_{hot} - T_{final}$ for the hot substance, and $T_{final} - T_{cold}$ for the cold substance.

## Common pitfalls

- **Wrong:** Using mass in grams instead of kilograms in SHC calculations
  - Why it fails: SHC units are J/(kg °C), so mass in grams will produce an incorrect energy value 1000x larger or smaller than the true value
  - Correct: Always convert mass from grams to kilograms by dividing by 1000 before substituting into $Q = mc\Delta T$
- **Wrong:** Calculating $\Delta T$ as initial minus final temperature for a substance being heated
  - Why it fails: $\Delta T$ measures the increase in temperature for heated substances, so it must be a positive value
  - Correct: $\Delta T = T_{final} - T_{initial}$ for heated substances; use the magnitude of the change for cooled substances
- **Wrong:** Stating liquids expand more than gases for the same temperature rise
  - Why it fails: Interparticle forces in gases are negligible, so gas particles move much further apart than liquid particles when heated
  - Correct: Recall the expansion order: solid < liquid < gas for equal temperature change, linked to interparticle force strength
- **Wrong:** Claiming practical SHC values are lower than true values due to heat loss
  - Why it fails: Heat loss means the energy absorbed by the test substance is less than the energy supplied by the heater, so calculated SHC will be higher
  - Correct: State that calculated SHC from practicals is higher than the true value, and suggest improvements like better insulation
- **Wrong:** Using the same SHC value for both substances in Extended energy transfer problems
  - Why it fails: Every substance has a unique specific heat capacity, so using the wrong $c$ value will produce an incorrect final temperature
  - Correct: Label $c$, $m$ and $\Delta T$ values for each substance clearly before substituting into the energy balance equation

## Cheatsheet

| Concept | Core Rule/Formula | Extended Add-on |
| --- | --- | --- |
| Thermal expansion | Solids, liquids and gases expand when heated at constant pressure; gases expand most and solids least; used in railway-track gaps and bimetallic strips | Explain the relative order (solid < liquid < gas) in terms of the motion and arrangement of particles |
| Internal energy | A rise in the temperature of an object increases its internal energy | A temperature rise corresponds to an increase in the average kinetic energy of all the particles |
| Specific heat capacity | Not required at Core | Define SHC as energy per unit mass per unit temperature rise; recall and use $Q = mc\Delta T$ (official form $c = \Delta E / (m\Delta\theta)$); $c_{water} = 4200$ J/(kg °C) |
| Measuring SHC | Not required at Core | $Q = P \times t$ for the energy supplied; insulate to reduce heat loss, so the measured $c$ comes out slightly too high |
| Multi-substance transfers | Not required at Core | Energy lost by the hot substance = energy gained by the cold substance (assuming no losses) |

## What's next

Now that you have mastered thermal expansion and specific heat capacity, you are ready to progress to other thermal physics topics in the CIE IGCSE Physics 0625 syllabus. Next, you will learn about the three mechanisms of thermal energy transfer (conduction, convection, radiation), which explain how heat moves between substances, and thermal processes associated with changes of state. This knowledge builds the foundation for combined thermal physics questions in both Core and Extended papers, which frequently test SHC alongside energy transfer and practical skills. Be sure to practice both single-substance Core calculations and multi-substance Extended problems to build confidence before your exam.

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