A 50.0 mL sample of dilute HCl and 50.0 mL of dilute NaOH are at 19.50oC to begi
ID: 1001791 • Letter: A
Question
A 50.0 mL sample of dilute HCl and 50.0 mL of dilute NaOH are at 19.50oC to begin with. After mixing the two in a styrofoam cup, the temperature rises to 21.21oC. Calculate the heat of the reaction. (HINT: the density of the total solution mixture can be assumed to be that of water 1.00 g/mL) C = 4.184 J/goC qrxn = - qH2O
From this data: Find qrxn for the process. (again, because the solutions are dilute, you can assume the solution has a density of 1.00 g/mL) (note, keep in mind the q you are solving for! Is the q of the reaction what you are measuring, or are you calculating the q of the surroundings?)
Explanation / Answer
total volume = volume of HCl + volume of NaOH
so
total volume = 50 + 50
total volume = 100 ml
now
mass = density x volume
mass of solution = 1 x 100
mass of solution = 100 g
now
heat = mass x specific heat x temp change
Q = m x s x dT
Q = 100 x 4.184 x ( 21.21 - 19.5)
Q = 715.464 J
so
heat of the reaction is -715.464 J
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