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

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

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multiply 2kg by latent heat of vap