2. A piece of Al(s) at 98°C is placed in 66g of H2O at 25°C. The...

2. A piece of Al(s) at 98°C is placed in 66g of H2O at 25°C. The Aluminum cools and after equilibrium the final temperature of Aluminum and water is 30°C. What was the mass of the aluminum? Cs(H2O)= 4.184J/(g°C)

Cs(Al)= 0.90J/(g.°C)

___ g

Expert Solution

heat lose of Al = heat gain of water

mct = mCt

m*0.9*(98-30) = 66*4.18*(30-25)

m = 22.54g

mass of aluminium = 22.54g

queen_honey_blossom answered 1 year ago

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