Question

of 27

A 13 g piece of metal at 73°C is placedi n 60 g of water at 22°C. The water temperature increases to 27°C. What is the specific heat of the metal? (the specific heat of water is

4.184 J/gºC)

A 2.10

B 1.89

C 0.476

D 49

The specific heat of the metal is 2.09899 J/g℃.

Explanation:

Given,

For Metal sample,

mass = 13 grams

T = 73°C

For Water sample,

mass = 60 grams

T = 22°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 22°C to 27°C and metal sample temperature changed from 73°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,

(13) (73°C - 27°C) (Cp) = (60) (27°C - 22℃) (4.184)

By solving, we get Cp =

Therefore, specific heat of the metal sample is 2.09899 J/g℃.