A 22.6 g piece of aluminum (which has a molar heat capacity of 24.03 J/mol°C) is heated to 85.2°C and dropped into a calorimeter containing water (the specific heat capacity of water is 4.18 J/g°C) initially at 23.4°C. The final temperature of the water is 27.8°C. Calculate the mass of water in the calorimeter.

Question

Quizplus · Accepted Answer

The correct answer is A.The heat lost by the aluminum is equal to the heat gained by the water.$$Q_{Al} = -Q_{water}$$$$mC_{Al}\Delta T_{Al} = -mC_{water}\Delta T_{water}$$$$(22.6 g)(24.03 J/mol°C)(85.2°C - 23.4°C) = -(mC_{water})(27.8°C - 23.4°C)$$$$(22.6 g)(24.03 J/mol°C)(61.8°C) = -(mC_{water})(4.4°C)$$$$34,200 J = -(mC_{water})(4.4°C)$$$$mC_{water} = \frac{34,200 J}{4.4°C} = 7,773 g$$Therefore, the mass of water in the calorimeter is 7,773 g.

A 226 G Piece of Aluminum (Which Has a Molar Heat