Question

A 62.5 g of iron with a heat capacity of .450 J/C is heated to 100 degree C. It is then transferred to a coffee cup calorimeter containing 52.7 g of water (specific heat of 4.184J/g C) initially at 20.63C. if the final temp of the system is 29.59 what was the heat capacity of the calorimeter?

Answer 1

now

heat lost by iron = mass x specific heat capacity x temp change

so

heat lost by iron = 62.5 x 0.45 x ( 100 - 29.59)

heat lost by iron = 1980.28 J

now

heat lost by iron = heat gained by water + calorimeter

heat gained by water = m x s x dT

so

heat gained by water = 52.7 x 4.184 x ( 29.59 - 20.63)

heat gained by water = 1975.65 J

now

heat gained by calorimter = 1980.28 - 1975.65

heat gained by calorimeter = 4.63

now

heat gained by calorimeter = heat capacity x temp change

so

4.63 = heat capacity x ( 29.59 - 20.63)

heat capacity = 0.5167 J / C

so

the heat capacity of the calorimeter is 0.5167 J / C