  ## BRAVO! That's the right answer!

Call the final temperature Tf. Then the change in temperature for the coffee is T1 = T1f - T1i Anything = change in anything = (final value) - (initial value) T1 = Tf - 95oC

The heat transferred from the coffee (or by the coffee ) is Q1 = c m1 T1 Q1 = [4186 J/(kg Co)] [0.6 kg] [Tf - 95oC]

Notice that [Tf - 95oC] will be negative which means Q1 will be negative. That means that Q1 represents heat flowing from or out of the coffee.

The change in temperature for the milk is T2 = T2f - T2i T2 = Tf - 10oC

The heat transferred from the milk (or to the milk) is Q2 = c m2 T2 Q2 = [4186 J/(kg Co)] [0.15 kg] [Tf - 10oC]

Notice that [Tf - 10oC] will be positive which means Q2 will be positive. That means that Q2 represents heat flowing into the milk.

Energy conservation -- or the "Work - Energy Theorem" or the "First Law of Thermodynamics" -- tells us that the sum of those two heats is zero; Q1 + Q2 = 0

[4186 J/(kg Co)] [0.6 kg] [Tf - 95oC] + [4186 J/(kg Co)] [0.15 kg] [Tf - 10oC] = 0

Both of these materials are essentially water so their two specific heats are the same. Since the specific heats are the same, we can immediately factor that out and reduce this equation to

[0.6 kg] [Tf - 95oC] + [0.15 kg] [Tf - 10oC] = 0

[0.6] [Tf - 95oC] + [0.15] [Tf - 10oC] = 0

0.6 Tf - 57oC + 0.15 Tf - 1.5oC = 0

0.75 Tf - 58.5oC = 0

0.75 Tf = 58.5oC

Tf = 78oC   