Answer:
The specific heat of the metal is 10.93 J/g°C.
Explanation:
Given,
For Metal sample,
mass = 6.98 grams
T = 91.29°C
For Water sample,
volume = 114.84 mL
T = 24.37°C.
Final temperature of mixture = 33.54°C.
When the metal sample and water sample are mixed,
The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.
Since, heat lost by metal is equal to the heat gained by water,
Qlost = Qgain
However,
Q = (mass) (ΔT) (Cp)
(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)
After mixing both samples, their temperature changes to 27°C.
It implies that
water sample temperature changed from 24.37°C to 33.54°C and metal sample temperature changed from 91.29°C to 33.54°C.
We have all values, but, here mass of water is not given. It can be found by using the formula
Density = Mass/Volume
Since, density of water = 1 g/mL
we get, Mass = 114.84 grams.
Since specific heat of water is 4.184 J/g°C.
Now substituting all values in
(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)
(6.98)(91.29 - 33.54)(Cp) = (114.84)(33.54 - 24.37)(4.184)
solving, we get,
Cp = 10.93 J/g°C.
the specific heat of the metal is 10.93 J/g°C.
