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 Q1 represents heat flowing into the milk.

Energy conservation -- or the Work - Energy Theorem -- 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

(c) 2002, Doug Davis; all rights reserved.