Find thermal equilibrium with an energy balance

Define what exchanges energy

The hot sample begins at 80 °C. A 100 g cold sample and its cup begin at 20 °C. Both samples have assigned specific heat 4 J/(g K); the cup has heat capacity 40 J/K. Predict the common final temperature before changing the hot mass. The cup is part of the energy account, even though it is not part of the cold sample mass.

This is an endpoint model. The samples and cup reach one final temperature with no reaction, phase change or mixing enthalpy. The selected external energy export is prescribed. No cooling rate, elapsed time or external-surroundings temperature is calculated. It reuses the chemistry calorimetry investigation without reusing your chemistry notebook answers.

Build the balance from signed changes

For each sample, the thermal energy change is q = mc(Tf − Ti). Tf and Ti are final and initial temperatures. Use the cup’s whole-object capacity: qcup = Ccup(Tf − 20). The hot sample cools, so its change is negative. The cold sample and cup warm, so theirs are positive.

Let qoutside be the energy gained by the external surroundings. Conservation for the larger account gives qhot + qcold + qcup + qoutside = 0. For a boundary containing only samples and cup, the same statement is that their energy change equals −qoutside. The signs change with how the boundary is described; the physical transfer does not.

The OpenStax calorimetry reference develops energy balances for thermal contact. Here the material properties and exported-energy controls are teaching assumptions, not laboratory readings.

Solve a capacity-weighted equilibrium

With 100 g hot sample and zero loss, each sample has capacity 400 J/K. The cold sample and cup together have 440 J/K. The equation is 400(Tf − 80) + 440(Tf − 20) = 0. Rearranging gives 840Tf = 40800, so Tf = 48.571 °C.

The arithmetic mean 50 °C would ignore the cup. The colder side has greater total capacity, so the equilibrium lies closer to its initial temperature. More generally for two objects exchanging energy without outside transfer, Tf = (C₁T₁ + C₂T₂)/(C₁ + C₂). C₁ and C₂ are positive heat capacities. This is a weighted average, valid under constant-capacity and unchanged-phase assumptions.

The hot sample releases 12.571 kJ. The cold sample gains 11.429 kJ and the cup gains 1.143 kJ. Displayed sums can differ slightly from rounding; the calculation uses unrounded values. Equal final temperature does not mean equal energy changes, because capacities and starting temperatures differ.

At zero loss, compare hot masses 50, 100 and 200 g. Predicted endpoints are 38.750, 48.571 and 58.710 °C. The larger hot sample supplies more energy per kelvin of its cooling. None of these results states how quickly equilibrium is reached.

Interpret the loss setting before calculating

The exported energy is qoutside = f Chot × 60 K, where Chot = mhot × 4 J/(g K) and f is the selected percentage divided by 100. This is a fraction of the hot sample’s reference energy above 20 °C. It is not a fraction of the actual hot-sample release after equilibration and is not a measured heat-leak coefficient.

At 100 g and 20%, qoutside = 0.20 × 400 × 60 = 4800 J. The balance becomes 840Tf = 40800 − 4800, giving 42.857 °C. The hot sample releases 14.857 kJ; the cold sample gains 9.143 kJ, the cup 0.914 kJ and the surroundings 4.800 kJ. Energy has crossed the smaller system boundary, not disappeared.

The surroundings need not share the reported common temperature. Their capacity and temperature are unspecified. A long-term experiment in contact with room air could approach a different eventual endpoint. A model of that process needs transfer rates and an external temperature boundary.

Use discrepancies to improve an experiment

If a measured endpoint is lower than the ideal prediction, possible explanations include heat export, an underestimated cup capacity, a biased thermometer or incorrect sample masses/properties. One endpoint alone may not identify which explanation is responsible. Propose a comparison: calibrate the thermometer, measure cup capacity separately, or vary insulation while keeping other conditions fixed. Record uncertainty and repeat measurements rather than adjusting every parameter until one number matches.

Transfer to an unfamiliar pair

Two solid bodies with capacities 120 J/K and 280 J/K begin at 70 °C and 10 °C. They reach a common temperature and export 1600 J to their surroundings. Calculate the endpoint and each body’s signed energy change. Check the larger energy account.

Check the equilibrium transfer

120(Tf − 70) + 280(Tf − 10) + 1600 = 0 gives 400Tf = 9600, so Tf = 24 °C. The initially hot body changes by 120(24 − 70) = −5520 J. The cold body gains 280(24 − 10) = 3920 J. The sum −5520 + 3920 + 1600 is zero. These numbers predict an endpoint only under the stated constant-capacity assumptions.