Chemistry · Lesson

Heat capacity, calorimetry and reaction enthalpy

Infer reaction heat from a measured temperature change using supplied heat capacities and an explicit apparatus ledger. Distinguish constant-pressure enthalpy from constant-volume internal energy.

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By the end, you can

Distinguish total heat capacity from mass-specific heat capacity.
Include solution and apparatus heat when inferring reaction heat.
State the conditions for q_p = ΔH and q_v = ΔU.

Temperature change needs a capacity

Heat capacity C relates a small heat input to temperature change under specified conditions, with units J K⁻¹ for the whole selected object. Mass-specific heat capacity c expresses this per mass, often J g⁻¹ K⁻¹. If c is approximately constant over the interval, q ≈ mcΔT for sensible heating under the relevant constraints.

Heat capacity depends on material, temperature, phase and conditions such as constant pressure or volume. A phase transition can absorb energy without the same temperature rise, so a single mcΔT expression does not describe an entire melting or boiling process. Use capacities supplied for the actual calculation.

Account for every relevant warming part

Choose the reacting material as system. If the solution and calorimeter warm, their heat gains can be written q_solution = mcΔT and q_cal = C_calΔT under the stated approximation. If heat losses and other transfers are negligible, q_reaction = −(q_solution + q_cal).

A warmer solution does not mean the reaction absorbed heat: the reaction can have released energy into those surroundings. Apparatus capacity matters. Omitting it is a deliberate approximation that needs justification or an estimate of its effect, rather than an automatic rule.

A supplied capacity gives a checkable line

An original hypothetical calorimeter model uses 100.0 g solution, c = 4.18 J g⁻¹ K⁻¹ and apparatus capacity C_cal = 50.0 J K⁻¹. Their combined calculated capacity is approximately 468 J K⁻¹. A 5.00 K temperature rise gives 2.34 kJ of heat gained by solution plus apparatus.

The graph shows that stipulated constant-capacity relationship, q_absorbed = (mc + C_cal)ΔT. It is not a measured calibration curve, and the waterlike value of c is a supplied exercise input rather than a property guaranteed for every solution. The reacting-system heat has the opposite sign in the simple insulated ledger.

Include solution and apparatus in the heat ledger

Solution + apparatus heat gained / kJ

Temperature rise ΔT / K

Calculated absorbed heat
— Stipulated capacity relation

Underlying data

Temperature rise ΔT / K	Solution + apparatus heat gained / kJ
1	0.468 · Calculated absorbed heat
3	1.404 · Calculated absorbed heat
5	2.34 · Calculated absorbed heat

Original constant-capacity model: 100.0 g × 4.18 J g⁻¹ K⁻¹ + 50.0 J K⁻¹ ≈ 468 J K⁻¹. Markers and dashed line calculate positive heat gained by solution plus apparatus. The reaction releases the corresponding heat in the stated insulated model.

Pressure and work conditions determine what heat measures

Enthalpy is H = U + pV. For a closed stationary process at constant pressure, with boundary work −pΔV and no other work, the reacting-system heat q_p equals ΔH. At constant volume, q_v equals ΔU when other work and energy transfers are absent or appropriately corrected. Constant volume alone does not exclude every possible kind of work.

Enthalpy H and its change ΔH refer to the whole specified system, so they scale with material amount. A molar reaction result divides the change by the stated stoichiometric reaction amount. For the model A → B consuming 0.0500 mol A on a 1:1 basis, total ΔH = −2.34 kJ under the stated constant-pressure conditions; the mean molar reaction enthalpy change is −46.8 kJ mol⁻¹ on that basis. These are a total change and a per-mole result, respectively.

Worked examples

Infer reaction heat from the full capacity

Use the hypothetical 100.0 g solution with c = 4.18 J g⁻¹ K⁻¹, C_cal = 50.0 J K⁻¹ and ΔT = +5.00 K. The model reaction A → B consumes 0.0500 mol A. Find heat released and the constant-pressure molar enthalpy change under PV-only and negligible-loss assumptions.

Solution heat = 100.0 × 4.18 × 5.00 = +2090 J.
Apparatus heat = 50.0 × 5.00 = +250 J.
Reacting-system heat = −(2090 + 250) = −2340 J = −2.34 kJ.
Divide by 0.0500 mol on the stated 1:1 reaction basis.

q_reaction = total ΔH = −2.34 kJ under the stated conditions. The mean molar reaction enthalpy change is −46.8 kJ mol⁻¹ on the stated reaction basis; the reacting system is exothermic.

Try it yourself

Does c in J g⁻¹ K⁻¹ already represent the whole solution’s heat capacity?
Show the answer and reasoning
No.
Multiply the mass-specific capacity by the solution mass.
In the insulated model, solution plus apparatus gain +2.34 kJ. What is q_reaction?
Show the answer and reasoning
−2.34 kJ.
The reacting material loses the energy gained by its model surroundings.
Can apparatus heat capacity always be omitted without changing the inferred reaction heat?
Show the answer and reasoning
No.
Its temperature change can represent a significant heat gain or loss.
Does constant volume alone guarantee q_v = ΔU if electrical work also enters the system?
Show the answer and reasoning
No.
Other work must be absent or accounted for in the energy balance.

What to carry forward

Choose the capacity and constraints for the calculation.
Include relevant apparatus heat or justify neglecting it.
Reacting-system heat has the opposite sign to its thermal surroundings in the simple ledger.
q_p = ΔH and q_v = ΔU each require stated work and transfer conditions.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.