A 100.0 ml volume of 1.0 mol/L hydrochloric acid was mixed with 100.0 ml of 1.0....

Question

A 100.0 ml volume of 1.0 mol/L hydrochloric acid was mixed with 100.0 ml of 1.0 mol/l sodium hydroxide at 23.5 Celsius in a coffee cup Calorimeter. After the reactants mix, the temperature increases to 30.6 Celsius. The specific heat capacity of water is 4.18 j/g Celsius. How much energy (in kJ ) is released in this reaction? Give the quantity and the unit ( e.g. 3.5 cm).

Answer 1

You can start off by organizing what info you're given from the problem--

HCl volume = 100mL

NaOH volume = 100mL

Initial temp = 23.5

Final temp = 30.6

specific heat: 4.18

This is almost enough information to solve the problem using the equation ∆H = mc∆t. All we're missing is the "m" (mass). But, we're given volume, and we know that density = mass divided by volume (d = m/v). Rearranging this to solve for mass (since we already know volume, and that the denisty of water is 1g/mL), we get:

m = dv (density times volume)

m = 200 mL x (1g/1mL)

m = 200 g of water (since the mL cancels out)

Now we can use the equation ∆H = mc∆T, solving for ∆H (where m is mass, c is specific heat, and ∆T is the change in temp, which is obtained when you do final temp minus the initial temp)

∆H = (200g) (4.18 J/g Celsius) (30.6-23.5 Celsius)

∆H = 5935.6 J

Convert this to kJ, and you get about 5.9 kJ