Question

A mass of 43.0 g of an unknown solid initially at 130.0 ∘C is added to an ideal constant pressure calorimeter containing 100.0 g of water (Cs,water=4.184J/(g⋅∘C)) initially at 20.0 ∘C. After the mixture reaches thermal equilibrium, the final temperature is recorded to be 36.49 ∘C. What is the specific heat capacity of the unknown solid?

Express your answer to three significant figures.

Answer 1

Since the temperature of water in the calorimeter increases so heat must be absorbed by the liberated heat by the solid

Heat lost by solid = Heat gained by water

mcdt = m'c'dt'

Where

m = mass of solid = 43.0 g

c = specific heat capacity of solid = ?

dt = change in temperature of solid = 130.0 - 36.49 oC = 93.51 oC

m' = mass of water = 100.0 g

c' = specific heat capacity of water = 4.184 J/g-oC

dt' = change in temperature of water = 36.49 - 20.0 oC = 16.49 oC

Plug the values we get c = (m'c'dt')/(mdt)

                                    = 1.716 J/g-oC