How many grams of water at 20°C are necessary to change 800g of water at 90°C to 50°C?

Answer 1

[mass cool H2O x specific heat H2O x (Tfinal-Tinitial)] + [mass cool H2O x specific heat H2O x (Tfinal-Tinitial)] = 0

Substitute all of the numbers; mass cool H2O @ 20C is the only unknonwn.

Answer 2

so its setup like
(20*1.00)*(90-50)+(20*1.00)*(90-50) = 1600g?