Calorimetry and Heat Balance

Introduction

Calorimetry measures temperature changes and uses heat capacities to infer energy transfers. In a well-insulated assembly, heat lost by a hot part or reaction is gained by the cooler parts. The central equation is a signed heat balance, not an assumption that every component experiences the same heat or that temperature change itself equals reaction enthalpy.

Core explanation

Choose an assembly containing the reaction, solution and calorimeter. If little heat escapes to the room during the measurement, conservation of energy gives q reaction + q solution + q calorimeter ≈ 0, with each q signed for its named component. If the solution and apparatus warm, their q values are positive, so the reaction term is negative. If a reaction cools the solution, the solution loses heat and the reaction generally absorbs it.

For nearly constant heat capacities and a common temperature change ΔT, write q solution = m solution c solutionΔT and q calorimeter = C calΔT. Therefore q reaction ≈ −(m solution c solution + C cal)ΔT. This value applies to the actual amount that reacted. To report molar reaction enthalpy, divide by the number of moles of reaction as the balanced equation is written, using the limiting reactant and stoichiometric coefficients where necessary.

The calorimeter constant must not be double-counted. Some instruments report a calibrated total capacity that already includes the solution or vessel under a standard setup; others give apparatus capacity separately. Read the wording and units. If C total is supplied for the whole assembly, q surroundings = C totalΔT. Adding mc again would overestimate the reaction heat.

Mixing hot and cold nonreacting liquids is another calorimetry problem. In an insulated cup, q hot + q cold + q cup ≈ 0. If cup heat is negligible and both liquids are the same substance, the masses and temperatures determine a weighted final temperature. One should not simply average initial temperatures unless the masses and heat capacities are equal.

Real experiments have limitations. Heat may leak to the room; stirring may add work; a thermometer responds with delay; dissolution or evaporation may contribute heat. A calibrated apparatus and extrapolated peak temperature can reduce error. The model q lost + q gained ≈ 0 is an approximation for a suitably chosen assembly, not a claim that every physical component is perfectly isolated.

The calorimeter measures a process heat under its conditions. Whether that heat represents ΔH or ΔU depends on pressure-volume constraints and other work modes. A coffee-cup experiment is usually near constant pressure and a sealed rigid bomb near constant volume. Heat balance determines q; thermodynamic interpretation determines which state-function change it estimates.

Step-by-step reasoning

1. Draw the reaction and every warmed or cooled component. 2. Compute a signed ΔT for each component or a shared final temperature. 3. Calculate each q using its appropriate heat capacity. 4. Set the sum of q terms approximately zero for the insulated assembly. 5. Divide reaction heat by reaction extent only after solving the balance.

Visual explanation

Draw a small reaction bubble inside solution and a cup. Arrows leave the bubble and enter both solution and cup. Label q rxn < 0, q sol > 0 and q cup > 0, with q rxn + q sol + q cup ≈ 0 below. A thermometer shows the common positive ΔT.

Real-world analogy

If one account pays two recipients, its loss equals their combined gains when no fee or leakage occurs. The solution and cup can both receive energy from a reaction, so measuring only one recipient undercounts the total transfer.

Real-world example

When acid and base are mixed in an insulated cup, the solution temperature rises. The solution's mass and specific heat estimate its absorbed heat; a calibrated cup constant accounts for the cup. The negative of their sum estimates the exothermic reaction heat for the amount mixed.

Why?

Why is reaction heat the negative of the warmed surroundings heat? Energy conserved within an approximately insulated assembly must leave one part when it enters another. The signs are opposite because each component is treated as its own system.

Common misconception

“A warmer solution means q reaction is positive.” It means the solution's q is positive. The reaction that warmed it has negative q under the reaction-as-system convention.

Worked example

A reaction warms 100 g of solution by 3.0 K. Take c = 4.18 J g⁻¹ K⁻¹ and C cal = 50 J K⁻¹. The solution gains 100(4.18)(3.0) = 1254 J; the apparatus gains 50(3.0) = 150 J. Thus q reaction ≈ −1404 J. If 0.0250 mol of reaction occurred, the molar process heat is −1404/0.0250 = −56,160 J mol⁻¹, or about −56.2 kJ mol⁻¹.

Quick check

1. If both solution and cup warm, should their heat terms be added before reversing the reaction sign? Answer: Yes. Both absorb energy, so the reaction loses their combined heat if external losses are negligible.

Exam focus

Write the full signed heat balance before using a shortcut. Verify whether a supplied calorimeter constant already includes the solution. Report a molar value only after determining actual reaction extent.

Advanced insight

Calorimeter calibration may use a known electrical energy input or a reference reaction. The calibration converts observed ΔT into an effective heat capacity for the whole instrument. Uncertainty in heat leakage, temperature reading and solution heat capacity propagates into the inferred reaction enthalpy.

Summary

Calorimetry uses conservation of energy: the reaction's heat plus heat gained or lost by solution and apparatus is approximately zero in an insulated assembly. Heat capacities convert measured ΔT to energy. Pressure and work conditions then determine whether the result estimates ΔH or ΔU.

Practice questions

1. A reaction causes solution q = +800 J and cup q = +100 J. Find q rxn with negligible loss. Answer: q rxn ≈ −900 J. 2. Equal masses of the same liquid at 20 °C and 40 °C mix in an ideal insulated cup. What is the final temperature? Answer: 30 °C, because equal masses and heat capacities lose and gain equal heat. 3. Why is the simple arithmetic average wrong when one liquid mass is three times the other? Answer: The larger mass has three times the thermal capacity and weights the final temperature more strongly.