Question

What is recorded as the final temperature (in oC) on a constant pressure calorimeter if 9.377 g of potassium hydroxide is dissolved in 30.431 mL of water originally at 21.094oC?

Answer 1

enthalphy of solution for KOH = -57.61 KJ/mol

mass of KOH =9.377 g
Molar mass of KOH = 56.1 g/mol
number of moles of KOH = mass/ molar mass = 9.377/56.1 = 0.167 mol

Heat released = number of moles * enthalphy of solution = 0.167*57.61 = 9.6 KJ = 9600 J

Specific heat capacity of water = 4.186 J/goC
mass of water = volume * density = 30.431 mL * 1 g/mL = 30.431 g

This heat is absorbed by water
Let final temperature be T oC
Q = m*C* delta T
9600 = 30.431 * 4.186 * (T- 21.094)
T-21.094 = 75.36
T=96.45 oC
Answer: 96.45 oC