Describe a liquid specific heat capacity experiment
Use liquid mass by difference, a representative temperature and a container energy correction.
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 c = ΔE/(mΔθ) and IVt.
- Distinguish sample mass from container-plus-sample mass.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
An empty vessel is 0.080 kg and the filled vessel 0.280 kg. What mass belongs to the liquid?
What you will learn
- Describe an insulated-liquid method using sample mass and mixed temperature.
- Calculate an apparent estimate and a supplied container/loss correction.
Determine the liquid sample and its temperature
Weigh the empty container and then container plus liquid; subtract to find liquid mass. In a teacher-supervised method, use a suitable insulated vessel, appropriately rated immersed heater, thermometer and an appropriate cover with openings for the apparatus. Follow the laboratory procedure for the heater's immersion and electrical safety; the vessel is not sealed to build pressure.
Record initial temperature and heater voltage, current and time. Suitable controlled stirring or mixing gives a more representative liquid temperature rather than one hot region near the heater. Keep the heater appropriately immersed throughout the approved run and record the final mixed temperature. A drawing and paper plan are not evidence of practical competence.
Include the container in the account
For 12 V, 2.0 A and 180 s, input is 4320 J. If liquid mass is 0.20 kg and its rise is 5.0 K, assigning all input to the liquid gives apparent c = 4320 J/(kg K). The container and other apparatus may also warm, and energy can reach the surroundings. A value resembling a familiar material does not prove the assumptions correct.
If a supplied container heat capacity is 60 J/K, its gain for a 5.0 K rise is 300 J. With a further stated surroundings loss of 220 J and no other destinations in this model, liquid gain is 4320 − 300 − 220 = 3800 J, giving c = 3800 J/(kg K). This is a corrected result for the supplied account, not a measured universal material value. Container heat capacity in J/K differs from liquid specific heat capacity in J/(kg K).
Sources for this lesson
Pause and explain
Container heat capacity is 60 J/K and its rise 5.0 K. What energy does it gain?
Worked example
Empty container mass is 0.080 kg; filled mass is 0.280 kg. Input is 4320 J, rise 5.0 K, container heat capacity 60 J/K and supplied loss 220 J. Find corrected liquid c.
Show the worked solution
- Liquid mass is 0.280 − 0.080 = 0.200 kg.
- Container gain is 60 × 5.0 = 300 J; liquid gain is 4320 − 300 − 220 = 3800 J.
- c = 3800/(0.200 × 5.0) = 3800 J/(kg K).
Answer 3800 J/(kg K) for the supplied corrected model.
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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.
Describe the liquid method's mass and temperature measurements and the purpose of insulation and mixing.
Give me a hint
Identify the sample and obtain a representative temperature interval.
Compare my reasoning
- Find liquid mass by filled minus empty vessel mass.
- Use the approved insulated setup and suitable mixing to record initial/final liquid temperatures.
- Measure heater voltage, current and time, then evaluate container heating and transfer out.
A liquid-only mass, representative interval and explicit energy account.
Look for these in your work
- Separate liquid from container.
- Explain mixing and insulation without claiming perfection.
For 12 V, 2.0 A, 180 s, liquid mass 0.20 kg and rise 5.0 K, find the all-input apparent c.
Give me a hint
Do not apply corrections that have not been supplied.
Compare my reasoning
- IVt = 4320 J.
- mΔθ = 0.20 × 5.0 = 1.0 kg K.
- Apparent c = 4320/1.0 = 4320 J/(kg K).
Apparent c = 4320 J/(kg K).
Look for these in your work
- Calculate input consistently.
- Label the all-input assumption.
Correct the same 4320 J input using container heat capacity 60 J/K, rise 5.0 K, and surroundings loss 220 J. Assume no other destinations.
Give me a hint
Subtract container gain and the stated loss before dividing.
Compare my reasoning
- Container gain is 300 J.
- Liquid gain is 4320 − 300 − 220 = 3800 J.
- c = 3800/(0.20 × 5.0) = 3800 J/(kg K).
3800 J/(kg K) under the supplied corrected account.
Look for these in your work
- Use the container's heat-capacity units.
- Keep the no-other-destinations condition explicit.
Common mistakes
- Use liquid mass by difference, not gross filled mass.
- Container heat capacity and liquid specific heat capacity have different units.
- Mixing and insulation improve a method without guaranteeing perfect uniformity or zero losses.
What can you explain now?
Why use suitable mixing during the liquid method?
Compare with the explanation
To obtain a more representative sample temperature.
Local liquid near the heater may be hotter than the rest; an unmixed local reading can misrepresent the sample.
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Make it stick
Describe liquid mass by difference, representative temperature and the distinction between apparent and corrected specific heat capacity.
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.