Question

In the following experiment, a coffee-cup calorimeter containing 100 mL of H2O is used. The initial temperature of the calorimeter is 23.0 ∘C. If 5.10 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution ΔHsoln of CaCl2 is −82.8 kJ/mol .

Answer 1

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

Explanation :

First we have to calculate the heat produced.

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

where,

`\Delta H` = enthalpy change = 82.8 kJ/mol

q = heat released = ?

m = mass of
`CaCl_2` = 5.10 g

Molar mass of
`CaCl_2` = 110.98 g/mol

`\text{Moles of }CaCl_2=\frac{\text{Mass of }CaCl_2}{\text{Molar mass of }CaCl_2}=(5.10g)/(110.98g/mole)=0.0459mole`

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

`82.8kJ/mol=(q)/(0.0459mole)`

`q=3.80kJ`

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

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

where,

q = heat produced = 3.80 kJ = 3800 J

m = mass of solution = 100 + 5.10 = 105.10 g

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

`T_1` = initial temperature =
`23.0^oC`

`T_2` = final temperature = ?

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

`3800J=105.10g* 4.18J/g^oC* (T_2-23.0)`

`T_2=31.6^oC`

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