Question

A 32.1 g iron rod, initially at 22.8 ∘C, is submerged into an unknown mass of water at 64.0 ∘C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 58.8 ∘C. What is the mass of the water? Express your answer to two significant figures and include the appropriate units.

Answer 1

Amount of heat lost by water = heat gained by iron

mcdt = m'c'dt'

Where

m = mass of water = ?

c = specific heat capacity of water = 4.186 J/goC

dt = change in temperature of water = initial - final = 64.0 - 58.8 = 5.2 oC

m' = mass of iron = 32.1 g

c' = specific heat capacity of iron = 0.450 J/goC

dt' = change in temperature of iron = final - initial = 58.8 - 22.8 = 36.0 oC

Plug the values we get

m = ( m'c'dt') / ( cdt)

= 23.9 g

Therefore the mass of water required is 23.9 g