Define and use specific heat capacity
Relate energy transfer to mass, specific heat capacity and temperature change, with consistent units and sign reasoning.
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 14. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use temperature differences, proportional reasoning, energy, mass and SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does specific heat capacity describe?
What you will learn
- Define specific heat capacity
- Use Q = mcΔT
- Distinguish a temperature change from an absolute temperature
Specific heat capacity describes energy per mass per rise
The specific heat capacity c of a substance is the energy required to raise the temperature of unit mass by one kelvin. Its unit is J kg⁻¹ K⁻¹. A large c means more energy is needed for the same mass and temperature rise.
For a temperature change ΔT, the energy transferred is Q = mcΔT. Use mass in kilograms, c in J kg⁻¹ K⁻¹ and ΔT in kelvins; the result is in joules. A temperature difference of 1 K has the same size as a difference of 1 °C.
Check the energy direction and temperature change
For heating, Q is positive for energy transferred to the object and its temperature rises. For cooling, energy leaves and the temperature falls; the magnitude is still mc times the magnitude of the temperature change.
Use the initial and final temperatures to find ΔT = Tfinal − Tinitial. Do not substitute an absolute temperature for ΔT, and convert the mass to kilograms before calculating.
Pause and explain
A 2.0 kg object has c = 500 J kg⁻¹ K⁻¹ and warms by 3.0 K. What energy is transferred?
Worked example
A 0.40 kg metal block with c = 900 J kg⁻¹ K⁻¹ is heated by 35 K. Calculate the energy supplied.
Show the worked solution
- Use Q = mcΔT.
- Q = 0.40 × 900 × 35.
- Give the energy in joules.
Answer Q = 12 600 J.
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Test this topic →From guided practice to a new situation
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.
A 0.25 kg sample with specific heat capacity 800 J kg⁻¹ K⁻¹ warms by 12 K. Find the energy supplied.
Give me a hint
Multiply mass, specific heat capacity and temperature change.
Compare my reasoning
- Q = mcΔT.
- Q = 0.25 × 800 × 12.
- State the result in joules.
Q = 2400 J.
Look for these in your work
- Equation selected
- Mass in kilograms
- Energy unit stated
A 1.5 kg object receives 9000 J and warms by 10 K. Calculate its specific heat capacity.
Give me a hint
Rearrange Q = mcΔT to make c the subject.
Compare my reasoning
- c = Q/(mΔT).
- c = 9000/(1.5 × 10).
- Give J kg⁻¹ K⁻¹.
c = 600 J kg⁻¹ K⁻¹.
Look for these in your work
- Equation rearranged
- Denominator includes mΔT
- Unit stated
Two equal-mass objects receive equal energy. Object A rises by 4 K and object B by 10 K. Compare their specific heat capacities.
Give me a hint
For fixed Q and m, use c = Q/(mΔT) to compare inverse temperature changes.
Compare my reasoning
- c is inversely proportional to ΔT when Q and m are fixed.
- cA/cB = ΔTB/ΔTA = 10/4.
- Object A has 2.5 times the specific heat capacity of object B.
cA = 2.5cB; the smaller temperature rise indicates the larger specific heat capacity.
Look for these in your work
- Fixed quantities identified
- Inverse ratio applied
- Conclusion explained
Common mistakes
- Using mass in grams without converting to kilograms
- Using the final temperature instead of ΔT
- Writing the heat-capacity unit without the per-kilogram and per-kelvin factors
What can you explain now?
What does specific heat capacity describe?
Compare with the explanation
Energy needed per unit mass for a one-kelvin rise
That is the definition of c.
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Make it stick
Which three quantities determine the energy change in Q = mcΔT?
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