Question

A 61.93 gram sample of iron (with a specific heat of 0.450 J/g °C) is heated to 100.0 °C. It is then transferred to a coffee cup calorimeter containing 40.6 g of water (specific heat of 4.184 J/ g °C) initally at 20.63 °C. If the final temperature of the system is 23.59, what was the heat absorbed (q) of the calorimeter? (total heat absorbed by the water and calorimeter = heat released by the iron)

Answer 1

- heat released by iron = heat absorbed by water + heat absorbed by calorimeter

Heat given by iron = mass x specific heat x difference in temeperature

= 61.93g x 0.450J/g.C x (23.59-100) C

= -2129.43 J

heat absorbed by water = mass x specific heat x difference in temeperature

= 40.6g x 4.184J/g.C x (23.59-20.63) C

=502.816 J

Substituting

- (-2129.43J ) = 502.816 + heat absorbed by calorimeter

Thus

heat absorbed by calorimeter = 1626.61 J