Question

A 2.32 g sample of a substance suspected of being pure gold is warmed to 71.9 ∘C and submerged into 15.9 g of water initially at 24.9 ∘C. The final temperature of the mixture is 26.7 ∘C.

What is the heat capacity of the unknown substance?

Answer 1

The heat lost by the unknown metal = heat gained by the water

The amount of heat lost by the metal = mass * specific heat capacity of metal * delta T

Given mass of the metal = 2.32 g

deltaT = [71.9 ∘C - 26.7 ∘C] = 45.2 ∘C

Let the specific heat capacity of metal = Cp

The amount of heat lost by the metal = mass * specific heat capacity of metal * delta T

                                                           = 2.32 g* Cp * 45.2 ∘C

Given mass of water = 15.9 g

the specific heat capacity of water = 4.184 J/∘C. g

deltaT = [26.7 ∘C- 24.9∘C] = 1.8 ∘C

Heat gained by the water = mass of water * specific heat capacity of water * delta T

                                          = 15.9 g * 4.184 J/∘C. g * 1.8 ∘C

Hence : The heat lost by the unknown metal = heat gained by the water

2.32 g* Cp * 45.2 ∘C = 15.9 g * 4.184 J/∘C. g * 1.8 ∘C

therefore Cp = [15.9 g * 4.184 J/∘C. g * 1.8 ∘C]/[2.32 g * 45.2 ∘C ]

                   = 1.14J/∘C. g