[06.01]you have 500 grams of water at 80° celsius. you want to lower the temperature of the water to as close to 60° celsius as possible. which of the following would get the 500 grams of water the closest to 60° celsius?

The heat lost by the 80 deg.C solution would equal the heat gained by the colder solution, and the final temperature would be 60 deg.C. x is the amount of colder solution added and y is the initial temp of the colder solution.

Heat lost by 80 deg.C solution = (500 g)(1.00 cal/g C)(80 deg.C - 60 deg.C) = 1000 cal

Heat gained by colder solution = (x)(60 - y)

By evaluating further, we will arrive at adding 100 grams of 50° Celsius water. This satisfies to the problem above.