Calorimetry Energy Balance

Introduction

A reaction in a calorimeter changes the temperature of its surroundings. To infer reaction heat, account for every part that warms or cools: usually the solution and sometimes the calorimeter vessel. In an approximately insulated experiment, heat gained by those parts equals heat lost by the reaction, and vice versa. The sign comes from this energy balance, not from the temperature reading alone.

Core explanation

For a well-mixed solution of mass m sol and specific heat capacity c sol, its heat change is q sol = m sol c sol ΔT. Here ΔT = T final − T initial, and a positive ΔT makes q sol positive. If a calorimeter's own heat capacity C cal is supplied separately, q cal = C cal ΔT. Under a negligible-environment-exchange approximation, q rxn + q sol + q cal = 0, so q rxn = −(m sol c sol + C cal)ΔT. This q rxn is for the actual amount reacted in that run, not automatically for one mole of reaction.

The units distinguish the two heat-capacity quantities. Specific heat capacity c sol may be in J g⁻¹ K⁻¹ and must be multiplied by grams of solution; calorimeter heat capacity C cal is already in J K⁻¹ and is multiplied only by ΔT. If C cal was calibrated to include the solution or another component, adding that component again would double count. Read how the problem defines the apparatus constant. A temperature difference in Celsius has the same numerical magnitude as a difference in kelvin.

Suppose 100.0 g of dilute aqueous solution is approximated with c sol = 4.18 J g⁻¹ K⁻¹, and the cup has C cal = 50.0 J K⁻¹. If temperature rises by 5.00 K, q sol = 2090 J and q cal = 250 J. Total surroundings gain 2340 J, so q rxn = −2340 J. A negative reaction heat corresponds to heat released by the reaction under this sign convention. If temperature falls, the surroundings heat terms are negative and q rxn is positive: the reaction absorbs heat.

To report heat per mole of reaction as written, calculate reaction extent ξ from the limiting reagent and divide q rxn by ξ. For example, if ξ = 0.0200 mol in the preceding run, heat per mole extent is −2340 J/0.0200 mol = −117 kJ mol⁻¹. If a displayed equation is multiplied by two, its per-mole-of-equation enthalpy doubles, so always state the balanced equation used. In a constant-pressure calorimeter, an appropriate measured heat can approximate the reaction enthalpy change under the experiment's conditions. A bomb calorimeter at constant volume instead relates directly to internal-energy change and may need further thermodynamic conversion.

An actual experiment exchanges some energy with air and can have temperature lag, incomplete mixing, phase changes or varying heat capacities. The simple insulated balance is a model whose assumptions should be stated. A temperature rise does not prove that all heat came from the named reaction if mixing, dissolution or side reactions also release heat.

Step-by-step reasoning

1. Define the system as the chemical reaction and list all surroundings terms included by the question. 2. Find solution mass, stated specific heat capacity, calorimeter heat capacity and ΔT. 3. Compute q sol and q cal with units and signs. 4. Use q rxn = −(q sol + q cal) under the stated insulated approximation. 5. If requested, divide by the calculated reaction extent to express heat per mole of balanced reaction.

Visual explanation

Draw a central reaction box with an outward heat arrow to two adjacent boxes: solution and cup. Above the solution write m c ΔT; above the cup write C cal ΔT. A bracket sums those gains, and a minus sign points back to q rxn. For an endothermic process, reverse the heat arrow and let ΔT be negative.

Real-world analogy

If one account pays two bills, the amount withdrawn equals the sum deposited in the two receiving accounts, with the opposite sign on the payer's balance. A calorimeter energy balance similarly gives the reaction's heat from the solution and vessel's measured heat changes. The analogy is about accounting, not about how molecular energy is transferred.

Real-world example

In a neutralisation experiment, acid and base are mixed in an insulated cup. The solution warms, and a calibrated cup also warms slightly. The measured temperature rise and both heat capacities estimate heat released by the reaction. The amount reacting comes from the acid–base stoichiometry, not simply the total mixture volume.

Why?

Why is q rxn the negative of the sum of surroundings terms? Energy transferred out of the reaction enters the solution and vessel when outside losses are negligible. Conservation of energy makes their signed heat changes add to zero under the selected system boundary.

Common misconception

“A positive temperature change means q rxn is positive.” Positive ΔT means the solution and cup gained heat. The reaction is the heat source, so its q is negative in the insulated exothermic case. State the system before assigning a sign.

Worked example

A reaction warms 80.0 g solution from 20.0 °C to 23.0 °C. Take c sol = 4.00 J g⁻¹ K⁻¹ and C cal = 40.0 J K⁻¹ for the cup alone. ΔT = +3.0 K. Solution heat gain is q sol = (80.0)(4.00)(3.0) = 960 J. Cup heat gain is q cal = (40.0)(3.0) = 120 J. Thus q rxn = −1080 J = −1.08 kJ for this run. If 0.0150 mol of reaction as written occurred, heat per mole extent is −1.08/0.0150 = −72.0 kJ mol⁻¹. No separate heat loss to room air is assumed.

Quick check

1. A solution absorbs +500 J and a cup absorbs +100 J. What reaction heat follows from the insulated balance? Answer: q rxn = −(500 + 100) = −600 J, indicating heat released by the reaction.

Exam focus

Write the signed energy balance before inserting numbers. Distinguish specific heat capacity from the vessel's total heat capacity, and check whether a supplied calibration constant already includes solution heat. Divide by reaction extent only after finding limiting stoichiometry.

Advanced insight

The measured temperature change is often extrapolated to the mixing time to correct for heat exchange with the surroundings during data collection. Such corrections improve an experimental estimate but do not change the core conservation equation; they refine the ΔT used in it.

Summary

Under an insulated calorimeter model, reaction heat is the negative sum of solution and vessel heat changes. Compute m c ΔT and C cal ΔT separately with signs, then scale the run's heat by reaction extent if a molar quantity is requested.

Practice questions

1. A 50.0 g solution with c = 4.00 J g⁻¹ K⁻¹ warms 2.00 K. What is q sol? Answer: q sol = 50.0 × 4.00 × 2.00 = +400 J. 2. If the cup's heat capacity is 25.0 J K⁻¹, what is q rxn for that run under insulation? Answer: q cal = +50.0 J, so q rxn = −(400 + 50.0) = −450 J. 3. If that run represents 0.0100 mol reaction extent, what is heat per mole extent? Answer: −450 J/0.0100 mol = −45.0 kJ mol⁻¹.