A 32.5 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 24.2°C.Ignoring significant figures,calculate the mass of water in the calorimeter.

Question

Quizplus · Accepted Answer

The heat lost by the aluminum equals the heat gained by the water:
(mass of aluminum)(molar heat capacity of aluminum)(change in temperature of aluminum) = (mass of water)(specific heat capacity of water)(change in temperature of water)

Rearranging the equation, we get:
mass of water = (mass of aluminum)(molar heat capacity of aluminum)(change in temperature of aluminum) / (specific heat capacity of water)(change in temperature of water)

Plugging in the given values into the equation, we get:
mass of water = (32.5 g)(24.03 J/°C·mol)(82.4°C - 22.3°C) / (4.18 J/g°C)(24.2°C - 22.3°C) ≈ 212 g

Therefore, the mass of water in the calorimeter is 212 g.

A 325 G Piece of Aluminum (Which Has a Molar Heat