Physics / Latent Heat of Vaporization
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EXPERIMENT 27 · PHYSICS LABORATORY

Latent Heat of Vaporization theory

Understand latent heat of vaporization: the equations, assumptions, supporting concepts and measurement method.

THE WORKING RELATIONSHIP
Q_net = C_sys (T_f − T_0) − m_v c_w (T_s − T_f) = m_v L_v ⇔ Y = L_v · X ⇔ L_v = S

Keep the physical quantities, graph variables and units consistent.

01Thermodynamics of Vaporization and Latent Enthalpy of Phase Change

Vaporization is the first-order thermodynamic phase transformation wherein liquid molecules overcome intermolecular cohesion to transition into the gaseous state. In liquid water, each molecule participates in an extensive, dynamic three-dimensional hydrogen-bonding network with an average binding energy of approximately 20 kJ mol⁻¹ per bond. Transforming one kilogram of liquid water into steam at 100 °C requires breaking the vast majority of these hydrogen bonds, which accounts for the extraordinarily high specific latent heat of vaporization of water (L_v ≈ 2.260 × 10⁶ J kg⁻¹ = 539.8 cal g⁻¹), one of the highest of all known molecular liquids. Thermodynamically, the latent heat of vaporization represents the change in specific enthalpy between the saturated vapor and saturated liquid states: L_v = Δh_vap = h_g − h_f. This enthalpy change comprises two distinct microscopic components: internal energy change Δu_vap (the work done against intermolecular attractive forces, often termed internal latent heat) and external expansion work P Δv_vap done against the surrounding atmospheric pressure as the specific volume expands from v_f = 0.00104 m³ kg⁻¹ to v_g = 1.673 m³ kg⁻¹. At 100 °C and 101.325 kPa, external work P (v_g − v_f) accounts for approximately 169.5 kJ kg⁻¹ (about 7.5% of L_v), while internal bond disruption accounts for the remaining 2,090.5 kJ kg⁻¹ (92.5%).

L_v = \Delta h_{\text{vap}} = \Delta u_{\text{vap}} + P(\nu_g - \nu_f) \approx 2.260 \times 10^6 \text{ J kg}^{-1}
What are the two physical components that constitute the latent heat of vaporization of water?

The latent heat of vaporization comprises: (1) internal latent heat, which is the energy required to overcome intermolecular hydrogen bonding attractions; and (2) external latent heat, which is the mechanical work performed against atmospheric pressure as the vapor expands to more than 1,600 times its liquid volume.

02The Clapeyron-Clausius Equation and Temperature Dependence of L_v

The thermodynamic relationship governing the coexistence of liquid and vapor phases along the saturation curve is expressed by the Clapeyron equation, formulated by Émile Clapeyron in 1834: dP/dT = L_v / [T (v_g − v_f)], where P is the saturation vapor pressure, T is the absolute temperature, and v_g, v_f are the specific volumes of vapor and liquid respectively. Neglecting the minuscule liquid volume v_f ≪ v_g and modeling the saturated vapor as an ideal gas (v_g = R_specific T / P), Rudolf Clausius simplified this into the celebrated Clausius-Clapeyron equation: d(ln P)/dT = L_v / (R_specific T²). Integrating under the assumption of constant L_v gives ln(P / P_0) = −(L_v / R_specific) (1/T − 1/T_0). Furthermore, according to Kirchhoff's thermochemical law, the latent heat of vaporization itself is not strictly invariant with temperature; its derivative along the saturation line is dL_v/dT = c_{p,\text{vap}} − c_{p,\text{liq}}. Because the specific heat of liquid water (c_w ≈ 4,186 J kg⁻¹ K⁻¹) is roughly double that of steam (c_p,steam ≈ 2,010 J kg⁻¹ K⁻¹), L_v decreases linearly with increasing temperature, declining from 2,501 kJ kg⁻¹ at 0 °C to 2,260 kJ kg⁻¹ at 100 °C, and vanishing entirely at the thermodynamic critical point (T_c = 373.95 °C, P_c = 22.06 MPa) where the meniscus between liquid and vapor disappears.

\frac{dP}{dT} = \frac{L_v}{T(\nu_g - \nu_f)} \approx \frac{P L_v}{R_{\text{spec}} T^2} \quad \text{and} \quad \frac{dL_v}{dT} = c_{p,\text{vap}} - c_{p,\text{liq}}
Why does the latent heat of vaporization of water decrease as temperature rises, and what happens to it at the critical point?

As temperature rises, thermal expansion separates liquid molecules, weakening intermolecular bonds prior to boiling, while the density of saturated vapor increases. The difference in enthalpy between liquid and vapor diminishes according to Kirchhoff’s equation dLv/dT = cp,vap − cp,liq. At the critical point (374 °C), the liquid and vapor densities become identical and Lv drops to zero.

03Condensation Calorimetry: Heat Exchange Conservation Equations

The method of mixtures applied to condensation calorimetry establishes a rigorous thermal energy balance inside the calorimeter. When a mass m_v of dry saturated steam at temperature T_s condenses in a calorimeter containing mass m_w of water at initial temperature T_0, two distinct heat release stages occur: (1) Phase change: m_v kilograms of steam condense isothermally into liquid water at T_s, liberating latent enthalpy Q_latent = m_v L_v; (2) Sensible cooling: the resulting m_v kilograms of condensed water cool from T_s down to the final equilibrium temperature T_f, releasing sensible heat Q_sens = m_v c_w (T_s − T_f). Simultaneously, the heat absorbed by the calorimeter assembly comprises: (3) Water warming: the initial water of mass m_w warms from T_0 to T_f, absorbing m_w c_w (T_f − T_0); (4) Vessel warming: the copper calorimeter vessel and stirrer of total water equivalent W_cal warm from T_0 to T_f, absorbing W_cal (T_f − T_0). Under adiabatic conditions, heat lost equals heat gained: m_v L_v + m_v c_w (T_s − T_f) = (m_w c_w + W_cal) (T_f − T_0) = C_sys (T_f − T_0). Isolating the latent contribution defines the net heat metric: Q_net = C_sys (T_f − T_0) − m_v c_w (T_s − T_f) = m_v L_v.

m_v L_v + m_v c_w (T_s - T_f) = C_{\text{sys}} (T_f - T_0) \implies Q_{\text{net}} = m_v L_v
In condensation calorimetry, what two distinct quantities of heat are released by the incoming steam?

First, the latent heat of vaporization released isothermally during the phase change from steam to liquid at 100 °C (Q = mv Lv). Second, the sensible heat released as the condensed water cools from 100 °C to the final equilibrium temperature Tf (Q = mv cw (Ts − Tf)).

04The Function of the Condensate Water Trap and Wet Steam Errors

In experimental steam calorimetry, the single largest source of systematic positive error is wet steam—the carryover of entrained microscopic liquid water droplets suspended in the vapor stream. If steam emerges directly from the boiling water surface through a simple tube, turbulent boiling entrains a fine mist of liquid droplets at 100 °C. If these liquid droplets enter the calorimeter, they contribute to the final measured mass m_v upon weighing, but they release zero latent heat of vaporization (they only release sensible heat as they cool from 100 °C to T_f). Consequently, the calculated latent heat L_v = [Q_gain − m_v c_w (T_s − T_f)] / m_v is severely depressed below its true value. If an entrained moisture fraction x_m = m_liquid / m_total is present, the apparent latent heat is related to the true latent heat by L_v,apparent = (1 − x_m) L_v,true. Even a 3% liquid carryover produces an unacceptable ~68 kJ kg⁻¹ deficit. To eliminate this defect, a specialized glass water trap (or Berthelot steam separator) is inserted immediately ahead of the calorimeter. The trap abruptly expands the flow channel, compelling droplet impingement against glass baffles and draining condensate downward, while only dry, super-saturated vapor flows upward into the calorimeter delivery nozzle.

L_{v,\text{apparent}} = (1 - x_m) L_{v,\text{true}} \quad \text{where } x_m = \frac{m_{\text{droplets}}}{m_{\text{total}}}
How does entrained liquid water (wet steam) affect the experimental calculation of latent heat of vaporization?

Liquid water droplets contribute to the measured condensed mass mv upon weighing, but they release no latent heat of vaporization (only sensible heat). Consequently, the heat released per gram of condensate appears falsely low, causing the calculated latent heat Lv to be systematically underestimated.

05Count Rumford's Radiation Compensation Technique

During the 5 to 10 minutes required to bubble several grams of steam into the calorimeter, heat exchange between the calorimeter vessel and the ambient laboratory environment introduces thermal drift. Sir Benjamin Thompson, Count Rumford, devised an ingenious experimental procedure to cancel this environmental heat exchange without requiring complex numerical cooling corrections. Rumford observed that if the initial water temperature T_0 is adjusted below the ambient room temperature T_room by exactly the same amount that the final equilibrium temperature T_f will rise above it (such that T_room − T_0 ≈ T_f − T_room), the rate of heat gained by the calorimeter from the room during the first half of the experiment will approximately equal the rate of heat lost by the calorimeter to the room during the second half. Because Newton's law of cooling dictates that heat transfer rate is proportional to temperature excess (dQ/dt ∝ T − T_room), the integral of environmental heat exchange over the symmetric thermal excursion evaluates close to zero: ∫ dQ_env dt ≈ 0. In our standardized configuration with room temperature at 20.0 °C and an initial chilled temperature T_0 = 15.00 °C (5 °C below ambient), steam is injected until T_f reaches 25 °C to 35 °C, maintaining the average temperature of the vessel very near ambient and minimizing radiation bias.

(T_{\text{room}} - T_0) \approx (T_f - T_{\text{room}}) \implies \int_0^{t_{\text{final}}} \frac{dQ_{\text{env}}}{dt} dt \approx 0
What is Rumford's method of radiation compensation, and how does it eliminate environmental heat exchange errors?

Rumford's method starts the experiment with the calorimeter water chilled as many degrees below room temperature as it is expected to rise above room temperature by the end. The heat absorbed from the room while the vessel is cooler than ambient cancels the heat lost to the room while it is warmer, making the net environmental heat exchange negligible.

06Barometric Pressure Corrections to the Boiling Temperature T_s

The temperature T_s of saturated steam in equilibrium with boiling water depends directly upon the prevailing atmospheric barometric pressure P_atm. The canonical boiling temperature of 100.00 °C applies strictly at standard sea-level atmospheric pressure of 760.00 mmHg (101.325 kPa = 1.01325 bar). In a physical laboratory, day-to-day barometric fluctuations typically range between 730 mmHg and 780 mmHg, which shifts the actual steam temperature. Over this meteorological span, Regnault's empirical relation provides the boiling point correction: T_s = 100.00 °C + 0.0373 (P_atm − 760) − 0.00005 (P_atm − 760)², where P_atm is the barometric reading in millimeters of mercury corrected for temperature and local gravity. For instance, on a low-pressure stormy day with P_atm = 745 mmHg, steam boils at 99.44 °C, while at high elevations (e.g., 1,000 meters above sea level, P ≈ 674 mmHg), water boils near 96.8 °C. Assuming T_s = 100.0 °C when the true temperature is lower alters the sensible heat subtraction m_v c_w (T_s − T_f), introducing a systematic offset into L_v. Recording the laboratory barometer is therefore a mandatory step in high-precision steam calorimetry.

T_s = 100.00 \, ^\circ\text{C} + 0.0373 (P_{\text{atm}} - 760) \quad (P_{\text{atm}} \text{ in mmHg})
Why must the ambient barometric pressure be measured when determining the latent heat of vaporization?

The boiling point of water is 100 °C only at standard atmospheric pressure (760 mmHg). Departures in barometric pressure shift the steam temperature by approximately 0.037 °C per mmHg. Measuring barometric pressure ensures the exact steam temperature Ts is used when calculating sensible heat.

07Evaluation of Water Equivalent and System Heat Capacity C_sys

Accurate calculation of the absorbed thermal energy Q_gain requires accounting for all physical components that warm alongside the water bath. The water equivalent W_cal of a calorimeter vessel is defined as the mass of water that would absorb the identical quantity of thermal energy as the calorimeter body for the same temperature rise: W_cal = m_cal · (c_material / c_water). For a specularly polished thin-walled copper vessel of mass m_cal = 75.00 g and specific heat c_cu = 385.0 J kg⁻¹ K⁻¹, its thermal capacity is C_cal = 0.0750 kg × 385.0 J kg⁻¹ K⁻¹ = 28.875 J K⁻¹. Dividing by c_w = 4,186 J kg⁻¹ K⁻¹ yields a water equivalent of W_cal = 28.875 / 4186 = 0.00690 kg = 6.90 g of water. Adding the wire stirrer (mass 15.0 g, brass, c = 380 J kg⁻¹ K⁻¹; C_stir = 5.70 J K⁻¹) and the submerged glass bulb of the thermometer (approximate heat capacity C_therm ≈ 1.5 J K⁻¹) gives a total apparatus thermal capacity of approximately 36.08 J K⁻¹. When filled with 200.00 g of pure water (C_water = 0.200 kg × 4,186 J kg⁻¹ K⁻¹ = 837.20 J K⁻¹), the aggregate heat capacity of the system is C_sys = 837.20 + 28.88 = 866.08 J K⁻¹. The aggregate capacity C_sys scales the observed temperature rise (T_f − T_0) into total absorbed joules.

C_{\text{sys}} = m_w c_w + m_{\text{cal}} c_{\text{cu}} + W_{\text{acc}} = 866.08 \text{ J K}^{-1}
What is the physical meaning of the water equivalent of a calorimeter, and why must it be added to the water mass?

The water equivalent represents the mass of water that would have the same thermal capacity as the metal calorimeter vessel and stirrer. Because the metal warms through the identical temperature rise as the water, its thermal capacity must be included to account for all heat absorbed.

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Connect the equations to a measurement.

The simulation evaluates equilibrium temperatures using the exact thermodynamic conservation relation T_f = (m_v L_v + m_v c_w T_s + C_sys T_0) / (C_sys + m_v c_w), where m_v is the condensed steam mass in kilograms, L_v = 2,260,000 J kg⁻¹ (2,260.00 kJ kg⁻¹) is the reference latent heat of vaporization, c_w = 4,186 J kg⁻¹ K⁻¹ is the specific heat capacity of water, T_s = 100.00 °C is the steam temperature, T_0 = 15.00 °C is the initial chilled water temperature, and C_sys = 866.08 J K⁻¹ is the aggregate thermal capacity. Equilibrium temperatures are rounded to 0.01 °C to simulate precision calorimetric thermometry, and ordinary least-squares regression of Q_net against m_v yields the reported latent heat of vaporization.