Define specific heat capacity and calculate energy changes
Use sample mass and temperature change to find an energy gain within one state.
Section P2.2.2 · Specific heat capacity
Cross-referenced to the 2028–2029 0653 section using original teaching from the related single-science courses. A 0653-specific objective-by-objective and Core/Supplement review is in progress. Written practice is self-checked and does not demonstrate supervised laboratory competence or award an official Cambridge grade.
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.
Sources for this lesson
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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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.
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.