Adiabatic Mixing of Acetone and Dichloromethane: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2015, Q6

Question 6: Adiabatic Mixing of Acetone and Dichloromethane (Part B — 25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Equal masses of two pure liquids at 298 K are mixed adiabatically at 1 bar with no work. Acetone + dichloromethane mix exothermically: the C–H···O=C hydrogen bond between the two liquids gives a negative excess enthalpy, the same behaviour as acetone–chloroform. The tabulated positive heat of mixing is therefore read as heat released per kg of solution, so the product warms above 298 K.

Quantity	Value
$C_p$ acetone / dichloromethane	2.173 / 1.193 kJ/kg·K
Solution $C_p$ (equal mass)	½(2.173+1.193) = 1.683 kJ/kg·K
Heat of mixing at 293 / 298 / 303 K	12.468 / 12.380 / 12.292 kJ/kg
Feed temperature	298 K, adiabatic, 1 bar, no work

Find. The temperature $T$ of the solution formed.

Figure 6 — Adiabatic mixer: equal masses of acetone and dichloromethane enter at 298 K; the exothermic heat of mixing raises the product temperature.

Approach. Adiabatic and work-free means the overall enthalpy change is zero. Follow a path that first heats the pure components from 298 K to the final $T$, then mixes them at $T$; setting the total to zero equates the sensible heat gained to the heat of mixing released (evaluated at $T$, using the tabulated temperature dependence).

Zero-enthalpy path. Per kg of solution, heating the (still-pure) components 298 → $T$ costs $C_{p,soln}(T-298)$ and mixing at $T$ releases $q_{mix}(T)$; adiabatic operation sets their sum to zero: $$C_{p,soln}\,(T-298)=q_{mix}(T).$$
Temperature dependence of the heat of mixing. The three data points are linear in $T$ with slope $\dfrac{12.292-12.468}{303-293}=-0.0176$ kJ/kg·K, so $q_{mix}(T)=12.380-0.0176\,(T-298)$.
Solve for the mixing temperature. Substituting with $C_{p,soln}=1.683$ and letting $u=T-298$: $$1.683\,u=12.380-0.0176\,u\;\Rightarrow\;u=\frac{12.380}{1.683+0.0176}=7.28\ \text{K}\;\Rightarrow\;T=\boxed{305.3\ \text{K}\;(32.1\ ^\circ\text{C}).}$$ The weak temperature dependence barely matters here — ignoring it gives 305.4 K — but including it is the rigorous route the three data points invite.

Assumption

The exam prints the heat of mixing as a positive number without stating a sign convention. It is taken here as heat released, because acetone–dichloromethane mixing is exothermic. If the same values were instead read as an endothermic $\Delta H_{mix}$ (heat absorbed), the balance becomes $1.683\,u=-(12.380-0.0176\,u)$, giving $u=-7.43$ K and $T\approx290.6$ K. State whichever convention you adopt; the method is unchanged.

Quantity	Result
Solution heat capacity	1.683 kJ/kg·K
Temperature rise	+7.3 K
Solution temperature	305.3 K (32.1 °C)

Concept & Theory

Topic. Adiabatic mixing with a non-zero heat (enthalpy) of mixing — an energy balance on a real solution.

Key relations. Adiabatic, work-free balance $\Delta H_{total}=0$; construct a path (heat pures → mix at $T$): $C_{p,soln}(T-T_0)=q_{mix}(T)$; linear interpolation of $q_{mix}(T)$ from tabulated values.

Why this works. Enthalpy is a state function, so any convenient path between the fixed initial (pure components at 298 K) and final (solution at $T$) states gives the same $\Delta H$; choosing "heat then mix" isolates a sensible-heat term and a mixing term whose sum must vanish under adiabatic operation. Because acetone–dichloromethane mixing is exothermic, that released heat is what raises the temperature.

Common pitfall. Getting the sign of the heat of mixing wrong (an exothermic solution warms, not cools), or evaluating $q_{mix}$ at the feed temperature instead of the final temperature — a small correction here, but conceptually the mixing occurs along the path at $T$.

Question 6 of 6: Adiabatic Mixing of Acetone and Dichloromethane

Question 6: Adiabatic Mixing of Acetone and Dichloromethane (Part B — 25 marks)