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.

$\Delta H=\frac{q}{n}$

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

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

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

$82.8kJ/mol=\frac{q}{0.0459mole}$

$q=3.80kJ$

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

$q=m\times c\times (T_2-T_1)$

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:

$3800J=105.10g\times 4.18J/g^oC\times (T_2-23.0)$

$T_2=31.6^oC$

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