Heat capacities and calorimetry

A warm object and a cup that also warms

A warm 0.0800 kg solid at 80.0 °C is placed in 0.150 kg of a water-side liquid at 20.0 °C. The cup and thermometer begin at 20.0 °C too. If the assembly is insulated, what final temperature should we expect? Does it matter that the cup absorbs energy along with the liquid?

Use an assigned constant specific heat capacity of 900 J/(kg K) for the solid and 4180 J/(kg K) for the liquid, plus a total cup-and-instrument capacity of 30.0 J/K. These are teaching inputs, not measurements of an identified sample. The liquid value is an illustrative approximation; real water properties depend on temperature, pressure and phase. Assume no reaction, dissolution energy, evaporation or phase change.

Select the whole sample–liquid–cup assembly as the system. At the stated fixed ambient pressure, with no external heat loss or non-volume work, its total enthalpy H = U + pV is unchanged. Here U is internal energy, p pressure and V volume. The component enthalpy changes can be added under the stated absence of mixing or reaction contributions. For condensed materials with negligible volume work, this is also the familiar thermal energy balance.

First follow the capacity definitions and units, the cup balance, the inverse measurement, and the variable-capacity, calibration and melting examples. Sections marked “Further” provide a return path through general response identities, material models and uncertainty. Published preparation is “Physical systems, states, boundaries, and environments” and “Units, standards, and physical constants”. Needed heat, work and enthalpy terms are introduced locally; the planned earlier thermodynamics subjects are not required live prerequisites.

Heat capacity needs a condition and a temperature

Heat capacity describes how a system’s energy changes when its temperature changes under specified conditions. In a heating experiment where heat is the relevant energy input, the differential ratio δQ/dT must carry those conditions, such as fixed pressure or fixed volume. It is not a universal property of every possible path through the same state. The symbol δQ emphasizes that heat is a transfer, not a stored state quantity.

For a closed fixed-composition system, the standard total capacities are C_V = (∂U/∂T)_V and C_p = (∂H/∂T)_p, with amount and other relevant coordinates held fixed. Their units are J/K. These state derivatives also let us determine capacities using an electrical heater, provided its work input and every other energy channel are included in the balance. One must not call an electrical input heat merely because it raises temperature.

With only pressure–volume work and suitable equilibrium states, fixed-volume heating has Q = ΔU, whereas constant-pressure heating has Q = ΔH. Other work, changing composition, a phase transition or a changing constraint can alter the inference from a temperature change. A heat capacity should therefore be reported with its material state, constraint and temperature range.

For a finite interval, E/ΔT is usually an average capacity under the specified energy balance. It equals a local derivative only when the capacity is effectively constant over that interval or an appropriate limiting procedure is used. An infinitesimal definition and a finite calorimeter reading are related, but they are not interchangeable without that check.

Total, molar and specific capacities count different amounts

Total heat capacity C in J/K belongs to the selected amount of material. Mass-specific heat capacity c in J/(kg K) is capacity per mass, and molar heat capacity C_m in J/(mol K) is capacity per amount of specified entities. For the same uniform material and constraint, C = mc = nC_m. If molar mass M is in kg/mol, C_m = Mc. Keep a pressure or volume subscript when it matters.

For a separate hypothetical sample with m = 0.0800 kg, assigned M = 0.0400 kg/mol and c = 900 J/(kg K), the amount is 2.00 mol, C = 72.0 J/K and C_m = 36.0 J/(mol K). These three capacity values describe the same sample with different counting conventions. Replicating the sample doubles C while leaving its uniform-state c and C_m unchanged.

A mole must refer to a stated entity. A molar value per mole of atoms differs from a value per mole of multi-atom formula units. A mass-specific value also cannot be inserted into nC_mΔT without conversion. In a mixture, capacity can depend on composition as well as temperature and pressure.

A temperature change of 1 K has the same numerical size as a change of 1 °C. Thus m cΔT may use a properly formed Celsius difference. Absolute-temperature factors, such as T in a thermodynamic response identity, still use kelvin. The number 20 °C cannot be substituted as 20 K.

Why constant pressure and constant volume differ

At fixed volume, pressure–volume work is zero, so heating a closed simple system with no other work directly changes U. At fixed pressure, thermal expansion can transfer work outward as well. Enthalpy H = U + pV includes the pressure–volume contribution, which is why C_p and C_V are different derivatives.

For a consistent ideal gas, U depends on temperature alone. Here n is amount of substance and R is the universal molar gas constant, approximately 8.31446 J/(mol K). At fixed amount, differentiating H = U + nRT gives C_p − C_V = nR, or C_p,m − C_V,m = R on a molar basis. The mass-specific difference is R/M. This is an ideal-gas relation, not a formula for every liquid or solid.

The ideal-gas energy difference remains ΔU = ∫C_V(T)dT for any equilibrium endpoint pair, even when the actual path is not constant volume. Similarly, ΔH = ∫C_p(T)dT for that ideal gas. The subscripts define the material functions; they do not require every use of those functions to follow the defining measurement path.

For solids and liquids, C_p and C_V can be close enough for a particular task, but closeness needs a quantitative check. Expansion, compressibility and temperature determine the difference for a simple homogeneous phase. A later return section gives that relation and its stability conditions. Near a phase boundary or strong anomaly, a casual equality can fail.

Balance the whole calorimeter

Return to the warm-solid experiment. The solid’s total capacity is m_sc_s = 72.0 J/K. The liquid contributes 627.0 J/K and the cup contributes 30.0 J/K, so the cold side has 657.0 J/K. If the final common temperature is T_f, write the signed changes as 72(T_f − T_s) + 657(T_f − T_c) = 0, with the capacities in J/K and temperatures in a consistent scale.

Solving gives T_f ≈ 299.07593 K = 25.92593 °C. The warm solid loses about 3893.333 J of enthalpy; the liquid gains about 3715.556 J and the cup gains about 177.778 J. The signed sum is zero within numerical rounding. Heat flows internally from warmer to cooler parts, while the selected insulated assembly has no heat exchange with the room.

The result is a capacity-weighted temperature, not a mass-weighted average. The final value lies between the initial temperatures because every assigned capacity is positive and there are no extra energy terms. This is a useful check before trusting the arithmetic.

If the cup were incorrectly omitted, the model would predict about 26.18026 °C, roughly 0.25433 K too high. The cup is an energy recipient, not a correction added after an otherwise complete balance. In practice, the effective addenda also include the thermometer, stirrer, heater and any other parts that track the measured temperature; their initial temperatures and coupling must be specified.

Infer an unknown capacity from a temperature change

A separate hypothetical experiment uses the same masses, initial temperatures, liquid model and calibrated cup, but the sample capacity is unknown. Suppose the observed final equilibrium temperature is 26.00 °C = 299.15 K. The liquid and cup warm through 6.00 K while the sample cools through 54.00 K.

The no-loss inference is c_s = (m_wc_w + C_cal)(T_f − T_c)/[m_s(T_s − T_f)]. Substitution gives c_s = 657 × 6/(0.0800 × 54) = 912.5 J/(kg K). A known liquid, known addenda and measured temperature changes have turned an energy balance into a property measurement.

If a separately established 36.0 J leaves this assembly during the transfer, the sample must supply that energy too. The numerator becomes 657 × 6 + 36 J, giving c_s ≈ 920.833 J/(kg K). Ignoring this particular positive heat loss biases the inferred sample capacity low. A different measurement design can have the opposite bias, so use the signed balance rather than memorizing a universal correction direction.

The initial sample temperature must describe the sample at the moment the selected transfer balance begins. Cooling during handling, a wet sample, evaporation, incomplete equilibration, an unmodeled reaction or dissolution, and a thermometer that follows only one region all change the inference. A neat final formula does not make those physical conditions true.

Integrate a capacity that changes with temperature

For fixed-pressure heating in one unchanged phase, ΔH = m∫ from T_i to T_f c_p(T)dT. The familiar m c_pΔT replaces this integral only when a constant c_p is adequate or an explicitly defined interval average is used. Plotting capacity against temperature makes the approximation visible: the energy per mass is the area under that curve.

For a separate assigned solid model, let c_p(T) = 700 + 1.20(T − 300 K) in J/(kg K), where the slope has units J/(kg K²). Heat 0.120 kg from 300 K to 360 K without a transition. The local specific capacity rises from 700 to 772 J/(kg K), while integration gives ΔH = 5299.2 J.

The interval-average specific capacity is 5299.2/(0.120 × 60) = 736.0 J/(kg K). On a total basis, the local endpoint capacities are 84.0 and 92.64 J/K, while ΔH/ΔT = 88.32 J/K. These are different, correctly defined quantities. The slope dH/dT gives the local total C_p; the chord ΔH/ΔT gives an average.

For this linear teaching function, the average equals the value at the midpoint temperature, 330 K. That coincidence is not a general rule for a curved capacity function or an interval containing a transition. Property tables and fitted functions should retain their phase, composition, constraint and temperature domain.

Calibrate the cell and its addenda

An electrical heater offers a controlled energy input: E_el = ∫VI dt, using voltage across the heater and the current through it. If both are constant, E_el = VIΔt. This energy crosses the selected boundary as electrical work. A temperature rise then constrains capacity through a balance that includes heat leaks, mechanical work and participating parts.

In a hypothetical blank calibration, 3.00 V and 2.00 A act for 120 s, supplying 720 J. The effective cell temperature rises by 24.0 K with negligible loss and a constant effective capacity, so C_eff = 720/24 = 30.0 J/K. At fixed pressure the supplied energy increases H; at fixed volume it increases U under the corresponding conditions. This explains how an adiabatic calorimeter can measure a heat capacity even though its actual boundary heat is zero.

A blank run identifies the thermal response of the cell without the sample. Sample and blank measurements must refer to compatible temperatures, constraints, hardware and heat-transfer conditions. Subtracting effective addenda is justified only to the extent that those contributions remain the same or are separately modeled. A sample with poor thermal contact can fail to follow the thermometer even when the empty cell calibrated well.

Real calibration also checks thermometer response and calibration, electrical power and time, leads, stirring or sensor heating, drift, and the energy assigned to each boundary. Reference materials provide temperature-dependent comparison values with their own documented scope and uncertainty. Calibration tests an instrument/model combination; it does not guarantee every later sample or temperature range.

Heat leak and measurement time can imitate a larger capacity

An insulated-looking calorimeter can still exchange heat through supports, leads, gas or radiation. A simple diagnostic model treats the assembly as one uniform temperature with constant capacity C_eff and a linear heat leak to an environment at T_env: C_eff dT/dt = P − G(T − T_env). Here P is delivered heater power in W and G is thermal conductance in W/K. Internal equilibration must be fast enough for the lumped-temperature approximation.

Starting at T_env with constant P and G gives T(t) − T_env = (P/G)[1 − exp(−Gt/C_eff)]. In the no-leak limit G → 0, the rise becomes Pt/C_eff. The timescale C_eff/G organizes this particular model; it is not a universal calorimeter relaxation time or an inferred material property.

For the separate assigned assembly C_eff = 90.0 J/K, P = 12.0 W, G = 0.0800 W/K and T_env = 293.15 K, after 180 s the rise is about 22.17843 K. Of the 2160 J supplied, about 1996.059 J remains as enthalpy increase and 163.941 J has left. Without the leak, the same model would rise by 24.0 K.

If one ignores the leak and computes Pt/ΔT, the apparent capacity is about 97.392 J/K, larger than the assigned 90.0 J/K. This bias differs from the inverse mixing example because the unknown occurs in a different balance. Actual correction may need measured pre- and post-heating drift, changing conductance, power variation or spatial gradients. A straight-line extrapolation is an experimental model to test, not an automatic recovery of the true capacity.

A phase transition is another energy term

A heat input can change phase amounts instead of raising temperature. At fixed pressure, an ideal equilibrium first-order melting transition of a pure material occurs at a definite T_m. The latent enthalpy per mass L describes the enthalpy difference of the endpoint phases: converting mass m requires mL in addition to sensible heating before and after the transition.

For a separate hypothetical material, heat 0.0500 kg from 290 K to 320 K at fixed pressure. Assign T_m = 300 K, solid c_p = 1000 J/(kg K), liquid c_p = 2000 J/(kg K), and L = 100 000 J/kg. Warming the solid requires 500 J; melting requires 5000 J; warming the liquid requires 2000 J. The total is 7500 J.

At a constant delivered rate of 20.0 W with negligible loss, melting starts after 25.0 s and ends after 275.0 s; the final state is reached after 375.0 s. During the plateau, temperature remains 300 K while the liquid mass fraction changes. The ordinary finite single-phase derivative C_p does not describe that two-phase conversion as one finite constant capacity. Dividing a nonzero transition energy by zero temperature rise is not a useful sensible-heat estimate.

Real transitions can broaden or shift with impurities, multiple components, pressure changes, finite heating rate, poor thermal contact, nucleation or hysteresis. A measured peak can also involve reaction, evaporation or relaxation. Identify phases and baselines before interpreting its area as a latent enthalpy. The phase-equilibrium and materials subjects develop those mechanisms beyond this assigned heating curve.

Differential scanning: compare against a reference

Differential scanning calorimetry compares energy inputs or calibrated heat-flow rates for a sample and a reference during a controlled temperature program. Different instrument designs measure or infer that difference differently. The raw signal must be connected to an energy-rate calibration and an appropriate baseline before it becomes a heat capacity or transition enthalpy.

In a simple matched-cell single-phase model, let β = dT/dt be a positive scan rate. If sample and reference backgrounds and losses match, the excess required energy rate is ΔP = m c_pβ. Thus c_p = ΔP/(mβ). This model assumes that the sample follows the controlled temperature and that phase change, reaction and changing backgrounds do not add an extra energy term.

For an assigned 20.0 mg sample with c_p = 850 J/(kg K), a 0.0200 J/K cell background and β = 0.100 K/s, the blank needs 2.00 mW and sample plus cell needs 3.70 mW. Their difference, 1.70 mW, gives the assigned sample capacity. Failing to subtract the background would attribute part of the cell response to the sample.

During a transition, integrating a properly calibrated excess energy rate over time, with a physically justified baseline, can yield transition energy. A peak height alone is not latent enthalpy, and a temperature difference between sample and reference is not automatically a calibrated heat-flow difference. Scan-rate dependence and heating/cooling comparisons can reveal kinetic or contact effects, but do not by themselves identify a mechanism.

Further: the general C_p − C_V relation

For a simple one-component homogeneous phase at fixed amount, define volumetric expansion α_p = (1/V)(∂V/∂T)_p and isothermal compressibility κ_T = −(1/V)(∂V/∂p)_T. The response identity is C_p − C_V = TVα_p²/κ_T. Molar form uses v_m = V/n: C_p,m − C_V,m = Tv_mα_p²/κ_T. Mass-specific form is c_p − c_v = Tα_p²/(ρκ_T), with density ρ.

A short derivation shows the conditions. Differentiating H = U + pV at fixed p gives C_p − C_V = [(∂U/∂V)_T + p](∂V/∂T)_p. For a consistent simple fluid, (∂U/∂V)_T = T(∂p/∂T)_V − p. Using (∂p/∂T)_V = α_p/κ_T yields the result. Amount, composition and other work coordinates stay fixed; the relevant derivatives must exist on the chosen homogeneous branch.

For an ordinary locally stable state with T > 0 and finite positive κ_T, the squared expansion coefficient makes C_p − C_V nonnegative. It can be zero when α_p = 0; strict inequality is not universal. This is a conditional equilibrium response result, not a rule to extend through every singular or two-phase state.

For an assigned material state at T = 300 K, ρ = 2500 kg/m³, α_p = 3.00 × 10⁻⁵ K⁻¹ and κ_T = 1.00 × 10⁻¹¹ Pa⁻¹, the difference is 10.8 J/(kg K). If c_p = 900 J/(kg K), then c_v = 889.2 J/(kg K), a 1.20% difference relative to c_p. The coefficient here is volumetric; using a linear length-expansion coefficient directly can give a wrong result. For a small isotropic expansion, the volumetric coefficient is approximately three times the linear one.

Further: why materials and temperature change the capacity

Capacity reflects how additional energy becomes accessible as temperature changes. Translation, rotation, vibration and electronic states can contribute, while interactions and changes of structure also affect the energy. It is not a fixed amount of heat stored inside a substance. The same temperature change can require different energy in different phases, compositions or temperature ranges.

A classical monatomic ideal gas with translational motion only has C_V,m = 3R/2 and C_p,m = 5R/2, approximately 12.472 and 20.786 J/(mol K). Molecular internal modes can add temperature-dependent contributions. Assigning one constant c_v or capacity ratio γ = C_p/C_V over a broad interval therefore requires a model or property-data justification.

For a classical harmonic solid at sufficiently high temperature relative to its mode scales, the familiar limit is C_V,m ≈ 3R ≈ 24.943 J/(mol K) per mole of atoms. Both kinetic and potential parts of the vibrational energy contribute. It is not a universal low-temperature value, not automatically C_p, and not automatically the value per mole of arbitrary multi-atom formula units.

At low temperatures, quantum occupation restricts which modes can absorb additional energy; vibrations and, in metals, electronic excitations can have different temperature dependence. Magnetic or structural ordering can produce anomalies too. The statistical-mechanics, quantum and materials subjects supply the detailed models. Here the practical lesson is to retain the correct state, counting basis and validity range rather than extending a familiar constant everywhere.

Further: uncertainty and model error belong in the inference

The unknown-capacity result depends on measured masses, temperature differences, cup capacity and the liquid property model. For c_s = A(T_f − T_c)/[m_s(T_s − T_f)] with A = m_wc_w + C_cal, a first-order standard uncertainty for uncorrelated inputs x_j is u_c² = Σ[(∂c_s/∂x_j)u_j]². Correlated inputs require covariance terms; the assumption of independence is part of the calculation.

For the separate inverse example, assign standard uncertainties of 0.0002 kg in sample mass, 0.0003 kg in liquid mass, 1.0 J/K in cup capacity, 0.10 K in initial sample temperature, 0.05 K in initial cold temperature and 0.03 K in final temperature. Treat the assigned liquid c_w as exact within this model. The sensitivity calculation gives u_c ≈ 9.826 J/(kg K), about 1.08% of the inferred 912.5 J/(kg K). This is a combined standard uncertainty, not an automatically chosen confidence interval.

Temperature differences can make correlations consequential. The final temperature appears in both numerator and denominator, so its sensitivity is ∂c_s/∂T_f = c_s[1/(T_f − T_c) + 1/(T_s − T_f)]. A common additive offset in all three thermometer readings cancels from both differences in this constant-property model. Counting that same offset as three independent errors would lose the cancellation. Scale errors, drift or temperature-dependent property lookups do not generally cancel in the same way.

A small propagated uncertainty does not cover an omitted heat leak, a sample that cooled before immersion, an inaccurate liquid property approximation or an unrecognized transition. Those are model and systematic-effect questions to investigate, correct or include with justified uncertainty. Published preparation “Error propagation and uncertainty budgets” develops the general methods; this section keeps the focus on the calorimetry inference.

Choose and check a measurement

Begin with the selected system, sample identity, phase, composition and mechanical constraint. Specify which temperature represents which part of the apparatus and when equilibrium is adequate. Decide whether the relevant property is total, mass-specific or molar, and whether the energy balance uses internal energy, enthalpy or an additional reaction or transition term.

Then identify every meaningful energy channel: thermal exchange with the environment, heater work, stirring and sensing work, evaporation or mass transfer, and the addenda that change temperature. Calibrate the participating hardware, choose a temperature interval small enough for the intended property model, and compare repeated or differently timed runs when they test a specific loss or equilibration assumption.

Check the signed balance, units, capacity basis, plausible final temperature and energy residual. For a smooth variable capacity, integrate it; for a transition, separate sensible and latent contributions. Use evaluated properties and reference materials within their documented conditions rather than treating one remembered number as universal.

Calorimetry links a controlled energy budget to a measured response. Its strength comes from making that link explicit. The first-law, process, entropy, phase-equilibrium and microscopic-model subjects supply their larger frameworks; this lesson establishes what must be held fixed, measured, calibrated and checked before a capacity value means what its units suggest.

Worked examples

Try it yourself

1. Can a capacity value be fully specified by writing only C=100 J/K?

   Show the answer and reasoning

   Not generally.

   The selected amount, state, constraint and temperature range matter; C_p and C_V are different derivatives.

2. Replicate the same homogeneous sample at the same state. Which capacities double?

   Show the answer and reasoning

   Total C doubles; c and C_m remain unchanged.

   Total capacity scales with amount. Specific and molar capacities divide out mass or amount under the stated bulk conditions.

3. A calorimeter rises from 20 °C to 26 °C. Is its rise 6 K or 299.15 K?

   Show the answer and reasoning

   6 K.

   299.15 K is the final absolute temperature. Offsets cancel in a temperature difference; thermodynamic T factors still need the absolute scale.

4. Why does omitting the initially cold cup raise the predicted opening final temperature?

   Show the answer and reasoning

   It removes a real energy recipient.

   For the same warm-sample energy decrease, the reduced cold-side capacity warms more. The omitted-cup model predicts about 0.25433 K too much warming.

5. In the specified inverse mixing model, does ignoring a known positive outward heat loss make the inferred sample c too large?

   Show the answer and reasoning

   No; it makes it too small.

   The sample must supply both cold-side gain and outward loss. Omitting the loss reduces the numerator for c_s.

6. An adiabatic cell warms under electrical input. Does Q=0 imply its heat capacity is zero?

   Show the answer and reasoning

   No.

   Electrical work changes its energy. The measured capacity follows from that energy balance and state derivative, not from arbitrary actual Q/ΔT.

7. In the linear variable-capacity example, is 736 J/(kg K) the local capacity at both endpoints?

   Show the answer and reasoning

   No.

   It is the interval average; local endpoint values are 700 and 772 J/(kg K). Its equality with the midpoint value is specific to the linear law.

8. Is C_p−C_V=nR valid for every solid and liquid?

   Show the answer and reasoning

   No.

   It is the ideal-gas result. The general simple-phase response relation uses expansion, compressibility and temperature.

9. For an ordinary smooth stable state with α_p=0 and finite κ_T>0, must C_p exceed C_V strictly?

   Show the answer and reasoning

   No; their difference can be zero.

   The response identity contains α_p². Stability/domain conditions do not turn a zero numerator into a positive one.

10. Does zero temperature rise while the assigned material melts mean no energy is needed?

    Show the answer and reasoning

    No.

    Its phase amounts change at the same T_m; the latent enthalpy term is mL. Ordinary finite single-phase capacity is not the right description.

11. Why does ignoring heat loss overestimate C in the specified constant-power heating experiment?

    Show the answer and reasoning

    It assigns all input energy to the observed temperature rise.

    Some supplied energy left the assembly, so dividing full Pt by the smaller rise yields an apparent capacity larger than the assigned C_eff.

12. Should the assigned scanning sample capacity be inferred from the full 3.70 mW sample-plus-cell input?

    Show the answer and reasoning

    No; subtract the matched 2.00 mW background first.

    The excess 1.70 mW is the sample contribution under the stated matched-cell single-phase model.

13. Would a shared constant offset added to all three inverse-calorimetry temperatures change the two temperature differences?

    Show the answer and reasoning

    No, within the constant-property model.

    The shared offset cancels in each difference. Independent noise, scale error, drift and temperature-dependent property corrections are different effects.

14. Does u_c≈9.826 J/(kg K) prove the inferred sample capacity is accurate to that amount in a real apparatus?

    Show the answer and reasoning

    No.

    That standard uncertainty follows the supplied independent-input model. Omitted heat loss, wrong properties or incomplete equilibration require separate investigation and justified treatment.

15. Does the classical solid limit 3R automatically apply per mole of every chemical formula unit?

    Show the answer and reasoning

    No.

    The stated harmonic high-temperature limit is per mole of atoms, for C_V under its model conditions. Multi-atom formula-unit counting and C_p need care.

What to carry forward

• A capacity value needs a counting basis, constraint, state and temperature range; heat itself is an energy transfer.
• A calorimeter measures a response of the selected assembly, including participating addenda and every relevant energy channel.
• Integrate a changing capacity. A finite energy/temperature ratio is generally an interval average, and phase conversion requires its own enthalpy term.
• Electrical calibration can determine capacity through a work/energy balance even when actual boundary heat is zero.
• C_p−C_V=nR is an ideal-gas result; the general simple-phase identity has explicit response and stability conditions.
• Calibration and a small propagated uncertainty do not excuse an incorrect phase, property law, heat-loss model or equilibration assumption.