In which case is more heat involved: mixing 100.0g of 90 degree celsius water and 80 degree celsius water or mixing 100g samples of 60 degree celsius water and 10 degree celsius water? assume no heat is lost to the environment.

The heat change by the each sample is proportional to its mass times the temp change. So if we use T for final temp, the energy lost by the 100g is:100(80-T) 
The heat gained by the 60g is 60(T-20) 
Set the two equal and do the algebra. T=60 
(I should have multiplied each by the heat capacity of water, but when I set the two terms equal, it would just cancel out.)