Choose a system boundary for thermal equilibration

Establish the model and evidence

A body can conserve its mass while gaining or losing internal energy. A closed subsystem is therefore not automatically isolated. Here the hot body, cold body and cup exchange energy internally, while their combined boundary is stipulated to exchange neither heat nor work with the surroundings.

For constant heat capacities and no phase change, each energy change is CΔT. Set the sum to zero and solve for the shared equilibrium temperature. The result is a heat-capacity-weighted average, not usually the arithmetic mean of initial temperatures. The cup belongs in the receiving capacity because it begins at the cold-body temperature and warms with that body.

The two plotted histories use a stipulated conductance to illustrate approach to equilibrium. They are not experimental cooling curves. Their precise time constant depends on the extra rate model; the final equilibrium temperature follows from energy conservation without requiring that rate law. Separate these two conclusions when explaining evidence.

At equilibrium the bodies still contain moving particles and exchange microscopic energy. What vanishes is the net thermal transfer between equal-temperature bodies. A local reduction in one body’s internal energy does not violate conservation when another body gains the same amount. This same boundary reasoning will later prevent incorrect entropy claims about a subsystem.

Thermal contact does not always mean direct solid-to-solid contact. Conduction transfers energy through microscopic interactions without bulk material flow. Convection transports energy with a moving fluid. Thermal radiation carries energy electromagnetically and can cross a vacuum. A warm object and a cooler object can each emit radiation while the net transfer is from warmer to cooler. These mechanisms can coexist; the stipulated conductance here combines the chosen contact pathway into one simplified rate law.

Worked evidence

With hot capacity 100 J/K at 350 K, cold capacity 100 J/K and cup capacity 20 J/K at 300 K, Tf=(100×350+120×300)/220≈322.727 K. The hot body loses about 2727.27 J; the cold body and cup gain that amount together.

Predict and investigate

Will doubling only the cold-body heat capacity move the equilibrium temperature toward the hot or cold initial temperature?

Capture cold capacity 100 J/K at 300 K, then 200 J/K at the same temperature. Compare final temperatures and all three signed energy changes. Explain why the plotted approach rate is a separate assumption.

Transfer to changed conditions

A 200 J/K object at 330 K meets a 100 J/K object at 300 K in an ideal isolated container of negligible heat capacity. Find Tf.

Compare the worked transfer

Tf=(200×330+100×300)/300=320 K. Energy changes are −2000 J and +2000 J.

Explain the boundary

Identify which boundaries are closed and which composite is isolated. Explain why the cup cannot be omitted simply because its temperature was not the main target.

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.