Account for heat in the samples, cup and surroundings

Choose the boundary before writing signs

Heat transferred into an object is positive; heat leaving it is negative. The hot sample releases energy, while the colder sample and cup gain it. If energy also leaves the samples and cup, include an external destination. Calling that transfer a loss does not mean energy ceases to exist.

The hot sample begins at 80 °C; the 100 g cold sample and cup begin at 20 °C. Both samples have assigned specific heat 4 J/(g K). The cup has heat capacity 40 J/K. It needs 40 J to warm by one kelvin regardless of the sample mass. The calculation assumes one final common temperature, no phase change, no reaction and no enthalpy of mixing.

Use a weighted balance

For each sample, q = mc(Tf − Ti). Here Tf is final temperature and Ti is its initial temperature. For the cup, q = Ccup(Tf − 20), where Ccup = 40 J/K. Conservation gives qhot + qcold + qcup + qoutside = 0.

At 100 g hot mass and zero external loss, the equation is 400(Tf − 80) + 400(Tf − 20) + 40(Tf − 20) = 0. Solving gives Tf = 40,800/840 = 48.571 °C. The arithmetic average 50 °C ignores the cup. A weighted average is needed because the objects have different heat capacities.

The hot sample releases 12.571 kJ. The cold sample gains 11.429 kJ and the cup gains 1.143 kJ, with small discrepancies in displayed sums possible from rounding. Keep unrounded values when checking conservation.

Interpret the loss control precisely

The external loss is prescribed as a fraction of the hot sample’s energy above 20 °C: qoutside = f × Chot × 60 K, where f is the selected percentage divided by 100. It is not a percentage of the actual hot-sample release, nor a fitted cooling rate.

For 100 g and 20%, qoutside = 0.2 × 400 × 60 = 4,800 J. Final temperature becomes 42.857 °C. The hot sample now releases about 14.857 kJ; the cold sample gains 9.143 kJ and the cup 0.914 kJ. Those gains plus 4.800 kJ outside balance the release. Increasing the hot mass changes both its heat capacity and the reference used to prescribe loss.

The cards show energy destinations, not relative sizes or temperatures of the surroundings. The widget does not calculate how fast equilibrium occurs. A physical calorimeter also needs calibration, stirring, temperature measurement and a justified estimate of external transfer.

Connect the balance to reaction measurements

A reaction calorimeter can infer reaction heat from the opposite heat change of the measured surroundings only when the relevant transfers and heat capacities are included. At constant pressure with only pressure-volume work, reaction heat corresponds to enthalpy change. This mixing model contains no reaction and cannot supply an unknown reaction enthalpy by itself.

Transfer: explain what would happen if the cup capacity were zero and both equal-mass samples had the same specific heat, with no external loss. Then explain why that answer cannot simply be reused for the current cup.

Check the calorimetry transfer

With only equal heat capacities at 80 and 20 °C, the final temperature is 50 °C. The real modeled cup is another initially cold energy recipient, so the displayed no-loss value is lower, 48.571 °C. Neither value requires energy to disappear.

OpenStax calorimetry provides reference heat-balance reasoning.