Thermal Properties and Temperature
Section: Thermal Physics | Syllabus: Cambridge IGCSE Physics (0625)
Thermal Expansion of Solids, Liquids and Gases
When most substances are heated, they expand (get bigger); when they are cooled, they contract (get smaller). This is called thermal expansion.
- heating a substance gives its particles more kinetic energy, so they move faster
- in a solid, the particles vibrate more strongly about their fixed positions
- in a liquid or gas, the particles move faster and, on average, spread further apart
- this increases the average distance between particles, so the substance expands
- cooling has the opposite effect - particles lose kinetic energy, move closer together on average, and the substance contracts
- at constant pressure, all three states of matter expand on heating, but by very different amounts:
- solids expand the least
- liquids expand more than solids
- gases expand the most
Everyday Applications and Consequences
- small gaps are left between sections of railway track so the metal can expand in hot weather without buckling
- bridges are built with expansion joints, which allow the bridge to expand and contract safely as the temperature changes throughout the year
- overhead power cables sag more in summer, when they are longer due to thermal expansion, and become tighter in winter
- if a structure is not given room to expand, thermal expansion can cause serious damage - for example, railway tracks can buckle, and pipes or window glass can crack, in very hot weather or if heated too quickly
Explaining the Relative Amounts of Expansion
Extended Only
- in a solid, particles are held in fixed positions by strong forces between them and can only vibrate about those positions - heating increases the vibrations only slightly, so the increase in average particle separation, and the expansion, is small
- in a liquid, the forces between particles are weaker, and particles are free to move around each other - heating lets them push slightly further apart on average, giving more expansion than in a solid
- in a gas, the forces between particles are extremely weak, and the particles are already moving freely and are far apart - heating increases their speed and the space between them by a much greater amount, giving far more expansion than in a solid or liquid
- this explains the order of expansion for the same rise in temperature: solid < liquid < gas
Internal Energy and Temperature
- the particles in an object are constantly moving or vibrating, and have both kinetic energy (from their motion) and potential energy (from the forces between them) - the total of all this energy is called the object's internal energy
- when an object is heated and its temperature rises, its internal energy increases
Specific Heat Capacity
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
- different substances need different amounts of energy to raise their temperature by the same amount - for example, a metal spoon left in a cup of tea heats up quickly, while the tea itself warms up much more slowly
Defining Specific Heat Capacity
- the specific heat capacity (c) of a substance is the energy required, per unit mass, per unit temperature increase
- Formula:
- c = (Δ E/m Δ θ)
- Where:
- c = specific heat capacity (J/(kg°C))
- Δ E = energy transferred (J)
- m = mass (kg)
- Δ θ = temperature change (°C)
- Formula:
- rearranged, this gives the energy transferred for a given temperature change: Δ E = mcΔ θ
- a substance with a high specific heat capacity (such as water, c ≈ 4200 J/(kg°C)) heats up and cools down slowly
- a substance with a low specific heat capacity (such as copper, c ≈ 385 J/(kg°C)) heats up and cools down quickly
| Substance | Approximate specific heat capacity (J/(kg°C)) | Behaviour when heated |
|---|---|---|
| Water | 4200 | Heats up and cools down slowly |
| Aluminium | 900 | Moderate rate of heating |
| Copper | 385 | Heats up and cools down quickly |
Investigating the Specific Heat Capacity of a Solid
- measure the mass, m, of the solid block
- insert an electrical heater and a thermometer into holes drilled in the block
- insulate the block (e.g. with cotton wool or foam) to reduce heat loss to the surroundings
- record the initial temperature, then switch on the heater and start a stopwatch
- record the voltage V and current I from the power supply
- after a measured time t, switch off the heater and record the highest temperature reached, to find the temperature change Δ θ
- calculate the energy transferred: Δ E = VIt
- calculate the specific heat capacity: c = (Δ E/m Δ θ)
Investigating the Specific Heat Capacity of a Liquid
- find the mass of liquid, m, by weighing an insulated container empty, then weighing it again after adding the liquid
- place an electrical heater, a thermometer and a stirrer into the liquid
- record the initial temperature, then switch on the heater and start a stopwatch, stirring continuously so the liquid heats evenly
- record the voltage V and current I from the power supply
- after a measured time t, switch off the heater and record the highest temperature reached, to find the temperature change Δ θ
- calculate the energy transferred and the specific heat capacity as before: Δ E = VIt, c = (Δ E/m Δ θ)
- stirring is essential for a liquid, but not needed for a solid, so the thermometer reads a temperature that is representative of the whole liquid, not just the region close to the heater
- the container should be well insulated (e.g. an insulated calorimeter or vacuum flask) to reduce heat losses to the surroundings, which would otherwise make the measured value of c too high
Worked Example
- Museli investigates the specific heat capacity of water. She heats 0.10 kg of water in an insulated container using an immersion heater connected to a supply of 10 V, drawing a current of 3 A. After 3 minutes 30 seconds (210 s), the temperature of the water has risen by 15°C. Calculate the specific heat capacity of the water, based on this experiment.
Museli's specific heat capacity experiment: 0.10 kg of water, 10 V, 3 A, 210 s, Δθ = 15°C
- Step 1: calculate the energy transferred
- Δ E = VIt = 10 × 3 × 210 = 6300 J
- Step 2: calculate the specific heat capacity
- c = (Δ E/m Δ θ) = (6300/0.10 × 15) = 4200 J/(kg°C)
- Step 1: calculate the energy transferred
Melting and Boiling
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