Separate heater input from the sample energy gain
Build a heating energy account before calculating c and explain biased estimates.
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
- Use energy = power × time.
- Calculate c from the energy received by a sample.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A 50 W heater runs steadily for 240 s. What is its input energy?
What you will learn
- Use a supplied steady heater power or IV to find input energy.
- Separate sample gain, apparatus gain and energy leaving the chosen system.
Follow all the energy destinations
For constant heater power P over time t, input energy is Pt. If the heater voltage and current are steady, P = IV. In a supplied heating account, input can increase the sample's internal energy, warm apparatus and leave for the surroundings. Use the stated boundary; energy entering the heater circuit is not automatically the sample-only gain.
A 50 W heater runs for 240 s, giving 12 000 J input. If 2000 J leaves for the surroundings and other apparatus gains 1000 J, the sample gains 9000 J. With m = 0.50 kg and Δθ = 20 K, c = 9000/(0.50 × 20) = 900 J/(kg K). The diagram's proportional bars show this supplied account, not newly measured values.
Understand apparent values and missing corrections
Assigning all 12 000 J to that sample gives 1200 J/(kg K), an overestimate in this particular account. Positive energy diverted elsewhere makes the sample-only gain smaller. Insulation reduces unwanted transfer but does not prove it is zero; the container, heater and probe can also receive energy.
When the corrections are unknown, report Pt/(mΔθ) as an apparent estimate under the assumption that all input warmed the sample. Do not silently call it the true material value. A smaller measured rise caused by a lagging probe can also raise that apparent c. Bounds need stated assumptions: other unknown energy inputs, non-uniform temperatures or inaccurate measurements prevent a universal guaranteed direction.
Sources for this lesson
Pause and explain
Input is 12 000 J, apparatus gain 1000 J and surroundings loss 2000 J. What enters the sample?
Worked example
Find sample c for input 50 W × 240 s, surroundings loss 2000 J, apparatus gain 1000 J, mass 0.50 kg and rise 20 K.
Show the worked solution
- Input is 12 000 J.
- Sample gain is 12 000 − 2000 − 1000 = 9000 J.
- c = 9000/(0.50 × 20) = 900 J/(kg K).
Answer 900 J/(kg K); all-input calculation would give an apparent 1200 J/(kg K).
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
Calculate c for the complete 12 000 J account with 2000 J lost, 1000 J gained by apparatus, m = 0.50 kg and Δθ = 20 K.
Give me a hint
Find sample energy before dividing by mass and interval.
Compare my reasoning
- Sample gain is 9000 J.
- mΔθ = 0.50 × 20 = 10 kg K.
- c = 9000/10 = 900 J/(kg K).
900 J/(kg K).
Look for these in your work
- Separate the destinations.
- Use the sample-only numerator.
An 8000 J input gives apparatus gain 1000 J and surroundings loss 2000 J. A 0.25 kg sample rises by 10 K. Find c.
Give me a hint
Use conservation before the heat-capacity equation.
Compare my reasoning
- Sample gain is 8000 − 1000 − 2000 = 5000 J.
- mΔθ = 0.25 × 10 = 2.5 kg K.
- c = 5000/2.5 = 2000 J/(kg K).
2000 J/(kg K).
Look for these in your work
- Subtract both supplied amounts.
- Keep sample mass and temperature interval.
Only Pt, sample mass and measured rise are known. Losses and apparatus heating have not been measured. Can you report a unique corrected c? Explain.
Give me a hint
Distinguish an all-input estimate from measured corrections.
Compare my reasoning
- Pt/(mΔθ) assumes every input joule warms the sample.
- Unknown destinations can make that numerator inappropriate.
- No unique corrected c follows without further measurements and assumptions.
No; an apparent all-input estimate is available, but a unique corrected c is not established.
Look for these in your work
- Label the estimate's assumption.
- Explain the missing evidence.
Common mistakes
- Input energy and sample-only gain can differ.
- Insulation is not proof of zero losses.
- Unknown corrections do not justify a unique corrected value.
What can you explain now?
Why does using all heater input overestimate c in the given account?
Compare with the explanation
Some input goes to the apparatus and surroundings.
The numerator is too large for the same sample mass and measured rise.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Write the sample energy account first, then explain why an apparent heat capacity may differ from a corrected value.
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.