Question

Part A

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 4.00 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.

Express your answer with the appropriate units.

Hints

Answer 1

molar mass of CaCl2= 40+71=111 g/mole, mass of CaCl2 used= 4 gm

moles of Cacl2= 4/111 =0.036, heat liberted from 0.036 moles of CaCl2= 82.8Kj/mole*0.036=2.98 Kj=2.98*1000J

this heat is transferred to water and CaCl2

mass of water= density of water* volume of water= 1 g/cc*100cc= 100gm

mass of CaCl2= 4 gm

total mass =100+4=104 gm

heat transferred= mass* specific heat* temperature difference

since there is 4 gm of Cacl2 in 100 gm of water, the specific heat can be assumed to be same as that of water

specific heat of water= 4.184 J/gm.deg.c

2.98*1000= 104*4.184*(T-23), T= final temperature

T-23= 2.98*1000/(4.184*104)= 6.84

T= 23+6.84=28.94 deg.c