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 3.50 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

we know that

moles = mass / molar mass

so

moles of CaCl2 = 3.5 / 111 = 0.03153

now

amount of heat produced = moles x heat of solution

so

amount of heat produced = 0.03153 x 82.8

amount of heat produced = 2.6108 kJ

amount of heat produced = 2610.8 J

now

mass of water = 100 g

mass of CaCl2 = 3.5 g

total mass = 103.5 g

now

we know that

heat = mass x specific heat x temp change

so

Q = m x s x dT

so

2610.8 = 103.5 x 4.184 x ( T - 23)

T = 29.04

so

the final temperature is 29.04 C