Question

When 1.365 g of anthracene, C14H10, is combusted in a bomb calorimeter that has a water jacket containing 500.0 g of water, the temperature of the water increases by 25.89°C. Assuming that the specific heat of water is 4.18 J/(g ∙°C), and that the heat absorption by the calorimeter is negligible, estimate the enthalpy of combustion per mole of anthracene.

Answer 1

The enthalpy of combustion per mole of anthracene : 7064 kj/mol(- sign=exothermic)

Further explanation  

The law of conservation of energy can be applied to heat changes, i.e. the heat received/absorbed is the same as the heat released  

Q in = Q out  

Heat can be calculated using the formula:  

Q = mc∆T  

Heat released by anthracene= Heat absorbed by water

Heat absorbed by water =

$\tt Q=500\times 4.18\times 25.89=54110.1~J$

mol of  anthracene (MW=178,23 g/mol)

$\tt \dfrac{1.365}{178.23}=0.00766$

The enthalpy of combustion per mole of anthracene :

$\tt \Delta H=-\dfrac{Q}{n}=\dfrac{54110.1}{0.00766}=-7063981.7~J/mol\approx -7064~kJ/mol$