Chemical Energy & Thermodynamics — Extension · Physical Science: Chemistry, Grade 11
Use calorimetry, Hess’s law, entropy, and free energy with explicit systems and signs.
Calorimetry and System Boundaries
Investigate
Where does energy go when water samples are mixed?
Use two dedicated sessions: first define the model, boundary, variables and comparison, then collect suitable records or analyze the complete supplied-data route; second complete calculations/graphs, compare alternatives, revise and prepare the report. Extend collection time where needed. Supplied-data work does not certify supervised laboratory minutes or mandated investigations.
Materials
- Two small containers
- Comfortably warm tap water androom-temperature water
- Thermometer if available; adult setup
- Provided-data option
Before you begin
Use only the listed household materials or analyze the supplied data. An adult should review setup. No flames, strong acids/alkalis, unknown chemicals, household mains, radiation sources, or high-speed projectiles are needed. Keep food-study materials separate from eating and wash hands afterward.
Plan a fair test
Compare equal 50 g samples at 20°C and 40°C; assume equal specific heat and small apparatus/loss effects.
Method
- Write a prediction and the physical/chemical mechanism behind it.
- Draw the setup or system boundary; label variables and units.
- Predict equilibrium temperature from heat balance.
- Mix carefully if equipment is available; record time-dependent temperature rather than a single immediate spot reading.
- Calculate heat gain/loss usingc=4.18 J/g°C; interpret discrepancy.
- Record repeated observations where possible; keep original readings and note resolution or uncertainty.
- Write a claim supported by specific numbers, explain the mechanism, and identify one limitation and a next test.
Example data — for analysis when materials are unavailable
Illustrative teaching data, not measurements from a real district laboratory. Analyze these when materials are unavailable and identify them as provided data. Do not claim to have performed a measurement you did not make.
| Trial | Initial cool(°C) | Initial warm(°C) | Final(°C) |
|---|---|---|---|
| 1 | 20 | 40 | 29.5 |
| 2 | 20 | 40 | 29.6 |
| 3 | 20 | 40 | 29.4 |
Analyze and explain
- Find ideal temperature and measured mean.
- Estimate warm-water loss and cool-water gain at 29.5°C.
- Name a missing energy term.
- Link the pattern to a labeled particle, structure, energy or amount model from the chapter.
- Distinguish supplied illustrative records from measurements actually collected.
- Identify uncertainty and propose a comparison that would test an alternative explanation.
What a strong explanation includes
- A stated prediction, conditions, and controlled variables or explicit model boundary.
- A labeled table/graph with units, calculations, and attention to repeated values or uncertainty.
- A claim citing numerical evidence and a scientifically accurate mechanism.
- A limitation, alternative explanation, and useful follow-up test.
- Preserve initial reasoning and an evidence-supported revision.
- Check units, species identity, system boundary and physical assumptions.
Parent / teacher guide: discuss after the investigation
Ideal final 30°C; supplied mean 29.5°C. Warm loses 50×4.18×10.5=2194.5 J while cool gains 1985.5 J. The 209 J difference can involve apparatus or surroundings and measurement limits; it is not destroyed energy. Preserve the original observations, intermediate working and a labeled mechanism. Report the relevant boundary and model assumptions, discuss an alternative explanation, and identify the next useful measurement; do not replace unexpected data with the illustrative record.
Compare this model with the learner’s actual evidence. Different measurements can support a different conclusion; check the reasoning and limitations, not matching wording.
Extend the chapter investigation by evaluating its evidence and proposing a better test.
Problem 1
Answer and explanation
Ideal final 30°C; supplied mean 29.5°C. Warm loses 50×4.18×10.5=2194.5 J while cool gains 1985.5 J. The 209 J difference can involve apparatus or surroundings and measurement limits; it is not destroyed energy. Preserve the original observations, intermediate working and a labeled mechanism. Report the relevant boundary and model assumptions, discuss an alternative explanation, and identify the next useful measurement; do not replace unexpected data with the illustrative record.
Problem 2
Answer and explanation
Ideal final 30°C; supplied mean 29.5°C. Warm loses 50×4.18×10.5=2194.5 J while cool gains 1985.5 J. The 209 J difference can involve apparatus or surroundings and measurement limits; it is not destroyed energy. Preserve the original observations, intermediate working and a labeled mechanism. Report the relevant boundary and model assumptions, discuss an alternative explanation, and identify the next useful measurement; do not replace unexpected data with the illustrative record.
Problem 3
Answer and explanation
Ideal final 30°C; supplied mean 29.5°C. Warm loses 50×4.18×10.5=2194.5 J while cool gains 1985.5 J. The 209 J difference can involve apparatus or surroundings and measurement limits; it is not destroyed energy. Preserve the original observations, intermediate working and a labeled mechanism. Report the relevant boundary and model assumptions, discuss an alternative explanation, and identify the next useful measurement; do not replace unexpected data with the illustrative record.
Problem 4
Answer and explanation
Ideal final 30°C; supplied mean 29.5°C. Warm loses 50×4.18×10.5=2194.5 J while cool gains 1985.5 J. The 209 J difference can involve apparatus or surroundings and measurement limits; it is not destroyed energy. Preserve the original observations, intermediate working and a labeled mechanism. Report the relevant boundary and model assumptions, discuss an alternative explanation, and identify the next useful measurement; do not replace unexpected data with the illustrative record.
Problem 5
Answer and explanation
Ideal final 30°C; supplied mean 29.5°C. Warm loses 50×4.18×10.5=2194.5 J while cool gains 1985.5 J. The 209 J difference can involve apparatus or surroundings and measurement limits; it is not destroyed energy. Preserve the original observations, intermediate working and a labeled mechanism. Report the relevant boundary and model assumptions, discuss an alternative explanation, and identify the next useful measurement; do not replace unexpected data with the illustrative record.
Problem 6
Answer and explanation
Ideal final 30°C; supplied mean 29.5°C. Warm loses 50×4.18×10.5=2194.5 J while cool gains 1985.5 J. The 209 J difference can involve apparatus or surroundings and measurement limits; it is not destroyed energy. Preserve the original observations, intermediate working and a labeled mechanism. Report the relevant boundary and model assumptions, discuss an alternative explanation, and identify the next useful measurement; do not replace unexpected data with the illustrative record.