A16.2 g piece of aluminum (which has a molar heat capacity of 24.03 j/°c·mol) is heated to 82.4°c and dropped into a calorimeter containing water (specific heat capacity of water is 4.18 j/g°c) initially at 22.3°c. the final temperature of the water is 25.7°c. ignoring significant figures, calculate the mass of water in the calorimeter.

The solution is:

q = m*c*dT 

First I converted 24.03J/°Cmol to grams 
(24.03J/°Cmol) * (1 mol Al/27g Al)= 0.89J/°Cg 

The final temperarure of water will be similar as the final temperature of aluminum 

q(aluminum) = 16.2g * 0.89J/°Cg * (25.7C - 82.4°C) = -817.5006 J 

q(water) = +817.5006 J = m * 4.18J/g°C * (25.7°C - 22.3°C) 

817.5006 = 14.212 m 
m = 57.52 grams of water or almost 58 grams

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