In a coffee-cup calorimeter experiment, 10.00 g of a soluble ionic compound was added to the calorimeter containing 75.0 g H2O initially at 23.2°C. The temperature of the water increased to 31.8°C. What was the change in enthalpy for the dissolution of the compound? Give your answer in units of joules per gram of compound. Assume that the specific heat of the solution is the same as that of pure water, 4.18 J ⁄ (g ⋅ °C).

In a coffee-cup calorimeter, the heat released by the dissolution process of the ionic compound is absorbed by the water. So, heat of dissolution ([tex]q_{dis}[/tex]) is equal to the heat of water ([tex]q_{water}[/tex]).

We can first calculate the heat absorbed by the water:

[tex]q_{water}[/tex]= m x Sh x ΔT

The specific heat of water(Sh) is known (4.18 J/g.ºC) and m is the mass of water in the calorimeter (75.0 g); ΔT is the change of temperature of water, from 23.2°C (initial temperature) to 31.8°C (final temperature). We introduce the data and calculate [tex]q_{water}[/tex]:

[tex]q_{water}[/tex]= 75.0 g x 4.18 J/g.ºC x (31.8°C - 23.2°C)

[tex]q_{water}[/tex]= 2696.1 J

As we know that heat of dissolution ([tex]q_{dis}[/tex]) is equal to the heat of water ([tex]q_{water}[/tex]) have already obtained [tex]q_{dis}[/tex], and we have to only divide it into the mass of compound (10.00 g). In this way, we obtain the enthalpy change of dissolution in J per gram:

[tex]q_{dis} =q_{water}[/tex]= ΔH/m= 2696.1 J/10.00 g= 269.61 J/g