To determine the value of the latent heat of water vaporization, the following assembly is performed (Figure 1).

A volume of 1500 ml of distilled water is introduced into the large beaker (2 l) and placed on the heating magnetic stirrer.

We squeeze on the base of the test tube (10 ml) a clamp, then we introduce into this test tube 8 ml of distilled water, then we mouth it with the thumb, we spill it and we immerse it in the large beaker.

The clamp, in the vertical position, in the axis of the specimen, is fixed at its other end, by another clamp to the gantry support.

Also insert the magnet bar and thermometer into the large beaker holding the thermometer with a clamp at the gantry stand and turn on the magnetic stirrer. Then measure the height h, between the bottom of the meniscus of the bubble, and the surface of the water in contact with the atmosphere.

To remedy the problem of measuring the height hi, it is necessary to fix a ruler of 30 cm, in vertical position, with a clamp outside of large beaker (2 l).

Experiments take place in an enclosure that is sufficiently insulated to prevent heat exchange with the outside world. This enclosure is called a calorimeter. The inner chamber and the calorimeter accessories (stirrer and thermometer) are involved in heat exchange, as their temperature varies from the initial to the final value (Figure 2).

To measure the calorimeter’s heat capacity K_cal, the following procedure is used: Weigh the empty calorimeter (m calorimeter), then add a quantity of cold water, noting the initial temperature T_fr of the cold water, and weigh the calorimeter again with the cold water when the temperature stabilizes to determine the mass of cold water introduced and the equilibrium temperature of the calorimeter and cold water T_m1. Heat water in a Pyrex® beaker (wait for boiling). Then pour in a mass of hot water of known initial temperature T_ch and take the mass of (calorimeter + cold water + hot water) and the mixing temperature Tm₂ when equilibrium is established between cold and hot water.

To calculate the enthalpy of fusion of ice L_f, we’ll carry out the following experimental protocol:

Weigh the empty calorimeter (m calorimeter), then add a quantity of distilled water (1/3V_cal), noting the initial temperature T_dit of the distilled water, and weigh the calorimeter again with the distilled water when the temperature stabilizes to determine the mass of distilled water introduced and the equilibrium temperature of the calorimeter and distilled water T_m3. Prepare an ice cube set at −23˚C (of mass heat capacity C water(s)) and immerse it immediately in the calorimeter water, quickly closing the lid. When the ice cube melts completely in the calorimeter, which has a total heat capacity of K_cal and contains a mass m of distilled water, or when the temperature seems to stabilize (15 to 20 min), note T_m4. Weigh the calorimeter again and deduce the mass of ice m ice.

The total pressure, P_tot(B), at point B in Figure 1 is equal to the sum of the partial pressures of air, P_air, and water vapour, P ( H 2 O ) g :

P t o t ( B ) = P a i r + P ( H 2 O ) g + P h 2 (4)

On the other hand, the total pressure, P_tot(A), at point A on Figure 1 is equal to the sum of the atmospheric pressure, P_atm, and the overpressure, P_h, corresponding to the height h between the bottom of the meniscus and the level where the water is in contact with the atmosphere:

P t o t ( A ) = P a t m + P h 1 (5)

Pascal’s principle [12] [13] [14] [15] implies that the pressures measured at points A and B in Figure 1 are equal:

In equation, this amounts to writing that:

P t o t ( A ) = P t o t ( B )

P t o t ( A ) = P ( A ) + P s u r f a c e   a r e a ( A )

with

P ( A ) = ρ w a t e r g h 1 and P s u r f a c e a r e a ( A ) = P a t m o s p h e r i c

and

P t o t ( B ) = P ( B ) + P s u r f a c e   a r e a ( B )

with

P ( B ) = ρ w a t e r g h 2 and P s u r f a c e   a r e a ( B ) = P a i r + P ( H 2 O ) g

After simplification:

ρ w a t e r g h 1 + P a t m o s p h e r i c = ρ w a t e r g h 2 + P a i r + P ( H 2 O ) g

This gives:

P a i r + P ( H 2 O ) g = ρ w a t e r g h + P a t m o s p h e r i c (6)

with h = h₁ – h₂.

At T₀ < 5˚C, the partial pressure of water vapor is negligible, so P ( H 2 O ) g tend around 0.

Hence, the relationship (6) becomes:

P a i r = ρ e a u g ( h 1 − h 2 ) + P a t m o s p h e r i c

P a i r = ρ w a t e r g h 0 + P a t m o s p h e r i c (7)

with h₁ – h₂ = h at T₀ < 5˚C.

At T₀ > 5˚C, the partial pressure of water vapor is not negligible, so P ( H 2 O ) g is given from (6)-(7), we deduce that:

P ( H 2 O ) g = ρ w a t e r g ( h i − h 0 ) (8)

Knowing that the density of water is equal to 10³ kg/m³; and the acceleration of gravity is equal to 9.81 m/s². Hence, the pressure exerted by a column of 1 cm of water is equivalent to 98 Pa.

P ( 1cmwatercolumn ) = ρ w a t e r g h = 10 3 × 9.81 × 1 × 10 − 2 = 0. 98   kg / cm ⋅ s 2 = 0. 98   Pa ( avec h ( cm ) )

And therefore the relation (8) becomes:

P ( H 2 O ) g = 0.98 × ( h i − h 0 )   Pa (9)

With the unit of h_i and h₀ is on cm.

Generally speaking, the water vapor pressure P ( H 2 O ) g exerted by an incompressible liquid depends only on:

- The height (or depth) h;

- The density of the liquid ρ;

- The acceleration of gravity g.

Water vapor pressure does not depend on the cross-section of the container.

The water vapor pressure P ( H 2 O ) g exerted by the liquid at a depth h is:

P ( H 2 O ) g = ρ × g × ( h i − h 0 )

From relationship (2):

ln ( P ( H 2 O ) g ) = A − Δ H T ° R T (2)

We conclude that:

ln ( ρ × g × ( h i − h 0 ) ) = A − Δ H T ° R T

Consequently, we deduce that the latent heat of vaporization L_v for an incompressible liquid depends only on:

- The height (or depth) h;

- The density of the liquid ρ;

- The acceleration of gravity g.

The latent heat of vaporization L_v does not depend on the cross-section of the container.

Table 1 gives the values of: T_i; 1/T_i; h₀; h_i; P ( H 2 O ) g et ln [ P ( H 2 O ) g ] .

The Vant’Hoff equation is a consequence of the second fundamental law of thermodynamics and the Clausius-Clapeyron formula [13] [14] [15] [16] . It is valid for any system of the described type, so also for the phenomena of fusion and sublimation. When the water vapor pressure curve is known,

P ( H 2 O ) g = 0.98 ( h i − h 0 )

It is very simple to determine the vaporization heat ΔH_T since it is directly proportional to d P ( H 2 O ) g / d T , in other words to the slope of the tangent to this curve, for the temperature under consideration. The practical determination of ΔH_T will be as follows:

The graphic representation of P ( H 2 O ) g in function of 1/T must therefore provide a line, whose angular coefficient is:

Δ ln ( P ( H 2 O ) g ) Δ ( 1 / T ) = − Δ H T ° R

Hence

Δ H T ° = − R Δ ln ( P ( H 2 O ) g ) Δ ( 1 / T )

with R = 8.3145 × 10⁻³ kJ/(mol∙K).

The curve ln ( P ( H 2 O ) g ) as a function of 1/T is shown in Figure 3.

From the curve of ln ( P ( H 2 O ) g ) in function of 1/T one can deduce the latent heat of water vaporization which is worth:

Δ H T ° = − R Δ ln ( P ( H 2 O ) g ) Δ ( 1 / T )

Δ H T   e x p ° = 41 .0 67   kJ / mol .

The value of the latent heat of vaporization of the water found experimentally is slightly different from their tabulated value ( Δ H T ° theorique = 40,657 kJ/mol)

because of the measurement errors of the manipulator during the measurements of the height between the bottom of the meniscus of the bubble, and the surface of the water in contact with the atmosphere at different temperatures [16] .

The relationship used to determine the relative difference in latent heat of vaporization is:

Relativedeviation ( L v ) = | L v   experimental − L v   theorique | L v   theorique × 100 = | 41.067 − 40.657 | 40.657 × 100 = 1.008 %

The value of the relative deviation of the latent heat of vaporization L_V is about 1.008%. A very plausible deviation, given the many possible sources of error.

Sources of error when determining the latent heat of vaporization L_V are:

- The accuracy of measurements of the height (or depth) h and therefore the accuracy of the corresponding heights;

- Heat exchanges between the system (water + large beaker, test tube, thermometer, stirrer) and the external environment (resulting in heat loss);

- Exchanges of matter (water in its gaseous state) between the system (water + large beaker. Test tube, thermometer, stirrer) and the external environment (resulting in a loss of water in its gaseous state);

- Exchanges of matter (water in gaseous state) within the system between the large beaker and the test tube (are they total or partial);

- Temperature reading on the thermometer (temperature stabilization);

- Height reading h_i on the ruler (ruler is better secured by a clamp on the stand support).

The mass latent heat of a substance’s change of state is the quantity of heat L required to effect, at constant temperature, the change of state of the body’s unit mass, with the pressure remaining constant and equal to the equilibrium pressure between the two states [3] [4] [10] [11] .

Q = m L L in J/kg

In this work, we will study a particular physical reaction corresponding to the following change of state of water:

H 2 O ( s ) ⇄ H 2 O ( l ) and its associated L_f.

We have seen in thermodynamics that an equilibrium between two phases of a pure body occurs at a given P and T, i.e. fixed. The reaction is isobaric and isothermal. In this case, the enthalpy of fusion by mass Δ_fusH represents the amount of heat Q per unit mass (J/kg) absorbed by the solid water to transform it into liquid water. If the atmospheric pressure P_atm under which we’re working is equal to the standard pressure P₀ = 1 bar, we speak of the standard enthalpy of fusion by mass, Δ_fusH^°.

- Global system S {hot water + cold water + calorimeter}

The cold system S_fr: {cold water introduced + calorimeter} will exchange a quantity of heat Q_fr with the calorimeter (cold water is initially introduced into the calorimeter).

- Hot system S_ch: {hot water}. Hot water will give off a quantity of heat Q_ch.

- The system studied is an isolated system (no exchange with the outside environment). The calorimeter is an adiabatic chamber.

The expression for the quantity of heat released or absorbed by each of these objects is:

- Quantity of heat received by the cold water: m_fr = 76 g; the temperature of the cold-water increases from T_fr = 23˚C to T_m1 = 22˚C. Therefore:

Q f r = [ m f r × C w a t e r ( l ) + K c a l ] × ( T m 2 − T m 1 )

- Quantity of heat released by hot water: m_ch = 85.5 g. Initial temperature of hot water: T_ch = 52˚C; Final temperature when equilibrium is reached: T_m2 = 35˚C. Taking into account the calorimeter:

Q c h = m c h × C w a t e r ( l ) × ( T m 2 − T c h )

- As the calorimeter is an adiabatic enclosure, everything inside is thermally insulated; the sum of heat quantities exchanged inside the calorimeter is zero:

Δ U = Q 1 + Q 2 = 0

The internal energy of a macroscopic system results from microscopic contributions:

U = E c ( microscopic ) + E p ( microscopic ) .

In this case, there is no exchange of energy with the external environment (neither in the form of work W, nor in the form of heat Q), so we can write:

Δ U = W + Q = 0

If the system is isolated (i.e. there is no exchange with the external environment), the internal energy remains constant, and the change in internal energy is zero, so ΔU = 0.

When the final state of equilibrium is reached:

Δ U = 0 i.e. Q 1 + Q 2 = 0 ;

The calorimetric equation is therefore:

[ K c a l + m f r C w a t e r ( l ) ] ( T m 2 − T m 1 ) + m c h C w a t e r ( l ) ( T m 2 − T c h ) = 0

From the calorimetric equation, we derive the expression for the calorimeter’s heat capacity K_cal in J/K, given that C_water(l) = 4.18 J/(g∙K) is given by the following equation:

K c a l = m w a t e r ( l ) [ − m f r + m c h ( T m 2 − T c h ) ] T m 1 − T m 2

A.N: K_cal = 150 J/K.

We often talk about the water value of the calorimeter. From a thermal point of view, this means equating the calorimeter with a fictitious mass µ of liquid water, and writing K_cal = µC_water(l). We deduce the calorimeter’s water value µ.

μ = K c a l C w a t e r ( l )

A.N: µ = 35.81 g.

The expression of the energy given up by the water initially present in the calorimeter, between the start of the experiment and the moment when the temperature of the mixture takes the value T_m4 is:

Q d i s = m d i s C w a t e r ( l ) ( T m 4 − T m 3 )

A.N: Q d i s = 108.6 × 4.18 × ( 6.5 − 20 )

Q d i s = − 6128.30   J

The energy released by the calorimeter during the same time interval is expressed as follows:

Q c a l = K c a l ( T m 4 − T m 3 )

A.N: Q d i s = 150 × ( 6.5 − 20 )

Q d i s = − 2025   J

The expression for the energy received by the ice cubes during the change of state where the temperature remains constant is given by:

Q i c e = m i c e [ C w a t e r ( s ) ( T f − T i c e ) + C w a t e r ( l ) ( T m 4 − T f ) ] .

A.N: Q i c e = 20 [ 2.09 ( 0 − ( − 23 ) ) + 2.18 ( 6.5 − 0 ) ]

Q i c e = 1504.8   J

The expression for the energy received by the water in the ice cubes when it has reached the final temperature of the mixture is as follows.

Q m e l t i n g = m i c e L f

Since the calorimeter is thermally insulated, we can assume that the energy received by the ice cubes is equal to the energy supplied by the calorimeter and the water it contains.

The first principle gives:

Δ U = Q d i s + Q c a l + Q m e l t i n g + Q i c e = 0

m i c e [ L f + C w a t e r ( s ) ( T f − T i c e ) + C w a t e r ( l ) ( T m 4 − T f ) ] +   [ m d i s C w a t e r ( l ) + K c a l ] ( T m 4 − T m 3 ) = 0

The literal expression for the latent heat of fusion of ice given the heat capacity of ice is C_water(s) = 2.09 J/(K∙g) which is written as:

L f = C w a t e r ( s ) ( T i c e − T f ) + C w a t e r ( l ) ( T f − T m 4 ) + [ m d i s C w a t e r ( l ) + K c a l ] ( T m 3 – T m 4 ) m i c e

A.N: L_f = 332.43 kJ/kg.

The value found is almost equal to the tabulated enthalpy of fusion of water L_f:

L_f = 6 kJ/mol or 334 kJ/kg, measured at room temperature.

The relationship used to determine the relative difference in latent heat of fusion is:

Relativedeviation ( L f ) = | L f   experimental − L f   theorique | L f   theorique × 100 = | 332.43 − 334 | 334 × 100 = 0.5 %

The value of the relative deviation of the latent heat of fusion LV is about 0.5%. A very small change, given the many possible sources of error.

Sources of error when determining the latent heat of fusion L_f are:

- The imperfect calorimeter (not completely adiabatic enclosure);

- The accuracy of the volumes of water taken from the graduated cylinder, and therefore the accuracy of the corresponding masses;

- Heat exchanges between the ice cube and the ambient air: the time required to transport the ice cube and place it in the calorimeter (resulting in heat loss);

- Temperature reading on the thermometer (temperature stabilization);

- Measuring the mass of the empty calorimeter, cold water and ice cube at equilibrium with the balance may not be accurate.

- If the mass of the ice cube is much greater than the mass of the cold water, it may increase the time taken to reach equilibrium H 2 O ( s ) ⇄ H 2 O ( l ) . Consequently, the increase in time to equilibrium results in heat loss.