Answer:
10.58 J/g-°C
Explanation:
To find the specific heat capacity of the metal, you need to know how much heat was lost when it reacted with water.
You know that there are 55.4 g of water, the initial temp. of water is 25°C, and the final temp. (the mixture's temp.) is 63.4°C.
You should also know that the specific heat capacity of water is 4.186 J/g-°C.
Plug this into the equation the q=mcΔT.
q = (55.4 g)(4.186 J/g-°C)(63.4°C - 25°C)
q = 8905.12896 J
If 8905.12896 J was gained by the water, then 8905.12896 J must have been lost from the metal.
You know that there are 23 g of the metal and that its initial temp. is 100°C.
Plug this information into q=mcΔT.
8905.12896 J = (23 g)(C)(63.4°C - 100°C)
C = 10.58 J/g-°C
*When you plug all of this into the calculator, it will result in a negative number but keep in mind that heat was LOST by the metal so 8905.12896 J is essentially negative. So the negative cancels out.*
