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MODULE 2 - MIXING PROPERTIES INTRODUCTION Homogeneous mixtures of different chemical species, particularly liquids, are formed in many ways in both natural and industrial processes. Moreover, “unmixing” processes are required for the separation of mixtures into their constituent species and for the purification of individual chemicals. The thermodynamic properties of mixtures are required for the analysis of these mixing and unmixing processes.

INTENDED LEARNING OUTCOMES After completing this chapter, the students must be able to: 1. Define a standard mixing process and develop the property changes that accompany it. 2. Relate properties of a mixture to the properties of its constituents as pure species. 3. Relate property changes of mixing to excess properties. 4. Treat in detail the heat effects of mixing and “unmixing” processes.

PROPERTY CHANGES OF MIXING Mixing processes are carried out in many ways, and each process results in a particular change of state, depending on initial and final conditions of temperature and pressure. For the rational study of mixtures, we must define a standard mixing process, much as we have done for standard property changes of chemical reaction. Experimental feasibility suggests mixing at constant T and P. Thus, we take as a standard mixing process one in which appropriate amounts of pure chemical species at T and P are mixed to yield a uniform mixture of specified composition, also at T and P. The pure species are said to be in their standard states, and their properties in this state are the pure-species properties Vi, Hi, Si, etc. 1This

conceptual illustration is not a practical device for making such measurements. In

practice, volume changes of mixing can be measured directly using calibrated glassware (e.g., graduated cylinders) and heats of mixing are measured calorimetrically Figure 11.1: Schematic diagram of experimental binary mixing process. Two pure species, both at T and P, are initially separated by a partition, withdrawal of which allows mixing. As mixing occurs, expansion or contraction of the system is accompanied by movement of the piston so that the pressure is constant.

The standard mixing process is represented schematically in Fig. 11.1 for mixing pure species 1 and 2.1 In this hypothetical device, n1 moles of pure species 1 are mixed with n2 moles of pure species 2 to form a homogeneous solution of composition x1

.

The observable phenomena accompanying the mixing process are expansion (or contraction) and a temperature change. Expansion is accommodated by movement of the piston so as to maintain pressure P, and temperature change is compensated by heat transfer to restore temperature T. When mixing is complete, the total volume change of the system (as measured by piston displacement d) is: Because the process occurs at constant pressure, the heat transfer Q is equal to the total enthalpy change of the system: Division of these equations by n1 + n2 gives:

And Q

Thus, the volume change of mixing ΔV and the enthalpy change of mixing ΔH are found from the measured quantities ΔVt and Q. Because of its experimental association with Q, ΔH is usually called the heat of mixing. Equations analogous to those for ΔV and ΔH can be written for any property, and they may also be generalized to apply to the mixing of any number of species:

where M can represent any intensive thermodynamic property of the mixture, for example, U, CP, S, G, or Z. Equation (11.1) is thus the defining equation for a class of thermodynamic properties, known as property changes of mixing. Property changes of mixing are functions of T, P, and composition and are directly related to excess properties. This result follows by combination of Eq. (10.85) with each of Eqs. (10.79) through (10.82), which provide expressions for the properties of an ideal solution: M

h

The first two terms on the right side of each equation represent a property change of mixing as defined by Eq. (11.1). Equations (11.2) through (11.5) can therefore be written: V

H

(11.6) .t

G

h

.7 .

For the special case of an ideal solution, each excess property is zero, and these equations become: 0

0

(11.10) .t

h

.7 .

These equations are alternative forms of Eqs. (10.79) through (10.82). They apply as well to the ideal-gas-state mixture as a special case of an ideal solution. Equations (11.6) through (11.9) show that excess properties and property changes of mixing are readily calculated one from the other. Although historically property changes of mixing were introduced first, because of their direct relation to experiment, excess properties fit more readily into the theoretical framework of solution thermodynamics. Because of their direct measurability, ΔV and ΔH are the property changes of mixing of major interest. Moreover, they are identical with the corresponding excess properties VE and HE. Figure 11.2 shows excess enthalpies HE for the ethanol/water system as a function of composition for several temperatures ranging from 30°C to 110°C. For the lower temperatures the behavior is exothermic, with heat removal required for isothermal

mixing. For the higher temperatures the behavior is endothermic, with heat addition required for isothermal mixing. At intermediate temperatures, regions of both exothermic and endothermic behavior appear. Data of this sort are often represented by polynomial equations in mole fraction, multiplied by x1x2 to ensure that the excess property goes to zero for both pure components.

Figure 11.3 illustrates the composition dependence of ΔG, ΔH, and TΔS for six binary liquid systems at 50°C and atmospheric pressure. The related quantities GE, HE, and TSE were shown for the same systems in Fig. 10.3. As with excess properties, property changes of mixing exhibit diverse behavior, but most systems have common features: 1. Each ΔM is zero for a pure species. 2. The Gibbs energy change of mixing ΔG is always negative. 3. The entropy change of mixing ΔS is positive. Feature 1 follows from Eq. (11.1). Feature 2 is a consequence of the requirement that the Gibbs energy be a minimum for an equilibrium state at specified T and P, as discussed in Sec. 12.4. Feature 3 reflects the fact that negative entropy changes of mixing are unusual; it is not a consequence of the second law of thermodynamics, which merely forbids negative entropy changes of mixing for systems isolated from their surroundings. For constant T and P, ΔS is observed to be negative for certain special classes of mixtures, none of which is represented in Fig. 11.3. For a binary system the partial excess properties are given by Eqs. (10.15) and (10.16) with M = ME. Thus,

. 悔

. ⦙

Figure 11.3: Property changes of mixing at 50°C for six binary liquid systems: (a)chloroform(1)/n-heptane(2);

(b)

acetone(1)/methanol(2);

(c)

acetone(1)/chloroform(2); (d) ethanol(1)/n-heptane(2); (e) ethanol(1)/chloroform(2); (f) ethanol(1)/water(2).

HEAT EFFECTS OF MIXING PROCESSES The heat of mixing, defined in accord with Eq. (11.1), is: ΔH = H −

h

(11.16)

Equation (11.16) gives the enthalpy change when pure species are mixed at constant T and P to form one mole (or a unit mass) of solution. Data are most commonly available for binary systems, for which Eq. (11.16) solved for H becomes: H = x1H1 + x2H2 + ΔH

(11.17)

This equation provides for the calculation of the enthalpies of binary mixtures from enthalpy data for pure species 1 and 2 and heats of mixing. Data for heats of mixing are usually available for only a limited number of temperatures. If the heat capacities of the pure species and of the mixture are known, heats of mixing are calculated for other temperatures by a method analogous to the calculation of standard heats of reaction at elevated temperatures from the value at 25°C. Heats of mixing are similar in many respects to heats of reaction. When a chemical reaction occurs, the energy of the products is different from the energy of the reactants at the same T and P because of the chemical rearrangement of the constituent atoms. When a mixture is formed, a similar energy change occurs because of changes in the interactions between molecules. That is, heats of reaction arise from changes in intramolecular interactions while heats of mixing arise from changes in intermolecular interactions. Intramolecular interactions (chemical bonds) are generally much stronger than intermolecular interactions (arising from electrostatic interactions, van der Waals forces, etc.) and as a result, heats of reaction are usually much larger in magnitude than heats of mixing. Large values of heats of mixing are observed when the intermolecular interactions in the solution are much different than in the pure components. Examples include systems with hydrogen bonding interactions and systems containing electrolytes that dissociate in solution.

Enthalpy/Concentration Diagrams An enthalpy/concentration (H–x) diagram is a useful way to represent enthalpy data for binary solutions. It plots enthalpy as a function of composition (mole fraction or mass fraction of one species) for a series of isotherms, all at a fixed pressure (usually 1 standard atmosphere). Equation (11.17) solved for H is directly applicable to each isotherm: H = x1H1 + x2H2 + ΔH Values of H for the solution depend not only on heats of mixing, but also on enthalpies H1 and H2 of the pure species. Once these are established for each isotherm, H is fixed for all solutions because ΔH has unique and measurable values for all compositions and temperatures. Because absolute enthalpies are unknown, arbitrary zero-point conditions are chosen for the enthalpies of the pure species, and this establishes the basis of the

diagram. The preparation of a complete H–x diagram with many isotherms is a major task, and relatively few have been published.2 Figure 11.4 presents a simplified H–x diagram for sulfuric acid(1)/water(2) mixtures, with only three isotherms.3 The basis for this diagram is H = 0 for the pure species at 298.15 K, and both the composition and enthalpy are on a mass basis. A useful feature of an enthalpy/concentration diagram is that all solutions formed by adiabatic mixing of two other solutions are represented by points lying on a straight line connecting the points that represent the initial solutions. This is shown as follows. Let subscripts a and b identify two initial binary solutions, consisting of na and nb moles or unit masses, respectively. Let subscript c identify the final solution obtained by adiabatic mixing of solutions a and b, for which ΔHt = Q = 0, and the total energy balance is:

(na + nb)Hc = naHa + nbHb On an enthalpy/concentration diagram these solutions are represented by points designated a, b, and c, and our purpose is to show that point c lies on the straight line that passes through points a and b. For a line that passes through both points a and b,

Ha = mx1a + k and Hb = mx1b + k

Figure 11.4: H–x diagram for H2SO4(1)/H2O(2). The basis for the diagram is H = 0 for the pure species at 298.15 K.

Substituting these expressions into the preceding material balance yields:

(na + nb)Hc = na(mx1a + k) + nb(mx1b + k) = m(na x1a + nb x1b) + (na + nb)k By a material balance for species 1,

(na + nb)x1c = na x1a + nb x1b Combining this equation with the preceding one gives, after reduction:

Hc = mx1c + k showing that point c lies on the same straight line as points a and b. This feature can be used to graphically estimate the final temperature when two solutions are mixed adiabatically. Such a graphical estimation is illustrated in Fig. 11.4 for the process of mixing 10 wt.% H2SO4 at 300 K (point a) with pure H2SO4 at 300 K (point b) in a 3.5:1 ratio, to yield a 30 wt% H2SO4 solution (point c). The graphical construction shows that after adiabatic mixing, the temperature will be nearly 350 K. Because the pure-species enthalpies H1 and H2 are arbitrary, when only a single isotherm is considered, they may be set equal to zero, in which case Eq. (11.17) becomes: H = ΔH = HE An HE–x diagram then serves as an enthalpy/concentration diagram for a single temperature. There are many such single-temperature diagrams in the literature, and they are usually accompanied by an equation that represents the curve. An example is the diagram of Fig. 11.5, showing data for sulfuric acid/water at 25°C. Again, xH2SO4 is the mass fraction of sulfuric acid and HE is on a unit-mass basis.

Figure 11.5: Excess enthalpies for H2SO4(1)/H2O(2) at 25°C.

The isotherm shown in Fig. 11.5 is represented by the equation HE = (−735.3 − 824.5 x1 + 195.2x12 − 914.6x13) x1(1 − x1) (A) The numbers adjacent to points on the graph come from this equation. The form of this equation is typical of models for excess properties: a polynomial in composition multiplied by x1 and x2. The x1x2 product ensures that the excess property goes to zero for both pure components. Alone, it would take the form of a symmetric parabola with a maximum or minimum at x1 = 0.5. Multiplication by a polynomial in x1 scales and biases the symmetric parabola. A simple problem is to find the quantity of heat that must be removed to restore the initial temperature (25°C) when pure water is mixed continuously with a 90% aqueous solution of sulfuric acid to dilute it to 50%. The calculation may be made with the usual material and energy balances. We take as a basis 1 kg of 50% acid produced. If ma is the mass of 90% acid, a mass balance on the acid is 0.9ma = 0.5, from which ma = 0.5556. The energy balance for this process, assuming negligible kinetic- and potential-energy changes, is the difference between the final and initial enthalpies:

Q = Hf E − (Ha E )(ma) = −303.3 − (−178.7)(0.5556) = −204.0 kJ where the enthalpy values are shown on Fig. 11.5. An alternative procedure is represented by the two straight lines on Fig. 11.5, the first for adiabatic mixing of pure water with the 90% aqueous solution of acid to form a 50% solution. The enthalpy of this solution lies at xH2SO4 = 0.5 on the line connecting the points representing the unmixed species. This line is a direct proportionality, from which HE = (−178.7/0.9)(0.5) = −99.3. The temperature at this point is well above 25°C, and the vertical line represents cooling to 25°C, for which Q = ΔHE = −303.3 − (−99.3) = −204.0 kJ. Because the initial step is adiabatic, this cooling step gives the total heat transfer for the process, indicating that 204 kJ⋅kg−1 are removed from the system. The temperature rise upon adiabatic mixing cannot be obtained directly from a diagram like this with a single isotherm, but it can be estimated if one knows the approximate heat capacity of the final solution. Equation (A) can be used in Eq. (11.14) and (11.15) to generate values of the partial excess enthalpies H -E H2SO4 and H-E H2O at 25°C. This produces the results shown in Fig. 11.6, where all values are on a unit-mass basis. The two curves are far from

symmetric, a consequence of the skewed nature of the HE curve. At high concentrations of H2SO4, H-EH2O reaches high values, in fact the infinite-dilution value approaches the latent heat of water. This is the reason that when water is added to pure sulfuric acid, a very high rate of heat removal is required for isothermal mixing. Under usual circumstances the heat-transfer rate is far from adequate, and the resulting temperature rise causes local boiling and sputtering. This problem does not arise when acid is added to water, because H-EH2SO4 is less than a third the infinite-dilution value of water.

Figure 11.6: Partial excess enthalpies for H2SO4(1)/H2O(2) at 25°C.

Heats of Solution When solids or gases are dissolved in liquids, the heat effect is called a heat of solution and is based on the dissolution of 1 mol of solute. If species 1 is the solute, then x1 is the moles of solute per mole of solution. Because ΔH is the heat effect per mole of solution, ΔH/x1 is the heat effect per mole of solute. Thus,

Δ˜H =

H

x

where Δ˜H is the heat of solution on the basis of one mole of solute.

Solution processes are conveniently represented by physical-change equations analogous to chemical-reaction equations. When 1 mol of LiCl(s) is mixed with 12 mol of H2O, the process is represented by: LiCl(s) + 12H2O(l) → LiCl(12H2O) The designation LiCl(12H2O) represents a solution of 1 mol of LiCl dissolved in 12 mol of H2O. The heat of solution for this process at 25°C and 1 bar is Δ˜H = −33,614 J. This means that the enthalpy of 1 mol of LiCl in 12 mol of H2O is 33,614 J less than the combined enthalpies of 1 mol of pure LiCl(s) and 12 mol of pure H2O(l). Equations for physical changes such as this are readily combined with equations for chemical reactions. This is illustrated in the following example, which incorporates the dissolution process just described. Example 11.3 Calculate the heat of formation of LiCl in 12 mol of H2O at 25°C. Solution 11.3 The process implied by the problem statement results in the formation from its constituent elements of 1 mol of LiCl in solution in 12 mol of H2O. The equation representing this process is obtained as follows:

Li + Cl2 → LiCl(s) ΔH°298 = −408,610 J LiCl(s) + 12H2O(l) → LiCl(12H2O) Δ˜H298 = −33,614 J ______________________________________________________ Li+ Cl2 + 12H2O(l) → LiCl(12H2O) ΔH°298 = −442,224 J

The first reaction describes a chemical change resulting in the formation of LiCl(s) from its elements, and the enthalpy change accompanying this reaction is the standard heat of formation of LiCl(s) at 25°C. The second reaction represents the physical change resulting in the dissolution of 1 mol of LiCl(s) in 12 mol of H2O(l), and the enthalpy change is a heat of solution. The overall enthalpy change, −442,224 J, is the heat of formation of LiCl in 12 mol of H2O. This figure does not include the heat of formation of the H2O. ___________________

Often heats of solution are not reported directly but must be determined from heats of formation by the reverse of the calculation just illustrated. Typical are data for the heats of formation of 1 mol of LiCl:4 LiCl(s)

−408,610 J

LiCl ⋅ H2O(s)

−712,580 J

LiCl ⋅ 2H2O(s)

−1,012,650 J

LiCl in 3 mol

− 429,366 J

LiCl ⋅ 3H2O(s)

−1,311,300 J

H 2O LiCl in 5 mol

−436,805 J

H2O LiCl in 8 mol

−440,529 J

H2O LiCl in 10 mol

−441,579 J

H2O LiCl in 12 mol

−442,224 J

H2O LiCl in 15 mol H2O

−442,835 J

Heats of solution are readily calculated from these data. The reaction representing the dissolution of 1 mol of LiCl(s) in 5 mol of H2O(l) is obtained as follows:

Li + Cl2 + 5H2O(l) → LiCl(5H2O)

ΔH°298 = −436,805 J

LiCl(s) → Li + Cl2 ΔH°298 = 408,610 J ____________________________________________________ LiCl(s) + 5H2O(l) → LiCl( 5H2O) Δ˜H298= −28,195 J

This calculation can be carried out for each quantity of H2O for which data are given. The results are then conveniently represented graphically by a plot of Δ˜H, the heat of solution per mole of solute, vs. ñ, the moles of solvent per mole of solute. The composition variable, ñ ≡ n2/n1, is related to x1:

ñ=

x n x n

n

=

n

x

x

whence

x1 =

ñ

The following equations therefore relate ΔH, the heat of mixing based on 1 mol of solution, and Δ˜H, the heat of solution based on 1 mol of solute:

Δ˜H =

H

x

= ΔH(1 + ñ ) or ΔH =

˜H

ñ

Figure 11.7 shows plots of Δ˜H vs. ñ for LiCl(s)and HCl(g) dissolved in water at 25°C. Data in this form are readily applied to the solution of practical problems. Because water of hydration in solids is an integral part of a chemical compound, the heat of formation of a hydrated salt includes the heat of formation of the water of hydration. The dissolution of 1 mol of LiCl⋅2H2O(s) in 8 mol of H2O produces a solution containing 1 mol LiCl in 10 mol of H2O, represented by LiCl(10H2O). The equations which sum to give this process are: Li +

Cl2 + 10H2O(l) → LiCl(10H2O)

LiCl ⋅ 2H2O(s) → Li +

Cl2 + 2H2 + O2

ΔH°298

= −441,579 J

ΔH°298 = 1,012,650 J

2H2 + O2 → 2H2O(l) ΔH°298 = (2)(−285,830) J ___________________________________________________________ LiCl ⋅ 2H2O(s) + 8H2O(l) → LiCl(10H2O) Δ˜H298 = −589 J

Example 11.4 A single-effect evaporator operating at atmospheric pressure concentrates a 15% (by weight) LiCl solution to 40%. The feed enters the evaporator at a rate of 2 kg·s–1 at 25°C. The normal boiling point of a 40% LiCl solution is about 132°C, and its specific heat is estimated as 2.72 kJ·kg−1·°C−1. What is the heat-transfer rate in the evaporator?

Solution 11.4 The 2 kg of 15% LiCl solution entering the evaporator each second consists of 0.30 kg LiCl and 1.70 kg H2O. A material balance shows that 1.25 kg of H2O is evaporated and that 0.75 kg of 40% LiCl solution is produced. The process is represented by Fig. 11.8.

Figure 11.8: Process of Example 11.4. The energy balance for this flow process is ΔHt = Q, where ΔHt is the total enthalpy of the product streams minus the total enthalpy of the feed stream. Thus the problem reduces to finding ΔHt from available data. Because enthalpy is a state function, the computational path for ΔHt is immaterial and is selected for convenience and without reference to the actual path followed in the evaporator. The data available are heats of solution of LiCl in H2O at 25°C (Fig. 11.7), and the calculational path shown in Fig. 11.9 allows their direct use. The enthalpy changes for the individual steps shown in Fig. 11.9 must add to give the total enthalpy change: ΔHt = ΔHta + ΔHt b+ ΔHt c+ ΔHtd The individual enthalpy changes are determined as follows: 

ΔHta: This step involves the separation of 2 kg of a 15% LiCl solution into

its pure constituents at 25°C. For this “unmixing” process the heat effect is the same as for the corresponding mixing process but is of opposite sign. For 2 kg of 15% LiCl solution, the moles of material entering are: 0. ⦙

悔 .3

000

= 7.077 mol LiCl

0.t⦙

000

t.0 ⦙

= 94.366 mol H2O

Thus the solution contains 13.33 mol of H2O per mole of LiCl. From Fig. 11.7 the heat of solution per mole of LiCl for ñ = 13.33 is −33,800 J. For the “unmixing” of 2 kg of solution,

ΔHta = (+33,800)(7.077) = 239,250 J

FIGURE 11.9: Computational path for process of Example 11.4. 

ΔHt b: This step results in the mixing of 0.45 kg of water with 0.30 kg of

LiCl(s) to form a 40% solution at 25°C. This solution comprises: 0.30 kg → 7.077 mol LiCl

and

0.45 kg → 24.979 mol H2O

Thus the final solution contains 3.53 mol of H2O per mole of LiCl. From Fig. 11.7 the heat of solution per mole of LiCl for ñ = 3.53 is −23,260 J. Therefore, ΔHt b = (−23,260)(7.077) = −164,630 J 

ΔHt c: For this step 0.75 kg of 40% LiCl solution is heated from 25 to 132°C.

Because CP is taken to be constant, ΔHt c = mCPΔT,

ΔHt c = (0.75)(2.72)(132 − 25) = 218.28 kJ = 218,280 J 

ΔHtd: In this step liquid water is vaporized and heated to 132°C. The

enthalpy change is obtained from the steam tables: ΔHtd = (1.25)(2740.3 − 104.8) = 3294.4 kJ = 3,294,400 J Adding the individual enthalpy changes gives: ΔH = ΔHta + ΔHt b+ ΔHt c+ ΔHtd = 239,250 − 164,630 + 218,280 + 3,294,400 = 3,587,300 J The required heat-transfer rate is therefore 3587.3 kJ⋅s−1. Although the enthalpy of vaporization of water dominates the overall heat requirement, the heat effect associated with the difference in heat of solution at the initial and final concentration is not negligible.

END OF THE MODULE PROBLEM EXERCISES DIRECTION: Solve the following problems. Round off your final answer to 3 decimal places. 1. If a liquid solution of HCl in water, containing 1 mol of HCl and 4.5 mol of H2O, absorbs an additional 1 mol of HCl(g) at a constant temperature of 25°C, what is the heat effect? 2. What is the heat effect when 20 kg of LiCl(s) is added to 125 kg of an aqueous solution containing 10-wt-% LiCl in an isothermal process at 25°C? 3. An LiCl/H2O solution at 25°C is made by adiabatically mixing cool water at 10°C

with a 20-mol-% LiCl/H2O solution at 25°C. What is the composition of the solution formed? 4. A 20-mol-% LiCl/H2O solution at 25°C is made by mixing a 25-mol-% LiCl/H2O solution at 25°C with chilled water at 5°C. What is the heat effect in joules per mole of final solution? 5. A liquid solution of LiCl in water at 25°C contains 1 mol of LiCl and 7 mol of water. If 1 mol of LiCl⋅ 3H2O(s) is dissolved isothermally in this solution, what is the heat effect? 6. What is the heat effect when 75 kg of H2SO4 is mixed with 175 kg of an aqueous solution containing 25-wt-% H2SO4 in an isothermal process at 300 K? 7. For a 50-wt-% aqueous solution of H2SO4 at 350 K, what is the excess enthalpy HE in kJ·kg−1? 8. A single-effect evaporator concentrates a 20-wt-% aqueous solution of H2SO4 to 70-wt-%. The feed rate is 15 kg·s−1, and the feed temperature is 300 K. The evaporator is maintained at an absolute pressure of 10 kPa, at which pressure the boiling point of 70-wt-% H2SO4 is 102°C. What is the heat-transfer rate in the evaporator? 9. What is the heat effect when sufficient SO3(l) at 25°C is reacted with H2O at 25°C to give a 50-wt-% H2SO4 solution at 60°C? 10. A mass of 70 kg of 15-wt-% solution of H2SO4 in water at 70°C is mixed at atmospheric pressure with 110 kg of 80-wt-% H2SO4 at 38°C. During the process heat in the amount of 20,000 kJ is transferred from the system. Determine the temperature of the product solution. 11. An insulated tank, open to the atmosphere, contains 750 kg of 40-wt-% sulfuric acid at 290 K. It is heated to 350 K by injection of saturated steam at 1 bar, which fully condenses in the process. How much steam is required, and what is the final concentration of H2SO4 in the tank?

12. Saturated steam at 3 bar is throttled to 1 bar and mixed adiabatically with (and condensed by) 45-wt-% sulfuric acid at 300 K in a flow process that raises the temperature of the acid to 350 K. How much steam is required for each pound mass of entering acid, and what is the concentration of the hot acid? 13. For a 35-wt-% aqueous solution of H2SO4 at 300 K, what is the heat of mixing ΔH in kJ·kg−1? 14. If pure liquid H2SO4 at 300 K is added adiabatically to pure liquid water at 300 K to form a 40-wt-% solution, what is the final temperature of the solution?

REFERENCE: Smith, J.M. et al., Introduction to Chemical Engineering Thermodynamics, 8th Edition, McGraw-Hill Companies,Inc.,Singapore,2018.