Question

In my Study Sheet I have a question that is as follows: What is the maximum temperature that can be reached in a foam cup calorimeter containing 105 mL of water at 20.12 °C following the addition and dissolving a small pellet of KOH weighing 0.215 g? (AH KOH=-57.6 J/mol) (Ans: 20.25 °C) (Strategy: heat given off by KOH = heat absorbed by water) I will like the class to help solve this problem in the discussion board. Don't just post your answer or calculations. Try to explain how you would approach it and set it up.

Answer 1

First we calculate the heat given off by dissolution of KOH in water

Given : mass of KOH pellet = 0.215 g

moles KOH = (mass KOH) / (molar mass KOH)

moles KOH = (0.215 g) / (56.1 g/mol)

moles KOH = 0.00383 mol

Heat given by dissolution of KOH = (moles KOH) * ($\dpi{100} \small \Delta$H KOH)

Heat given by dissolution of KOH = (0.0383 mol) * (-57.6 kJ/mol)

Heat given by dissolution of KOH = -0.221 kJ

Heat given by dissolution of KOH = -221 J     (negative sign indicates heat release)

Heat absorbed by water = Heat given by dissolution of KOH

Heat absorbed by water = 221 J

Now we find the final temperature of water

volume of water = 105 mL

mass of water = (volume of water) * (density of water)

mass of water = (105 mL) * (1.00 g/mL)

mass of water = 105 g

Total mass of solution = (mass of water) + (mass KOH)

Total mass of solution = (105 g) + (0.215 g)

Total mass of solution = 105.215 g

we know that,

heat absorbed by solution = (mass of solution) * (specific heat of solution) * (final temp. - initial temp.)

221 J = (105.215 g) * (4.18 J/g.^oC) * (final temp. - 20.12 ^oC)

(final temp. - 20.12 ^oC) = (221 J) / [(105.215 g) * (4.18 J/g.^oC)]

(final temp. - 20.12 ^oC) = 0.50 ^oC

final temp. = 20.12 ^oC + 0.50 ^oC

final temp. = 20.62 ^oC