A quantity of `1.534 g` of naphthanlene `(C_(10) H_(g))` is burned in constant-volume bomb calorimeter. Consequently, the temperature of the water rises form `20.00^(@)C` to `25.00^(@)C`. If the quanity of water surrounding the calorimeter is exactly `3000g` calculate the heat capcity of combustion of one mole of naphthalene (molar heat of combustion)
Strategy : First calculate the heat changes for the water and the bomb calroimeter. using Finally, divide the value by the number of moles of naphthlene to calculate the molar heat of combusiton. Remember ot change `2.75 kJ^(@)C^(-1)` to `2.75 xx 1000 j^(@)C C^(-1)`

Explanation

`q_(water) = m_(water) C_(water) Delta V`
`= (3000 g) (4.184 J g^(-1) 1^(@)C^(-1)) (25^(@)C - 20^(@)C)`
`= 6.28 xx 10^(4) J`
`q_(bomb) = C_(bomb) Delta V`
`= (2.75 xx 1000 J^(@)C^(-1)) (25^(@)C - 20^(@)C)`
`= 1.38 xx 10^(4) J`
From Eq., we write
`q_(reaction) = - (6.28 xx 10^(4) J + 13.8 xx 10^(4) J)`
` = - 7.66 xx 10^(4) J`