Question

What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter? NaOH (s) → Na+ (aq) + OH– (aq) ∆H = -44.5 kJ/mol

Answer 1

Answer : The final temperature of the solution in the calorimeter is,
`31.0^oC`

Explanation :

First we have to calculate the heat produced.

`\Delta H=(q)/(n)`

where,

`\Delta H` = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of
`NaOH` = 1.52 g

Molar mass of
`NaOH` = 40 g/mol

`\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=(1.52g)/(40g/mole)=0.038mole`

Now put all the given values in the above formula, we get:

`44.5kJ/mol=(q)/(0.038mol)`

`q=1.691kJ`

Now we have to calculate the final temperature of solution in the calorimeter.

`q=m* c* (T_2-T_1)`

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water =
`4.18J/g^oC`

`T_1` = initial temperature =
`20.1^oC`

`T_2` = final temperature = ?

Now put all the given values in the above formula, we get:

`1691J=37.02g* 4.18J/g^oC* (T_2-20.1)`

`T_2=31.0^oC`

Thus, the final temperature of the solution in the calorimeter is,
`31.0^oC`