Question

a 0.500 gram sample of CaCl2 is dissolved in 100.0 g of water with both substances starting at 25.0 C. calculate the final temperature of the solution assuming the solution has no heat loss to the surroundings and assuming that the solution has a specific heat of 4.18 J/g C

Answer 1

$Qdis+Qsol=0$

The heat of solution of CaCL₂ at 25 °C

Qsol=ms\times Cs\times(T-Ti)

Qsol=ms×Cs×(T−Ti) is qdis=-83.3Kj/mol

$M ( CaCL2)=1\times40+2\times35.45=110.9 g/mol$

$Qdis=qdis\times\frac{m}{M}=-83.3\times\frac{0.5}{110.9}=-0.38Kj=-380j$

$Qsol=ms\times Cs\times(T-Ti)$

$Qsol=ms\times Cs\times(T-Ti)=ms×Cs×(T−Ti)+Qdis=0$

$T=Ti-Qdis\times(ms\times Cs)^-1=25-\frac{-380}{(100+0.5)\times4.18}=25-\frac{-380}{420.09}=25-(-0.9)=25.9°C$