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 4.90 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

Given that;

Volume of water = 100 ml

We know that the density of water is 1.0g/ml so mass of water is 100g.

And here 4.90 g CaCl2

Mole sof CaCl2 = amount of CaCl2/ molar mass

= 4.90 g/ 110.98 g CaCl2

= 0.044 moles CaCl2

According to the problem; The heat of solution ΔHsoln of CaCl2 is −82.8 kJ/mol.

Heat produced by CaCl2 = −82.8 kJ/mol * 0.044 moles CaCl2

= - 3.656 kJ

= 3656 J

Heat capacity of water , C= 4.184 J/ g-c

Heat = mass of solution ; m* C* temperature change

3656 J = (100 +4. 90) g * 4.184 J/ g-C* temperature change

Temperature change or t final – t initial = 8.33 C

T final = 8.33 C + t initial

= 8.33 C + 23.0 ∘C

=31.33 ∘C