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 8.70 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

Molecular weight of CaCl2= 8.7 gm

Moles of CaCl2= mass/Molecular weight , Molecular weight of CaCl2= 111 gms

Moles of CaCl2= 8.7/111=0.0783 moles

Mass of water =100*1 =100 gm ( density of water = 1 gm/cc)

Mass of solution = mass of water + mass of CaCl2= 100+8.7= 108.7 gms

heat of solution = -82.8 Kj/mol

1mol of CaCl2 gives rise to 82.8 Kj= 82.8*1000 joules=82800 joules

0.0783 moles give rise to 0.0783*82800 joules=6483.24joules

Cp = specific heat of solution , same as that of water= 4.184, m =mass of solution, delT= temperature difference

From mCpdelT= 6483.24

108.7*4.184*delT)= 6483.24

delT= 6483.24/(108.7*4.184)=14.25

T- 23 =14.25 , T = 37.25 deg.c