Bomb Calorimeter, gross heat of combustion

In the following experiment 1.1651g of benzoic acid (6318 cal/gm) produced a net corrected temperature rise of 3.047C, also 8 cm (2.3 cal/cm) of fuse wire were consumed in the firing. How does one go about to compute the gross heat of combustion, in cal/gram. Then using this standardization how does one find the gross heat of combustion of this following unknown, in which 1.354g and 7 cm of fuse wire, with a temperature rise of 2.987C were consumed in the firing.

Solution:
With benzoic acid:
Total heat released = heat released by benzoic acid + heat released by wire
= 1.1651 g × 6318 cal/g + 8 cm × 2.3 cal/cm
= 7379.5018 cal

Heat capacity of the bomb = 7379.5018 cal ÷ 3.047 oC
= 2421.89 cal/oC

With the unknown:
Total heat released by the unknown and the wire = 2421.89 cal/oC × 2.987 oC
= 7234.188 cal

Heat released by the unknown = total heat released - heat released by the wire
= 7234.188 cal - 7 cm × 2.3 cal/cm
= 7218.088 cal

Gross heat of combustion of unknown
= 7218.088 cal ÷ 1.354 g
= 5330.94 cal/g

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