The key idea
Home of the two most misused ideas on the paper: specific heat capacity and the difference between boiling and evaporation. The calculation is routine; the marks die in the explanations.
Understand it · 01
Distinguish temperature from stored energy
Temperature and internal energy answer different questions. Temperature is linked to the average kinetic energy of particles. Internal energy is the total microscopic kinetic and potential energy of the particles in a system. A large bath and a small cup at the same temperature therefore need not have the same internal energy. The amount of material matters as well as its temperature and state.
Heating usually increases a substance’s temperature, but that is not the only possible outcome. During melting or boiling of a pure substance at constant pressure, the temperature can remain constant even while energy enters. Energy is then changing the arrangement and separation of particles instead of raising their average kinetic energy.
Understand it · 02
Solve a specific-heat-capacity calculation
Use E = mcΔθ when the energy transfer produces a temperature change without a change of state. The symbol Δθ means final temperature minus initial temperature, not the final thermometer reading alone. Use mass in kilograms and a specific heat capacity in J/(kg °C), or the equivalent J/(kg K). A temperature difference of one degree Celsius has the same size as a difference of one kelvin.
In a heating experiment, an electrical heater transfers energy E = Pt. Insulation reduces energy lost to the surroundings, while good thermal contact helps the measured temperature represent the sample. If you assume that all electrical energy heated the sample when some escaped, your calculated specific heat capacity will be too large. The container can also absorb part of the supplied energy.
Understand it · 03
Read a heating curve as an energy story
On a temperature–time graph with constant heating power, a rising section shows increasing temperature within one state. A flat section indicates a change of state under the idealised conditions shown. A steeper rising section means a larger temperature increase each second. For the same mass and heating power, that corresponds to a smaller specific heat capacity, provided heat losses are comparable.
A perfectly horizontal plateau is an idealisation for a pure substance at fixed pressure. Real experimental curves can be rounded by impurities, heat loss or the thermometer responding slowly. Do not read a horizontal line as “the particles stop moving”. Motion continues while energy changes the interactions between particles. Specific latent heat calculations should only be used when required by your own specification or explicitly supplied in the question.
Understand it · 04
Explain cooling and expansion without shrinking particles
Evaporation happens at a liquid’s surface over a range of temperatures. The particles that escape tend to have higher energies than the average. The average kinetic energy of those remaining can fall, so the liquid cools unless its surroundings replace the energy. Increasing the surface area exposes more particles, and moving air carries vapour away; both can increase the evaporation rate.
Boiling occurs throughout a liquid when bubbles of vapour form at its boiling temperature for the prevailing pressure. Thermal expansion is another distinct process: particles usually become more widely separated as temperature rises. A bridge expansion gap accommodates the resulting change in dimensions. Neither evaporation nor ordinary thermal expansion requires particles themselves to become larger.
Key definitions
- Specific heat capacity
- The energy required to raise the temperature of 1 kg of a substance by 1 °C.
- Melting point
- The temperature at which a solid changes to a liquid, with no temperature change while the change of state occurs.
- Evaporation
- The escape of sufficiently energetic particles from a liquid's surface. It can occur below the boiling point and can cool the remaining liquid.
Equations, units & conditions
Thermal energy
E = m × c × Δθ
Units: J
Worked examples, step by step
Example 1
How much energy is needed to heat 2.0 kg of water from 20 °C to 70 °C? (c of water = 4200 J/(kg °C))
Follow the method
- Δθ = 70 − 20 = 50 °C.
- E = mcΔθ = 2.0 × 4200 × 50.
E = 420,000 J = 420 kJ
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Starter · 3 suggested marks
1. Find specific heat capacity
A 0.35 kg metal block receives 6300 J and warms from 18 °C to 58 °C. Assuming no energy escapes, calculate its specific heat capacity.
Show solution and self-check: question 1
- Δθ = 58 − 18 = 40 °C.
- c = E/(mΔθ) = 6300/(0.35 × 40).
450 J/(kg °C)
Self-check · one mark per point
- Find the 40 °C temperature change.
- Rearrange E = mcΔθ correctly.
- Calculate 450 J/(kg °C).
Build · 3 suggested marks
2. Account for energy losses
A 60 W heater warms 0.20 kg of water by 12 °C in 210 s. Take c = 4200 J/(kg °C). Calculate the energy transferred to the water and the energy transferred elsewhere.
Show solution and self-check: question 2
- The water gains E = mcΔθ = 0.20 × 4200 × 12 = 10 080 J.
- The heater supplies Pt = 60 × 210 = 12 600 J.
- Other transfers total 12 600 − 10 080 = 2520 J.
Water: 10 080 J; elsewhere: 2520 J.
Self-check · one mark per point
- Calculate the water’s energy increase.
- Calculate the heater’s total energy supply.
- Subtract to find 2520 J.
Common mistakes & how to avoid them
- Using the full temperature instead of the temperature change in E = mcΔθ. It is Δθ — final minus initial.
- Assuming energy input must raise temperature during a change of state. For a pure substance changing state at constant pressure, energy changes particle arrangements and potential energy while the temperature remains constant.
- Confusing evaporation with boiling. Boiling occurs throughout a liquid at its boiling temperature for the prevailing pressure. Evaporation takes place at the surface over a range of temperatures.
- Explaining evaporation's cooling effect without particle energy: escaping particles tend to have above-average energy, reducing the average kinetic energy of those remaining unless energy is replaced from the surroundings.
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
Why does temperature stay constant during melting?
During melting of a pure substance at constant pressure, incoming energy changes particle arrangements and increases potential energy rather than average kinetic energy. Temperature therefore stays constant while solid and liquid coexist, even though the sample continues gaining internal energy.
A flat temperature trace does not mean the heater has stopped working. Follow the energy into the substance: particles become able to rearrange as the solid changes to liquid, so energy is associated with overcoming the interactions maintaining the solid structure. After the solid has melted, further heating can raise the liquid’s temperature. Impurities, variable pressure and experimental heat losses can make a real plateau less ideal. State the pure-substance, constant-pressure conditions rather than applying a perfectly flat line to every material mixture.
Why does evaporation cool a liquid?
Evaporation can cool a liquid because escaping particles tend to carry more energy than the average. Their departure lowers the average kinetic energy of the remaining particles unless energy is replaced from the surroundings. The temperature change depends on that overall energy balance.
Particles within a liquid have a spread of energies; they do not all travel at one identical speed. Some particles at the surface have enough energy to escape the attractions holding them in the liquid. Losing these particles can reduce the remaining average energy. That is why an evaporating liquid can draw energy from an object in contact with it. Do not say that evaporation creates “cold energy” or that every escaping particle was the single fastest particle: the explanation concerns an energetic group and an average.
What factors increase the rate of evaporation?
Increasing temperature, exposing a larger surface area, or moving air across the surface can increase evaporation. Temperature affects particle energies; area changes how many particles are near the surface; air movement removes vapour. A fair comparison changes one factor at a time.
To investigate air movement, use identical shallow dishes containing the same liquid, with matching initial masses and temperatures. Place a fan near one dish while keeping other conditions comparable, and measure mass lost over equal times. Repeated measurements help distinguish a real difference from reading uncertainty. Do not compare a wide warm dish in a draught with a narrow cold dish in still air and attribute the result solely to the fan: temperature and area would both be alternative explanations.
Compare evaporation rates
Two dishes lose 1.8 g and 3.6 g in 12 minutes under otherwise matched conditions. Compare their average mass-loss rates.
- Still-air dish: 1.8/12 = 0.15 g/min.
- Moving-air dish: 3.6/12 = 0.30 g/min.
The second average evaporation rate is twice the first.
Why do heat losses make a measured specific heat capacity too high?
If electrical energy escapes to the surroundings but you assume all of it heats the sample, the calculated specific heat capacity is too large. You divide an overestimated sample-energy input by its measured mass and temperature rise. Insulation reduces this systematic error.
The calculation c = Pt/(mΔθ) treats Pt as the sample’s energy gain. A real heater also warms its leads, the container and the air, so the sample receives less than that amount. A smaller measured temperature rise then produces a larger calculated value of c. Record the initial and final temperatures carefully and ensure good heater and thermometer contact. Repeating the same poorly insulated setup may reduce random variation, but it does not automatically remove this consistent energy-accounting error.
Include the energy loss
A 0.40 kg block warms by 25 °C while receiving 6000 J electrically, of which 1000 J escapes. Compare the naive and corrected specific heat capacities.
- Naive c = 6000/(0.40 × 25) = 600 J/(kg °C).
- Corrected c = 5000/(0.40 × 25) = 500 J/(kg °C).
Ignoring the loss overestimates c by 100 J/(kg °C).
Do I need to convert Celsius to kelvin in specific heat capacity calculations?
A temperature change has the same numerical value in kelvins and degrees Celsius because the two scales have equal-sized intervals. An absolute temperature is different: converting a reading requires adding approximately 273. Use the temperature difference in E = mcΔθ.
Write both readings before subtracting. If a sample warms from 15 °C to 45 °C, its rise is 30 °C. The approximate kelvin readings are 288 K and 318 K, also separated by 30 K. Adding 273 to the temperature difference would be a major error because that offset cancels when the readings are subtracted. Kelvin is especially important in absolute-temperature gas relationships; Celsius differences remain entirely suitable for specific-heat calculations when the units of c are consistent.
What is the difference between boiling and evaporation?
Boiling forms vapour bubbles throughout a liquid at its boiling temperature for the prevailing pressure. Evaporation takes place at the surface over a range of temperatures, including below boiling. Both produce gas, but their locations and conditions are different.
Water left in an open dish can disappear without ever reaching 100 °C, which shows that evaporation does not require boiling. During boiling, vapour bubbles can form within the liquid and reach the surface. A boiling temperature must be linked to pressure: 100 °C is the familiar value for pure water at standard atmospheric pressure, not a universal temperature under every condition. In an answer comparing the processes, give paired distinctions such as “surface versus throughout” and “range of temperatures versus boiling temperature at a specified pressure”.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- Describe thermal expansion of solids, liquids and gases and its applications
- Define specific heat capacity and use E = mcΔθ
- Describe melting and boiling in terms of energy input without temperature change
- Distinguish boiling from evaporation
- Explain the factors that affect the rate of evaporation and its cooling effect
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
- Cambridge IGCSE Physics 0625 — 2026–2028 syllabus
- Pearson Edexcel International GCSE Physics 4PH1 — specification
Reference this page
Dr Desouky Physics Academy. IGCSE Physics: Thermal properties of matter. Updated 2026-09-25. https://www.drdesouky.com/revision/igcse-physics/thermal-properties-of-matter
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