Rearrange heat capacity equations and interpret cooling
Find c, mass or temperature change while preserving signs and the material boundary.
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
- Use ΔE = mcΔθ within one state.
- Convert kilojoules to joules and calculate temperature differences.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which rearrangement finds mass from a known energy gain?
What you will learn
- Rearrange c = ΔE/(mΔθ) to find an unknown quantity.
- Distinguish signed energy change from a positive amount released on cooling.
Choose the unknown before substitution
Use c = ΔE/(mΔθ), m = ΔE/(cΔθ), or Δθ = ΔE/(mc). For 24 kJ received by 0.80 kg with a 30 K rise, c = 24 000/(0.80 × 30) = 1000 J/(kg K). Division by the whole product matters: dividing by mass and then multiplying by temperature gives a different expression.
If a 0.30 kg sample with c = 4200 J/(kg K) receives 12 600 J, its rise is 12 600/(0.30 × 4200) = 10 K. Add that rise to its initial temperature to find the final reading. Neither the rise nor a numerical c is an absolute internal-energy total. State the same-phase, constant-c assumptions.
Signs refer to the named sample
Cooling from 40°C to 25°C has signed Δθ = −15 K. For positive mass and c, the sample's signed ΔE = mcΔθ is negative: it loses energy. A question asking for the amount released often wants the positive magnitude, so identify the convention in the answer rather than presenting a negative amount without explanation.
For m = 0.20 kg and c = 1000 J/(kg K), that cooling gives ΔE = −3000 J and 3000 J released in the ideal single-state account. Losses or additional transfers require their own account. A negative sample energy change does not require a negative specific heat capacity or mean energy is destroyed.
Pause and explain
A 0.20 kg sample with c = 1000 J/(kg K) cools by 15 K. Which statement is correct?
Worked example
A 0.80 kg sample gains 24 kJ for a 30 K rise. Find c in consistent units.
Show the worked solution
- Convert 24 kJ to 24 000 J.
- Divide by mΔθ = 0.80 × 30 = 24 kg K.
- c = 24 000/24 = 1000 J/(kg K).
Answer 1000 J/(kg K).
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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 c for a 0.80 kg sample that gains 24 kJ and warms by 30 K.
Give me a hint
Convert energy units and divide by the product.
Compare my reasoning
- 24 kJ = 24 000 J.
- mΔθ = 24 kg K.
- c = 24 000/24 = 1000 J/(kg K).
1000 J/(kg K).
Look for these in your work
- Use joules.
- Keep both denominator factors.
A sample with c = 1000 J/(kg K) gains 6000 J and warms by 15 K. Find its mass.
Give me a hint
Use m = ΔE/(cΔθ) and divide by the whole denominator.
Compare my reasoning
- The denominator is 1000 × 15 = 15 000 J/kg.
- m = 6000/15 000.
- The mass is 0.40 kg, or 400 g.
0.40 kg.
Look for these in your work
- Isolate the required unknown.
- Attach a mass unit.
A 0.30 kg sample with c = 4200 J/(kg K) cools from 35°C to 25°C within one state. Find signed ΔE and the positive released amount.
Give me a hint
Subtract final minus initial, then interpret the sign.
Compare my reasoning
- Δθ = 25 − 35 = −10 K.
- ΔE = 0.30 × 4200 × (−10) = −12 600 J.
- The sample releases an amount 12 600 J in this ideal account.
ΔE = −12 600 J; 12 600 J released.
Look for these in your work
- Preserve the sample's sign.
- Distinguish signed change from released magnitude.
Common mistakes
- Divide by the entire mΔθ product.
- A temperature rise differs from the final temperature.
- A signed cooling loss is negative; its released magnitude is positive.
What can you explain now?
What is Δθ for cooling from 40°C to 25°C?
Compare with the explanation
−15 K.
Final minus initial is 25 − 40; the sign refers to the sample's change.
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Make it stick
Rearrange the heat-capacity equation and describe a cooling result using an explicit sign convention.
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.