Case Study
Passage with linked questions
Case Set 1
Case AnalysisPassage
A chemistry teacher demonstrates the concept of internal energy by performing two experiments on the same adiabatic system containing water in an insulated beaker. In Experiment 1, she rotates paddle wheels inside the water using mechanical work of 2 kJ. In Experiment 2, starting from the original state, she uses an electrical immersion rod to supply 2 kJ of electrical work to the same system. She measures the temperature change in both cases and finds them identical. She asks the class to explain why this happens and what thermodynamic principle it illustrates. The class also discusses what would happen if the walls were thermally conducting instead of insulating.
Question 1: What type of system is the insulated beaker described in the passage, and what is the value of q for each experiment?
- The insulated beaker is an adiabatic system — its walls do not allow heat transfer between system and surroundings, so q = 0 for both experiments.
- Since no heat flows in or out, any change in internal energy arises entirely from the work done on the system: ΔU = wad.
- For both experiments, q = 0; the system exchanges energy only through work (mechanical or electrical), not through heat.
Question 2: Why is the temperature change identical in both experiments despite using different types of work?
- Internal energy U is a state function — its value depends only on the current state of the system, not on the path or method used to reach that state.
- Joule's experiments (1840–50) established this: a given amount of adiabatic work, regardless of its type (mechanical or electrical), produces the same change in internal energy and hence the same temperature change.
- Since ΔU = wad = 2 kJ in both cases, and the same system is used, the temperature rise ΔT is identical in both experiments, confirming U is path-independent.
Question 3: If the insulating walls were replaced with thermally conducting copper walls and 2 kJ of mechanical work were done while heat q = 0.5 kJ was simultaneously lost to the surroundings, calculate ΔU and identify the type of system. Write the applicable form of the first law.
- With thermally conducting walls and sealed (no matter exchange), the system becomes a closed system — it can exchange energy (heat and work) but not matter with surroundings.
- Applying the first law: ΔU = q + w. By IUPAC convention, heat lost by the system is negative: q = -0.5 kJ; work done on the system is positive: w = +2 kJ.
- Therefore ΔU = -0.5 + 2 = +1.5 kJ. The internal energy increases by 1.5 kJ, less than in the fully adiabatic case because 0.5 kJ of the supplied work's energy equivalent is lost as heat.
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