In the following experiment, a coffee-cup calorimeter containing 100 mL of H2O i

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 8.80 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of the solution of delta H_soln of CaCl2 is -82.8 kJ/mol. 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 8.80 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of the solution of delta H_soln of CaCl2 is -82.8 kJ/mol.

Explanation / Answer

Moles of CaCl2 = given mass / molar mass = 8.80 g / 111 g = 0.079 mol

Heat capacity of solution = 0.079 mol * 82.8 x 10^3 J/mol = 6.541 x 10^3 J/mol

Heat capacity of the solution = Heat capacity of water (4.184 J/gC)

heat = m * c * delta T

6.541 x 10^3 = 100 g * 4.184 J/gC * deltaT

deltaT = 15.63 C

Final temperature is 23 + 15.63 = 38.63 degree C

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