Specific Heat

Question

When 23.6 g of calcium chloride were dissolved in 300 mL of water in a calorimeter, the temperature of the water rose from 25.0 degree celsius to 38.7 degree celsius. What is the heat energy change in kcal for this process? (Specific heat of h2O = 1.00 cal/g degree celsius

) Using your number as a conversion factor, how much heat energy is produced when one mole of calcium chloride dissolves in water? Report your answer in kJ/mol. *1 mole of calcium chloride = 110.96 g calcium chloride*

Answer 1

– Step 1

First, let us calculate the heat absorbed
by the calorimeter.
q_{cal = (1258 J/^oC)(38.7^oC-25^oC)}

 

Step 2

Molecular Mass= 111.1g

 

(23.6CaCl₂)x(1molCaCl₂/111.1g)=0.212molCaCl₂

 

 

 

Step 3 Now we must calculate the heat per mole of

calcium chloride.

q_cal=-q_rxn

 

ΔH=(q_rxn/ molCaCL₂) = -17.2KJ/0.212mol = -81.1 KJ/mol

 

 

 

 

 

 

Answer 2

q = mC∆T

q = heat

m = mass = 300 g

C = specific heat = 1 cal/g/deg

∆T = change in temperature = 38.7 - 25 = 13.7 degrees

q = (300 g)(1 cal/g/deg)(13.7 degrees)

q = 4110 cal and converting to kcal ... 

q = 4.11 kcal (3 significant figures)

NOTE:  In high school chemistry, the mass is usually taken as 300 g (mass of water assuming a density of 1g/ml). In college or advanced chemistry, m = 300 g + 23.6 g to include the mass of the solute.  You can repeat this calculation using m =  323.6 g if you wish.

 

To calculate heat of solution in kJ/mole:

moles CaCl2 dissolved = 23.6 g x 1 moles/110.96 g = 0.213 moles

heat = 4.11 kcal x 4.184 kJ/kcal = 1719 kJ (convert kcal to kJ)

heat of solution = 1719 kJ/0.213 moles = 80.7 kJ/mole (3 significant figures)