A piece of metal of mass 23 g at 100 ◦C is placed in a calorimeter containing 55.4 g of
water at 25◦C. The final temperature of the
mixture is 63.4
◦C. What is the specific heat
capacity of the metal? Assume that there is
no energy lost to the surroundings.
Answer in units of J
g ·
◦ C
.

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.*

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