Answer:
Specific heat capacity of aluminium is 0,863 J/g°C
Explanation:
<em>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</em>
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 <em>(1)</em>
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
<em><u>Q = 3661 J</u></em>
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
<em>C = 0,863 J/g°C</em>
I hope it helps!
