Distinguish uncertainty from omitted thermal pathways

Establish the model and evidence

Inferring a material property requires a model of where energy went. The synthetic hot sample cools into water and a cup. Its true specific heat is supplied so that you can test the inference, but the inference equation deliberately ignores an adjustable heat loss. A second control introduces a final-temperature sensor offset.

If the combined calorimeter were isolated, energy conservation would give mc(350−Tf)=(800+Ccup)(Tf−290). Solve for c rather than inserting an unexplained formula. The denominator uses the sample’s cooling interval, while the numerator uses the receiver’s warming interval. They are not equal temperature changes.

The displayed interval combines an assigned ±0.10 K reading bound and ±5 J/K cup-capacity bound using conservative endpoints. Both effects move the inferred ratio in known directions over this range. These bounds do not include an unknown sensor zero offset or unmodeled heat leakage. Therefore a narrow interval can miss the supplied true value without violating its own stated arithmetic.

Compare competing explanations instead of declaring that a disagreement proves the material changed. Heat loss lowers the measured final temperature; a positive sensor offset can partly hide that effect. A single final reading may not distinguish those causes. An independent thermometer check or a controlled comparison of insulation and waiting time would supply different evidence. No heating apparatus is required here.

Worked evidence

With no loss or sensor offset, Tf=(50×350+850×290)/900=293.333 K. The inferred c is 850×3.333/[0.100×56.667]≈500 J/(kg K). Omitting the cup would give a different inference despite the same measured temperature.

Predict and investigate

Predict how heat loss and a positive final-temperature offset move the inferred specific heat, separately.

Capture zero loss/zero offset, then 200 J loss/zero offset, then the same loss with +0.20 K offset. Compare the inferred values and reported bounds. Identify an independent check that separates compensating errors.

Transfer to changed conditions

A 0.20 kg sample cools by 20 K while the receiver gains 1600 J. Assuming an isolated boundary, infer c; how would an uncounted 400 J loss change the true required c?

Compare the worked transfer

Ignoring loss gives 400 J/(kg K). Including a 400 J transfer out requires sample loss 2000 J, hence c=500 J/(kg K).

Explain the boundary

State exactly which uncertainty sources are included. Explain why agreement with a reference value can occur through cancellation of errors.

Sources and authorship

Physical relationship reference. Current AP Physics 2 framework, checked 30 October 2026. The numerical cases, text and computed representations are original. All browser evidence is synthetic or calculated, not a record of completed practical work. No hazardous apparatus is required.