Question

Compute the change in temperature when 1.6 g of sodium chloride is dissolved in 250.0 mL of water initially at 24 C. The heat capacity of water is 4.184 J/mL/K. Neglect the heat capacity of the NaCl.
Species ΔfH (kJ/mol)
NaCl(aq) -407.3
NaCl(s) -411.2

Answer 1

First of all, remember the equation:

\[\Delta Hn=-mc \Delta T\]
The net reaction here is NaCl (s) --> NaCl (aq). We can find ΔH by finding ΔHf(products)-ΔHf(reactants) = -407.3-(-411.2) = +3.9 kJ/mol = +3900 J/mol.

We can find moles since we're given mass, so just divide by the molar mass of NaCl: 1.6/58.5 = 0.0274 mol

The left side of the equation represents the dissolution reaction, while the right side represents the effects on the solution (i.e. on the water). The question gave the heat capacity of water per mL, so we can use volume instead of mass in the original equation. Also note that J/mL/K = J/mL/C.

We're given the initial temperature and the heat capacity already, so all we need to do is plug everything in and solve for the final temperature:

(3900)(0.0274) = -(250)(4.184)(Tf-24) --> Tf = 23.9 C

This answer seems reasonable since the dissolution reaction is slightly endothermic, meaning the temperature of the surroundings should decrease somewhat.