Heat capacity, sensible heat and latent heat calculations

Cp and Cv, temperature-dependent heat capacity and mean Cp, sensible and latent heat, Watson and Trouton estimates, and enthalpy paths through a phase change.

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Why it matters

Every heater, cooler, condenser, reboiler and evaporator duty is a sum of sensible heat (temperature change) and latent heat (phase change). Getting the heat capacity right – including its temperature dependence – and the latent heat at the right temperature fixes utility consumption, exchanger size and the steam or cooling-water bill.

Key ideas

Heat capacity. The heat needed per unit temperature rise. Extensive heat capacity (J/K) depends on the amount; specific (per kg, J/(kg·K)) and molar (per mol, J/(mol·K)) heat capacities are intensive.

C_p and C_v. At constant pressure, heat added equals the enthalpy change: dH = C_p·dT. At constant volume, it equals the internal-energy change: dU = C_v·dT. For an ideal gas C_p − C_v = R; for liquids and solids C_p ≈ C_v. Process equipment usually runs at (nearly) constant pressure, so C_p and enthalpy are used.

Temperature dependence. C_p rises with temperature and is tabulated as a polynomial, e.g. C_p / R = A + B·T + C·T² + D·T⁻² (T in K) – the constants come from your data book. Over a wide range, integrate; over a narrow range, use a mean heat capacity C_p,mean = ΔH / ΔT for that interval.

Sensible heat. Heat that changes temperature without phase change: Q = n·∫C_p dT = n·C_p,mean·(T₂ − T₁). A temperature difference is the same in K and °C.

Latent heat. Heat absorbed or released at constant T and P during a phase change: fusion (melting), vaporisation (boiling) and sublimation. Latent heat of vaporisation falls as temperature rises and becomes zero at the critical point.

Estimating latent heat.

  • Trouton's rule: ΔH_vap / T_b ≈ 88 J/(mol·K) at the normal boiling point (rough; not for water, alcohols or H-bonded liquids).
  • Watson correlation: converts a known latent heat to another temperature using the critical temperature.
  • Clausius–Clapeyron from two vapour-pressure points (see the vapour-pressure topic).

Enthalpy is a state function. To find the heat for a process involving phase change at a temperature other than the tabulated one, build any convenient path (heat liquid → vaporise at the normal boiling point → heat vapour) – the result is path-independent. For water, read enthalpies directly from steam tables.

Mixtures. For ideal mixtures (gases, similar liquids), C_p,mix = Σ x_i·C_p,i (mole basis) or Σ w_i·c_p,i (mass basis).

Formulas

  • Q = m·c_p·ΔT – Q (J), m (kg), c_p (J/(kg·K)), ΔT (K).
  • Q = n·∫ C_p dT from T₁ to T₂ – n (mol), C_p (J/(mol·K)), T (K).
  • For C_p / R = A + B·T + D·T⁻²: ΔH = R·[A·(T₂ − T₁) + (B/2)·(T₂² − T₁²) − D·(1/T₂ − 1/T₁)] – J/mol.
  • C_p,mean = ΔH / (T₂ − T₁)
  • C_p − C_v = R – ideal gas, R = 8.314 J/(mol·K).
  • Q_latent = n·ΔH_vap (or m·λ).
  • ΔH_vap,2 = ΔH_vap,1 × [(T_c − T₂) / (T_c − T₁)]^0.38 – Watson; T in K.
  • ΔH_vap / T_b ≈ 88 J/(mol·K) – Trouton's rule.
  • C_p,mix = Σ y_i·C_p,i

Worked examples

Example 1 (standard, variable C_p). Nitrogen at 100 mol/s is heated from 25 °C to 500 °C at constant pressure. For N₂, C_p / R = 3.280 + 0.593 × 10⁻³·T + 0.040 × 10⁵·T⁻² (T in K; data-book values). Find the heater duty and the mean C_p.

  1. T₁ = 298.15 K, T₂ = 773.15 K.
  2. A term: 3.280 × (773.15 − 298.15) = 1558.0 K.
  3. B term: (0.593 × 10⁻³ / 2) × (773.15² − 298.15²) = 0.2965 × 10⁻³ × 508 868 = 150.9 K.
  4. D term: −0.040 × 10⁵ × (1/773.15 − 1/298.15) = −4000 × (0.0012934 − 0.0033540) = 8.2 K.
  5. ΔH = 8.314 × (1558.0 + 150.9 + 8.2) = 8.314 × 1717.1 = 14 276 J/mol.
  6. Duty = 100 mol/s × 14 276 J/mol = 1.428 × 10⁶ W; mean C_p = 14 276 / 475 = 30.06 J/(mol·K).

Answer: Q ≈ 1.43 MW; C_p,mean ≈ 30.1 J/(mol·K)

Example 2 (GATE level, phase change). 1000 kg/h of liquid benzene (M = 78.11) at 25 °C is converted to vapour at 120 °C and 1 atm. Data: mean C_p (liquid, 25–80 °C) = 0.146 kJ/(mol·K); ΔH_vap at the normal boiling point 80.1 °C = 30.77 kJ/mol; mean C_p (vapour, 80–120 °C) = 0.100 kJ/(mol·K); T_c = 562.2 K. Find the heat duty, and estimate ΔH_vap at 25 °C with the Watson correlation.

  1. Molar flow: n = 1000 / 78.11 = 12.80 kmol/h.
  2. Path: heat liquid 25 → 80.1 °C: 0.146 × 55.1 = 8.04 kJ/mol.
  3. Vaporise at 80.1 °C: 30.77 kJ/mol.
  4. Heat vapour 80.1 → 120 °C: 0.100 × 39.9 = 3.99 kJ/mol.
  5. Total ΔH = 8.04 + 30.77 + 3.99 = 42.80 kJ/mol; duty = 12.80 × 10³ mol/h × 42.80 kJ/mol = 5.48 × 10⁵ kJ/h = 152 kW.
  6. Watson: ΔH_vap(298.15 K) = 30.77 × [(562.2 − 298.15) / (562.2 − 353.25)]^0.38 = 30.77 × (1.2637)^0.38 = 33.6 kJ/mol – larger at lower temperature, as expected.

Answer: Duty ≈ 5.48 × 10⁵ kJ/h (≈ 152 kW); ΔH_vap(25 °C) ≈ 33.6 kJ/mol

Common mistakes

  • Using C_v instead of C_p for a constant-pressure process.
  • Using °C inside a C_p polynomial written for K (or vice versa).
  • Multiplying a C_p evaluated at one temperature by a large ΔT when C_p varies strongly.
  • Using a latent heat at the wrong temperature – correct it with Watson or a path.
  • Forgetting latent heat when a stream crosses its boiling point.
  • Mixing per-kg and per-mol data in one calculation.

For GATE CH

Expect NAT questions on heater or cooler duties with a C_p polynomial, mean heat capacity over an interval, multi-step heating through a phase change (ice to steam, liquid to superheated vapour), and Watson or Trouton estimates of latent heat. Practise building the enthalpy path and keeping units per mol or per kg consistent.

Quick check

  1. How much heat raises 5 kg of water (c_p = 4.18 kJ/(kg·K)) from 25 °C to 75 °C?
  2. For an ideal gas with C_p = 29.1 J/(mol·K), what is C_v?
  3. Does ΔH_vap increase or decrease as temperature rises toward T_c?
  4. Estimate ΔH_vap of a liquid boiling at 350 K by Trouton's rule.
  5. Why can enthalpy changes be calculated along any convenient path?

Answers: 1. 1045 kJ; 2. 20.8 J/(mol·K); 3. it decreases, reaching zero at T_c; 4. about 30.8 kJ/mol; 5. enthalpy is a state function.

See it move

All Chemical animations
Heat Capacity and Heat Calculations

Adjust the mass of water and temperature change to see how sensible heat varies. Observe how the heat required changes with different inputs.

Equations used
  • q = m·c·ΔT — q: Sensible heat (J), m: Mass (kg), c: Specific heat capacity (J/(kg·K)), ΔT: Temperature change (K)

Interview questions

All Process Calculations interview questions

Try answering each one aloud before you open it.

  1. 1.What is heat capacity and how is it different from specific heat capacity?Concept

    Heat capacity is the amount of heat required to change the temperature of a substance by one degree Celsius. It is an extensive property, meaning it depends on the amount of substance. Specific heat capacity, on the other hand, is the heat capacity per unit mass of a substance, making it an intensive property. It is used to compare the thermal properties of different materials regardless of their mass.

  2. 2.Explain the concept of sensible heat and latent heat.Concept

    Sensible heat is the heat absorbed or released by a substance during a change in temperature that does not involve a phase change. Latent heat is the heat absorbed or released during a phase change, such as melting or boiling, without a change in temperature. Sensible heat can be measured by a change in temperature, while latent heat is associated with a change in state.

  3. 3.Why is specific heat capacity important in process calculations?Application

    Specific heat capacity is crucial in process calculations because it helps determine the amount of energy required to heat or cool a substance. This information is essential for designing heating and cooling systems, optimizing energy consumption, and ensuring safety in chemical processes. It allows engineers to predict how a substance will respond to thermal energy changes.

  4. 4.What happens to the heat capacity of a substance as it undergoes a phase change?Application

    During a phase change, the heat capacity of a substance effectively becomes infinite because the temperature remains constant while heat is absorbed or released. This is due to the latent heat involved in the phase transition, which requires energy without a change in temperature. As a result, the heat capacity is not defined in the usual sense during a phase change.

  5. 5.How does latent heat matter in the design of heat exchangers?Application

    A condensing or boiling stream transfers a large amount of heat (its latent heat) at an almost constant temperature, so per kg it carries far more duty than a stream that only changes temperature. That is why saturated steam is the usual heating medium: the temperature driving force stays steady and is set by the steam pressure. Designers must split such exchangers into zones (desuperheating, condensing, subcooling) because the heat-transfer coefficients and temperature profiles differ in each.

  6. 6.What is the role of heat capacity in determining the thermal stability of a chemical process?Application

    Heat capacity plays a critical role in determining the thermal stability of a chemical process. A high heat capacity means the system can absorb more heat without a significant temperature rise, which can prevent runaway reactions. Conversely, a low heat capacity might lead to rapid temperature changes, increasing the risk of thermal instability. Engineers must consider heat capacity to ensure safe and stable process operations.

  7. 7.Calculate the amount of heat required to raise the temperature of 5 kg of water from 25°C to 75°C. (Specific heat capacity of water = 4.18 J/g°C)Numerical

    To calculate the heat required, use the formula: Q = m·c·ΔT. Here, m = 5000 g (since 5 kg = 5000 g), c = 4.18 J/g°C, and ΔT = 75°C - 25°C = 50°C. Therefore, Q = 5000 g × 4.18 J/g°C × 50°C = 1,045,000 J.

  8. 8.A 2 kg block of ice at 0°C is melted to water at 0°C. Calculate the heat absorbed. (Latent heat of fusion for ice = 334 J/g)Numerical

    To calculate the heat absorbed, use the formula: Q = m·L. Here, m = 2000 g (since 2 kg = 2000 g) and L = 334 J/g. Therefore, Q = 2000 g × 334 J/g = 668,000 J.

  9. 9.Explain why water is often used as a coolant in industrial processes.Application

    Water is often used as a coolant because it has a high specific heat capacity, meaning it can absorb a lot of heat before its temperature rises significantly. This property makes it effective at removing heat from systems, preventing overheating. Additionally, water is abundant, non-toxic, and relatively inexpensive, making it a practical choice for many industrial applications.

  10. 10.What would happen if a substance with a low specific heat capacity is used in a process requiring temperature stability?Application

    Using a substance with a low specific heat capacity in a process requiring temperature stability could lead to rapid temperature fluctuations. This is because such a substance would absorb or release heat quickly with only a small change in temperature. This could result in process instability, potential safety hazards, and inefficiencies in maintaining the desired operating conditions.

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