Specific heat capacity calculations and units

FREE PHYSICS REVISION · UNIT 3 · C4.4.2

Pearson BTEC Level 3 AAQ Applied Science · Issue 7 (May 2026)

What you will learn

ΔQ = mcΔT; c in J kg⁻¹ K⁻¹; rearrangements and graph interpretation.

Specification mapping: C4.4.2, C4.4.3

Prerequisites: C4.4.1

Physical principles

Specific heat capacity c is the energy required per kilogram per kelvin increase in temperature within a phase. Its unit is J kg⁻¹ K⁻¹. The required relationship is ΔQ = mcΔT. It assumes c is approximately constant over the temperature range; no phase change is included in that term.

Rearrange to c = ΔQ/(mΔT), m = ΔQ/(cΔT) or ΔT = ΔQ/(mc). Convert grams to kilograms and energy to joules if c uses SI units. A Celsius temperature difference is numerically equal to its kelvin difference, so it is valid for ΔT. Final temperature is initial temperature plus the calculated rise.

For cooling, ΔT and the sample’s energy change can be negative if using signed changes. In a question asking the magnitude of energy released, quote a positive magnitude and state the direction of transfer. A plot of energy gained against temperature rise has gradient mc for fixed mass; c follows by dividing the gradient by mass. Check whether the plotted energy is actual sample gain or electrical input.

Scientific diagram and interpretation

A 1.0 kg sample with constant c = 400 J kg⁻¹ K⁻¹ gains E = 400ΔT J. At 20 K rise it has gained 8000 J. The slope is total heat capacity, 400 J K⁻¹.
A 1.0 kg sample with constant c = 400 J kg⁻¹ K⁻¹ gains E = 400ΔT J. At 20 K rise it has gained 8000 J. The slope is total heat capacity, 400 J K⁻¹.

Equations and relationships

ΔQ = mcΔT; c in J kg⁻¹ K⁻¹

Worked examples and applications

Heating water

Use supplied c = 4.20 × 10³ J kg⁻¹ K⁻¹, m = 0.250 kg and rise 20.0 K. ΔQ = 0.250×4200×20.0 = 21000 J = 21.0 kJ using three significant figures in the supplied quantities.

Find temperature rise

A 2.0 kg object with c = 500 J kg⁻¹ K⁻¹ gains 30 kJ. ΔT = 30000/(2.0×500) = 30 K. Starting at 15 °C, it reaches 45 °C if there is no state change.

Find mass

For ΔQ = 18000 J, c = 900 J kg⁻¹ K⁻¹ and ΔT = 20 K, m = 18000/(900×20) = 1.0 kg.

Common misconceptions

Specific heat capacity is per kilogram; heat capacity of a whole object is mc in J K⁻¹.

Examination technique

Distinguish supplied material data from universal constants and quote mass in kilograms.

For an explanation, connect the physical cause to its effect. For a calculation, state the model, rearrange before substitution, retain intermediate digits, and finish with units. For a data question, refer to the actual readings and distinguish a trend from proof of a cause.

Check your understanding

Five questions. Choose an option, then Check Answer. Read the feedback and use Try Again to reset. Use Tab between controls, arrow keys within radio groups, and Enter or Space on buttons. These local practice answers are not saved.

Question 1 of 5 · C4.4.2

The unit of c is…

Question 2 of 5 · C4.4.2

m = 2 kg, c = 400 J kg⁻¹ K⁻¹, ΔT = 5 K require…

Question 3 of 5 · C4.4.2

Rearranging for c gives…

Question 4 of 5 · C4.4.2

An energy-against-ΔT slope for fixed mass is…

Question 5 of 5 · C4.4.2

A 25 °C temperature rise is…

Take Your Learning Further

Practise

Build Your Confidence

Reinforce what you’ve learned with our topic-based workbook. Every workbook page links back to its matching online lesson, so help is always at hand.

Explore Workbooks – Physics

Master

Get Exam Ready

Challenge yourself with exam-style practice papers and detailed mark schemes. Apply your knowledge, identify areas to improve and build exam confidence.

Practice Papers – Physics