A 12.1 gram piece of aluminum whose temperature is 81.7 celsius is added to a sample of water at 23.4 celsius in a constant-pressure calorimeter. The final temperate of water and aluminum is 25.7 celsius. Calculate the mass of the water in the calorimeter. The specific heat of AL is 0.900 J/(g celsius) and the specific heat of water is 4.184 J/(g celsius).

Please solve with steps thank you!

Question

A 12.1 gram piece of aluminum whose temperature is 81.7 celsius is added to a sample of water at 23.4 celsius in a constant-pressure calorimeter. The final temperate of water and aluminum is 25.7 celsius. Calculate the mass of the water in the calorimeter. The specific heat of AL is 0.900 J/(g celsius) and the specific heat of water is 4.184 J/(g celsius). 

Please solve with steps thank you!

cuanchi · Answer

you can use the formula  for the water and the aluminum E=m x Ce x delta T, where the  the energy exchange are the same (different sign) and you are given the two Ce and one mass. You have to consider the delta T of the metal is = 25.7-81.7 and the delta T of the water is = 25.7-23.4
m1 x Ce1 x deltaT1 = m2 xCe2 x deltaT2  => 
m2 = m1 x Ce1 x deltaT1 /(Ce2 x deltaT2)