Question

metal weighing 6.98 g was heated to 91.29 °C and then put it into 114.84 mL of water (initially at 24.37 °C). The metal and water were allowed to come to an equilibrium temperature, determined to be 33.54 °C. Assuming no heat lost to the environment, calculate the specific heat of the metal.

Answer 1

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.