Define specific heat capacity and calculate energy changes
Use sample mass and temperature change to find an energy gain within one state.
Section 2.2.2 · Specific heat capacity
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028, section 2.2.2. These lessons explain average particle kinetic energy, calculate specific heat capacity and describe solid/liquid methods. Written calculations and method analysis are self-checked; human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Calculate final minus initial temperature.
- Convert grams to kilograms and distinguish readings from intervals.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A sample warms from 20°C to 40°C. Which interval belongs in mcΔθ?
What you will learn
- Define specific heat capacity as energy per unit mass per unit temperature increase.
- Use ΔE = mcΔθ with consistent units and a stated same-state model.
A material property per mass and temperature interval
Specific heat capacity c is the energy needed per unit mass per unit increase in temperature. A value of 900 J/(kg K) means a 1 kg sample needs 900 J for a 1 K temperature increase in the stated conditions. J/(kg °C) gives the same numerical value because a 1°C interval equals a 1 K interval. This is a definition about an interval, not the absolute Celsius reading.
The selected equation is c = ΔE/(mΔθ). Rearranging gives ΔE = mcΔθ, where m is the sample mass and Δθ is final minus initial temperature. Use kilograms with c in J/(kg K), and express the energy in joules. More mass or a larger interval requires proportionally more energy when c is constant in this model.
Apply the model to the sample's energy
For a 0.50 kg sample with c = 900 J/(kg K), warming from 20°C to 40°C gives Δθ = 20 K and ΔE = 0.50 × 900 × 20 = 9000 J. Do not use the final 40°C alone or add 273 to the interval. The temperature difference is unchanged when both readings are converted to kelvin.
Use a single-state interval with c treated as constant and the energy actually received by the named sample. The equation does not describe the energy used for a constant-temperature phase change. A heater's total input can also warm apparatus or escape to the surroundings; later lessons separate that input from the sample gain.
Pause and explain
A 200 g sample has c = 900 J/(kg K) and warms by 30 K. What is its gain?
Worked example
A 0.50 kg sample has c = 900 J/(kg K). Find its energy gain from 20°C to 40°C, within one state.
Show the worked solution
- Δθ = 40 − 20 = 20 K.
- Use ΔE = mcΔθ with kilograms.
- ΔE = 0.50 × 900 × 20 = 9000 J.
Answer 9000 J, or 9.0 kJ, gained by the sample.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find the gain for 0.50 kg, c = 900 J/(kg K), warming from 20°C to 40°C.
Give me a hint
Calculate the interval first and retain the kilogram unit.
Compare my reasoning
- The temperature interval is 20 K.
- ΔE = 0.50 × 900 × 20.
- The sample gains 9000 J within the supplied same-state model.
9000 J.
Look for these in your work
- Use the interval.
- Include sample mass and consistent units.
A 250 g water sample has supplied c = 4200 J/(kg K) and warms from 18°C to 33°C without changing state. Find ΔE.
Give me a hint
Convert grams and subtract the initial reading.
Compare my reasoning
- 250 g = 0.250 kg.
- Δθ = 33 − 18 = 15 K.
- ΔE = 0.250 × 4200 × 15 = 15 750 J.
15 750 J.
Look for these in your work
- Convert mass correctly.
- Use the energy-change model within one state.
A pure liquid receives energy while boiling at constant temperature. A student substitutes Δθ = 0 and concludes zero energy input. Explain the mistake.
Give me a hint
The sensible-temperature-change equation has a stated same-state scope.
Compare my reasoning
- During the specified boiling plateau the temperature does not rise.
- Energy is still used to change liquid into gas.
- mcΔθ does not calculate the phase-change input; no numerical value is determined here.
The equation does not give the boiling energy; zero temperature rise is not zero input.
Look for these in your work
- State the model's limit.
- Do not invent a phase-change energy from missing data.
Common mistakes
- Use mass in kilograms, not unconverted grams.
- Use temperature change, not the final reading.
- Do not apply mcΔθ to a phase-change plateau.
What can you explain now?
Is a 20°C temperature rise also a 20 K interval?
Compare with the explanation
Yes.
The Celsius and kelvin scales have equal-sized intervals; their offset cancels in a difference.
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Make it stick
Define c, convert sample mass, use a temperature interval and state where mcΔθ applies.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.