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A sample of aluminum is put into a calorimeter (see sketch at right) that contains of water. The aluminum sample starts off at and the temperature of the water starts off at . When the temperature of the water stops changing it's . The pressure remains constant at . Calculate the specific heat capacity of aluminum according to this experiment. Be sure your answer is rounded to significant digits.

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Answer:

Specific heat capacity of aluminium is 0.869J/g°C

Explanation:

Values: 51.1g of Al, 150.0g of water, Aluminium sample: 92.1°C; Water: 18.0°C, final temperature: 22.9°C.

The heat absorbed for the water is the same relased for the aluminium, the formula is:

[tex]m_{H_2O}*dT_{H_2O}*C_{H_2O}=-m_{Al}*dT_{Al}*C_{Al}[/tex]

Where m is mass, dT is change in temperature and C is specific heat of each compound (4.18J/g°C for water)

Replacing:

150.0g×(22.9°C-18.0°C)×4.18J/g°C = -51.1g×(22.9°C-92.1°C)×C(Al)

3072.3J = 3536.12g°C×C(Al)

0.869J/g°C = C(Al)

Specific heat capacity of aluminium is 0.869J/g°C

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