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