In the following experiment, a coffee-cup calorimeter containing 100 m L of H 2

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 6.50g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution ?Hsoln 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 6.50g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution ?Hsoln of CaCl2 is ?82.8 kJ/mol.

Explanation / Answer

Moles of CaCl2 = mass/molar mass of CaCl2

= 6.50/110.985 = 0.058566 mol

Heat released = moles x -DH

= 0.058566 x 82.8 = 4.849 kJ = 4849 J


Let T be the final temperature

Mass of water = volume x density of water

= 100 x 1.00 = 100 g

Mass of solution = 6.50 + 100 = 106.5 g

Heat absorbed = mass x specific heat x temperature change of solution

= 106.5 x 4.184 x (T - 23.0)


Heat released = heat absorbed

4849 = 106.5 x 4.184 x (T - 23.0)

Final temperature T = 33.9 deg C

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