A 59.1 g sample of aluminum is put into a calorimeter that contains water. The aluminum sample starts off at a higher temperature, and the temperature of the water starts off at a lower temperature. When the temperature of the water stops changing, it reaches thermal equilibrium. The pressure remains constant.

Calculate the specific heat capacity of aluminum according to this experiment. Be sure your answer is rounded to the correct number of significant digits.

(Note: Include the missing values for the mass of water, initial temperatures, final temperature, and pressure to complete the question.)

To calculate the specific heat capacity of aluminum from the given experiment, we would need the initial and final temperatures of the aluminum and water, as well as the mass of water. Without these details, it's impossible to provide a direct answer.

Explanation:

The question pertains to calculating the specific heat capacity of aluminum from an experiment where a piece of aluminum changes temperature when placed in water, reaching thermal equilibrium. Unfortunately, critical numerical details necessary for the calculation (initial and final temperatures of the aluminum and water, and the mass of the water) are missing from the question. Proper calculation would involve using the formula q = mcΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. These values would allow us to solve for the specific heat capacity of aluminum by equating the heat lost by the aluminum to the heat gained by the water, acknowledging that these should be equal in magnitude but opposite in direction.