2.2CH2 Thermal Physics
Thermal Properties and Temperature — Cambridge O Level Physics 5054 Revision Notes & Topical Questions
10 min read 12 parts Lesson 2 of 3
Part 1 of 9Thermal expansion
- Expansion
- most solids, liquids and gases increase in volume when their temperature rises.
- The particles vibrate or move with larger amplitude and push slightly further apart;
- the particles themselves do not get bigger.
- Relative amounts
- gases expand the most, liquids expand less and solids expand the least for the same temperature rise.
- Consequences and uses
- expansion must be allowed for in engineering.
| Application or problem | Explanation |
|---|---|
| Gaps in railway lines | Rails expand in heat and would buckle without gaps |
| Expansion joints in bridges | Allow the structure to lengthen safely |
| Overhead cables sag in summer | The cables expand and become longer |
| Bimetallic strip in a thermostat | Two metals expand by different amounts, so the strip bends and switches the circuit |
| Liquid-in-glass thermometer | The liquid expands up a narrow tube when heated |
- Bimetallic strip
- it is made of two metals joined together;
- the one that expands more is on the outside of the curve when heated.
- Used in fire alarms, thermostats and circuit breakers.
- it is made of two metals joined together;
- Pouring hot water into a glass
- the inside expands first, causing stress that can crack ordinary glass.
Part 2 of 9Measuring temperature
- Temperature
- a measure of how hot or cold an object is, and a measure of the average kinetic energy of its particles.
- Liquid-in-glass thermometer
- it uses the expansion of a liquid such as alcohol or mercury in a narrow tube.
- Alcohol is used for low temperatures because its freezing point is much lower than that of mercury.
- Thermocouple
- two different metals joined together produce a small voltage that depends on the temperature difference between the junctions.
- It can measure a wide range of temperatures, including very high ones, and responds quickly.
- Features of a thermometer
- they affect how good it is.
| Feature | Meaning |
|---|---|
| Range | The lowest to highest temperatures it can measure |
| Sensitivity | The change in reading for a change in temperature |
| Linearity | Equal temperature changes give equal changes in the property |
| Response time | How quickly it follows temperature changes |
- Improving sensitivity
- use a narrower tube or a larger bulb in a liquid-in-glass thermometer.
- Calibration
- to calibrate a thermometer, mark the level at two fixed points and divide the interval evenly.
- Place the thermometer in pure melting ice and mark the liquid level as 0 °C.
- Place it in steam above boiling water at standard pressure and mark the level as 100 °C.
- Divide the distance between the marks into 100 equal intervals.
Part 3 of 9Specific heat capacity
- Thermal capacity
- the energy needed to raise the temperature of an object by 1 °C.
- Specific heat capacity
- the energy required to raise the temperature of 1 kg of a substance by 1 °C (or 1 K).
- Its unit is J/(kg °C) and for water it is about 4200 J/(kg °C).
- Equation
- the thermal energy change is ΔQ = mcΔθ.
- m is the mass in kg, c is specific heat capacity and Δθ is the temperature change.
- Meaning of a high value
- water needs a lot of energy to heat up, so it is good for cooling systems and central heating.
- Metals have lower values, so they heat up quickly.
| Substance | Specific heat capacity J/(kg °C) |
|---|---|
| Water | 4200 |
| Ice | 2100 |
| Aluminium | 900 |
| Iron | 450 |
| Copper | 390 |
Part 4 of 9Specific latent heat
- Latent heat
- the energy absorbed or released when a substance changes state at constant temperature.
- Specific latent heat of fusion
- the energy needed to change 1 kg of solid to liquid at its melting point without a temperature change.
- Specific latent heat of vaporisation
- the energy needed to change 1 kg of liquid to gas at its boiling point without a temperature change.
- Equation
- Q = ml, where l is the specific latent heat in J/kg.
- For water, fusion is about 3.3 × 105 J/kg and vaporisation about 2.3 × 106 J/kg.
- Why vaporisation is larger
- the particles must be completely separated against the attractive forces and push back the atmosphere.
- Heating curve
- a graph of temperature against time for heating ice.
- It rises, flattens at the melting point, rises again through the liquid, flattens at the boiling point and then rises as a gas.
- Measuring l
- use an electric heater in a melting solid or boiling liquid and calculate E = IVt divided by the mass changed.
- Compare the mass of liquid collected with and without the heater to remove heat-loss errors.
Part 5 of 9Using the energy equations together
- Multi-stage problems
- add the energy for each stage.
- Heat the solid to its melting point using Q = mcΔθ.
- Melt it with Q = mlf.
- Heat the liquid to its boiling point with Q = mcΔθ.
- Boil it with Q = mlv.
- Mixing and cooling
- in an insulated mixture, the energy lost by the hot object equals the energy gained by the cold one.
- Cooling by evaporation
- fastest particles leave the surface and the remaining liquid cools, a latent heat effect.
- A refrigerator removes heat from its contents as the working fluid evaporates.
Part 6 of 9Temperature, internal energy and kelvin conversion
- Internal energy
- the total energy of the particles in an object, which rises when its temperature increases.
- Temperature rise and particles
- it means that the average kinetic energy of all the particles has increased.
- Kelvin and Celsius
- T (in K) = θ (in °C) + 273.
- 100 °C is 373 K and 0 °C is 273 K, and a rise of 1 K equals a rise of 1 °C.
- Relative size of expansion
- for the same temperature rise gases expand most, liquids less and solids least.
Part 7 of 9Boiling compared with evaporation
- Melting and boiling points of water
- at standard atmospheric pressure, ice melts at 0 °C and water boils at 100 °C.
- Changes without temperature change
- during melting, solidification, boiling and condensation, the energy transferred changes the particle arrangement, not the temperature.
| Feature | Boiling | Evaporation |
|---|---|---|
| Temperature | Only at the boiling point | At any temperature |
| Where | Throughout the liquid, with bubbles | At the surface only |
| Speed | Fast | Slow |
| Energy source | Needs continuous heating | Takes energy from the liquid and surroundings |
- Evaporation in particles
- the more energetic particles near the surface escape, leaving particles of lower average kinetic energy, so the liquid cools.
- Factors that speed up evaporation
- higher temperature, larger surface area and moving air over the surface that carries vapour away.
- Wet washing dries faster on a warm, windy day when spread out.
- Latent heat in particle terms
- energy is needed to overcome the forces between particles when a solid melts or a liquid boils.
- Boiling needs more energy than melting because the particles must be separated completely.
Part 8 of 9Measuring specific heat capacity of a liquid and a solid
- Liquid method (immersion heater)
- a known mass of liquid is heated electrically.
- Pour a measured mass of water into an insulated, lidded container.
- Place an immersion heater and a thermometer in the water, and record the starting temperature.
- Switch on the heater and the stopwatch, and stir gently.
- Record the current and voltage, and the temperature after a measured time.
- Calculate c = IVt ÷ (mΔθ).
- Solid method
- use a metal block with holes for a heater and thermometer, as in the earlier experiment.
- Add a drop of oil to the thermometer hole for good thermal contact.
- Reducing error
- insulate and lid the container, stir the liquid and take readings after the temperature has risen well above the room temperature.
Part 9 of 9Worked examples
How much energy is needed to heat 2.5 kg of water from 20 °C to 80 °C? Use c = 4200 J/(kg °C). [3]3 marks
- ΔQ = mcΔθ = 2.5 × 4200 × (80 − 20).
- ΔQ = 2.5 × 4200 × 60.
- ΔQ = 6.3 × 105 J.
A 0.50 kg copper block at 20 °C is heated with a 60 W heater for 100 s. Find the final temperature if all the energy heats the block (c = 390 J/(kg °C)). [4]4 marks
- Energy = Pt = 60 × 100 = 6000 J.
- Δθ = E ÷ (mc) = 6000 ÷ (0.50 × 390) = 30.8 °C.
- Final temperature = 20 + 30.8 = 50.8 °C.
Calculate the energy needed to melt 0.40 kg of ice at 0 °C (lf = 3.3 × 105 J/kg). [2]2 marks
- Q = ml = 0.40 × 3.3 × 105.What the symbols mean: E or Q = energy to change state without a change of temperature (J) · m = mass (kg) · L = specific latent heat (J kg−1)
- Q = 1.32 × 105 J.
- Q = ml = 0.40 × 3.3 × 105.
A kettle of power 2000 W turns 0.10 kg of boiling water to steam. Assuming no losses, find the time (lv = 2.3 × 106 J/kg). [3]3 marks
- Energy = ml = 0.10 × 2.3 × 106 = 2.3 × 105 J.
- t = E ÷ P = 2.3 × 105 ÷ 2000.
- t = 115 s.
Why is the measured specific heat capacity in a school experiment usually too large? [2]2 marks
- Some of the electrical energy is lost to the surroundings, not stored in the block.
- Using the full energy supplied in the equation with a smaller temperature rise gives a larger c.
A 1.0 kg block of aluminium is heated by a 100 W heater for 90 s and its temperature rises by 10 °C. Find the specific heat capacity of aluminium from the experiment and say why it is probably too high. [4]4 marks
- Energy = Pt = 100 × 90 = 9000 J.
- c = E ÷ (mΔθ) = 9000 ÷ (1.0 × 10) = 900 J/(kg °C).
- Some of the energy was lost to the surroundings, so the energy that really heated the block was less.
- Using all 9000 J therefore tends to give a value of c that is too high.
Key terms
| Term | Meaning |
|---|---|
| Specific heat capacity | The energy needed to raise the temperature of 1 kg of a substance by 1 °C. |
| Specific latent heat | The energy needed to change the state of 1 kg of a substance at constant temperature. |
| Thermal expansion | The increase in size of a material when its temperature rises. |
| Thermocouple | A sensor made of two metals that gives a voltage depending on temperature. |
| Fixed point | A reproducible temperature, such as the melting point of ice, used to calibrate thermometers. |
| Sensitivity | The change in a thermometer reading per unit change in temperature. |
| Bimetallic strip | Two metals joined together that bend when heated because they expand by different amounts. |
Avoid these mistakes
- Saying temperature rises during melting — it stays constant while energy changes the particle arrangement.
- Using Q = mcΔθ for a change of state — use Q = ml because the temperature does not change.
- Quoting c in J/g without conversion — use kg in the equation.
- Saying atoms expand when heated — the spacing between particles increases, not their size.
- Forgetting insulation in experiments — heat loss gives an error, so insulate and stir.
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