a 32.0 gram sample of a metal 85.0°C was added to a 65.0 g of water at 15°C. The final temperature of the metal water mixture was 27.0°C. Calculate the specific heat capacity of the metal using the data.

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.

Given,

For Metal sample,

mass = 32 grams

T = 85°C

For Water sample,

mass = 65 grams

T = 15°C.

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 15°C to 27°C and metal sample temperature changed from 85°C to 27°C.

Since, Specific heat of water = 4.184 J/g°C

Let Cp be the specific heat of the metal.

Substituting values,

(32)(85°C - 27°C)(Cp) = (65)(27°C - 15℃)(4.184)

By Solving,

Cp = 1.7583 J/g°C.

Therefore, specific heat of the metal sample is 1.7583 J/g°C.