Answer:
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
