A 23.9 g sample of iridium is heated to 89.7 oC, and then dropped into 20.0 g of water in a coffee-cup calorimeter. The temperature of the water went from 20.1 oC to 22.6 oC. Calculate the specific heat of iridium. (specific heat of water = 4.18 J/g oC).

The specific heat capacity of iridium = 0.130 J/g°C

Explanation:

Assuming no heat losses to the environment and to the calorimeter,

Heat lost by the iridium sample = Heat gained by water

Heat lost by the iridium sample = mC ΔT

m = mass of iridium = 23.9 g

C = specific heat capacity of the iridium = ?

ΔT = change in temperature of the iridium = 89.7 - 22.6 = 67.1°C

Heat lost by the iridium sample = (23.9)(C)(67.1) = (1603.69 C) J

Heat gained by water = mC ΔT

m = mass of water = 20.0 g

C = 4.18 J/g°C

ΔT = 22.6 - 20.1 = 2.5°C

Heat gained by water = 20 × 4.18 × 2.5 = 209 J

Heat lost by the iridium sample = Heat gained by water

1603.69C = 209

C = (209/1603.69) = 0.130 J/g°C