Question

300.0 g of copper is heated to 100.0*C and transferred quickly to a calorimeter containing 400.0 grams of water initially at 25.0*C. If the final temperature is 29.4*C, calculate the specific heat of copper. The specific heat of water is 4.18 J/g-*C. What assumptions must be made about the calorimeter? How is the first law of thermodynamics and law of conservation of energy used in this experiment.

Answer 1

Assumption is that the calorimeter could not absorbs any heat

The heat lost by copper = heat absorbed by water in the calorimeter

     mcdt = m'c'dt'

Where

m = mass of copper = 300.0 g

c = specific heat capacity of copper = ?

dt = change in temperature of copper = 100.0 - 29.4 oC = 70.6 oC

m' = mass of water = 400.0 g

c' = specific heat capacity of water = 4.186 J/(goC)

dt' = change in temperature of water in the calorimeter = 29.4 - 25.0 oC = 4.4 oC

Plug the values we get

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

    = 0.348 J/(goC)

Therefore the specific heat capacity of copper is 0.348 J/(goC)