Internal energy and specific heat capacity
Follow the energy into a sample, predict its temperature rise and judge a real measurement.
Core: increasing temperature increases internal energy. Supplement / Extended: average particle kinetic energy, specific heat capacity, calculations and experiments. This is Cambridge IGCSE, separate from Egyptian Baccalaureate / Thanaweya Amma and not Year 3.
Core 2.2.2.1 · Supplement 2.2.2.2–4 · PDF pp.171–176 / printed pp.163–168.
Temperature is a reading; internal energy is stored energy
A spoon left in warmer soup gains internal energy as its temperature rises. Internal energy is the total microscopic kinetic and potential energy of the particles in a sample. Energy transfers by heating because of a temperature difference; “heat” is not a substance stored inside it.
Worked example
Compare equal masses of the same liquid in the same state. The sample at 60 °C has greater internal energy than the one at 25 °C. If it cools, energy transfers to its surroundings.
Think first: Can temperature alone tell you which of two different objects has more internal energy?
No. Amount of material, substance and state also matter. A temperature reading is not a measurement of total internal energy.
Worked calculation: find an unknown c
A heater transfers 9000 J into a 500 g sample as it warms from 22 °C to 32 °C. Ignore other energy transfers.
- Convert mass: 500 g = 0.500 kg; rise = 32 − 22 = 10 °C. [1 mark]
- Rearrange: c = E/(mΔθ) = 9000/(0.500 × 10). [1 mark]
- c = 1800 J/(kg °C). [1 mark]
Check by reversing: 0.500 × 1800 × 10 = 9000 J. This check helps you find errors; it is not an extra mark. If 9000 J were electrical input with unaccounted losses, this would be an apparent value, not a guaranteed material constant.
Try the reverse: find a temperature rise
A 0.30 kg sample with c = 2000 J/(kg °C) gains 3600 J. Δθ = E/(mc) = 3600/(0.30 × 2000) = 6 °C. Starting at 19 °C gives a final temperature of 25 °C, not 6 °C.
A measurement needs an energy budget
For a solid, secure the heater and thermometer with good thermal contact. For a liquid, subtract empty-container mass and stir gently. Both need measured mass, energy input and temperature rise, suitable insulation, repeat readings and stated limitations.
A joulemeter is the simplest energy measurement. If using VIt instead, connect the ammeter in series and the voltmeter in parallel across the heater. Use only teacher-approved low-voltage equipment; keep connections dry and never touch a hot heater or run an immersion heater uncovered.
Improvement is not perfection: insulation reduces loss, stirring reduces temperature differences, and a more responsive thermometer reduces lag. A careful evaluation names the specific error and its likely effect.
Heating and heat-loss explorer
Predict the temperature rise
Supplement / Extended. An imaginary liquid has c = 2000 J/(kg °C), starts at 20 °C and stays liquid. The container’s energy gain is neglected. A fixed fraction of input escapes; this simplified model is not a claim that real heat loss is constant.
Use E = Pt, then energy entering liquid = E × (1 − loss fraction), then Δθ = energy entering liquid / (mc).
Round the rise to two decimal places if needed. Changing a control clears your prediction. This model keeps no lasting score.
Practise, then explain
Four multiple-choice questions, four numerical responses and four written tasks. Internal-energy comparisons are Core; particle kinetic energy, specific heat capacity and its measurement are Supplement / Extended. Numerical answers are checked automatically; written answers use a self-assessment guide. The suggested marks are for this original practice, not an official Cambridge mark scheme.
Practice stays only while this page is open. Refreshing or leaving clears these responses. Copy useful written work into your notes before leaving.
A metal spoon warms from 20 °C to 60 °C in soup, without changing state. What happens to its internal energy?
Optional hint
Consider the energy transferred from the warmer soup to the cooler spoon.
Source clarifications
The original PDF is unchanged. Source Q6 gives c = 3000 J/(kg °C) and concludes the liquid is not water. We retain the calculation but qualify the conclusion: it conflicts with the expected value under the assumptions, yet one result without uncertainty and calibration checks does not prove identity. Ordinary heat loss alone would normally shift this estimate upward, not explain its low value.
The source apparatus Figure 24.1 informed the practical. To avoid ambiguity in its electrical sketch, the connection rule is stated explicitly: ammeter in series, voltmeter across the heater. Our diagram instead shows the energy budget. Subtracting the empty container mass does not correct for energy used to warm that container.
Temperature alone cannot compare the total internal energy of arbitrary objects. Broad climate effects are not explained by specific heat capacity alone. All model data here are illustrative, and c is treated as constant over a small temperature interval without a state change.
Source alignment: supplied Physics-0625 Volume A, PDF pp.171–176 / printed pp.163–168, Figure 24.1; official Cambridge 2026–2028 syllabus, section 2.2.2. Original teaching and practice; suggested marks are not an official Cambridge mark scheme. Existing source-download access remains unchanged.