A59.1 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.

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!

the answer to this question is 3.