Question

A 59.1 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 the correct number of significant digits.

Answer 1

Answer:

Specific heat capacity of aluminium is 0,863 J/g°C

Explanation:

Values: 250g of water, aluminum sample starts off at 91.3°C, water starts off at 16.0°C, temperature of the water stops changing it's 19.5°C

In this problem, the heat produced for the aluminium is the same consumed by water. The heat consumed by water is:

Q = C×m×ΔT (1)

Where C is specific heat of water (4,184J/g°C), m is mass of water (250,0g) and ΔT is change in temperature of water (19,5°C-16,0°C = 3,5°C)

Replacing:

Q = 4,184J/g°C×250,0g×3,5°C

Q = 3661 J

Using (1), it is possible to obtain specific heat of aluminium, thus:

Q / (m×ΔT) = C

Where Q is heat (3661J), m is mass (59,1g) and ΔT is change in temperature (91,3°C - 19,5°C( = 71,8°C

Replacing:

3661J / (59,1g×71,8°C) = C

C = 0,863 J/g°C

I hope it helps!