A0.10 kg piece of copper at an initial temperature of 95°c is dropped into 0.20 kg of water contained in a 0.28 kg aluminum calorimeter. the water and calorimeter are initially at 15°c. what is the final temperature of the system when it reaches equilibrium? show your work.

(cp of Copper = 387J / kg times degrees C; cp of Aluminum = 899 J / kg times degrees C; cp of Water = 4186J / kg times degrees C)
 Use the law of conservation of energy and assuming no heat loss to the surroundings, then 
 Heat given up by copper = heat absorbed by water + heat absorbed by calorimeter 
 Working formula is 
 Q = heat = MCp(delta T) 
 where 
 M = mass of the substance 
 Cp = specific heat of the substance 
 delta T = change in temperature 
 Heat given up by copper = 0.10(387)(95 - T) 
 Heat absorbed by water = 0.20(4186)(T - 15) 
 Heat absorbed by calorimeter = 0.28(899)(T - 15) 
 where 
 T = final temperature of the system 
 Substituting appropriate values, 

 0.10(387)(95 - T) = 0.20(4186)(T - 15) + 0.28(899)(T - 15) 
 38.7(95 - T) = 1088.92(T - 15) 
 3676.50 - 38.7T = 1088.92T - 16333.8 
 1127.62T = 20010.3 
 T = 17.75 C 

Why the rest of the question?