BRAVO! That's the right answer! Call the final temperature T

_{f}. Then thechangein temperature for thecoffeeisT _{1}= T_{1f}- T_{1i}

_{}Anything =changein anything = (final value) - (initial value)T

_{1}= T_{f}- 95^{o}CThe heat transferred from the

coffee(orbythecoffee) isQ _{1}= c m_{1}T_{1}Q

_{1}= [4186 J/(kg C^{o})] [0.6 kg] [T_{f}- 95^{o}C]Notice that [T

_{f}- 95^{o}C] will benegativewhich means Q_{1}will benegative.That means that Q_{1}represents heat flowingfromorout ofthecoffee.The

changein temperature for themilkisT _{2}= T_{2f}- T_{2i}T

_{2}= T_{f}- 10^{o}CThe heat transferred from the

milk(ortothemilk) isQ _{2}= c m_{2}T_{2}Q

_{2}= [4186 J/(kg C^{o})] [0.15 kg] [T_{f}- 10^{o}C]Notice that [T

_{f}- 10^{o}C] will bepositivewhich means Q_{2}will bepositive.That means that Q_{1}represents heat flowingintothemilk.Energy conservation -- or the Work - Energy Theorem -- tells us that the sum of those two heats is zero;

Q _{1}+ Q_{2}= 0[4186 J/(kg C

^{o})] [0.6 kg] [T_{f}- 95^{o}C] + [4186 J/(kg C^{o})] [0.15 kg] [T_{f}- 10^{o}C] = 0Both of these materials are essentially

waterso 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] [T _{f}- 95^{o}C] + [0.15 kg] [T_{f}- 10^{o}C] = 0[0.6] [T

_{f}- 95^{o}C] + [0.15] [T_{f}- 10^{o}C] = 00.6 T

_{f}- 57^{o}C + 0.15 T_{f}- 1.5^{o}C = 00.75 T

_{f}- 58.5^{o}C = 00.75 T

_{f}= 58.5^{o}C

T_{f}= 78^{o}C

(c) 2000, Doug Davis; all rights reserved.