$0.325\;g$ of benzoic acid are burnt by oxygen in a bomb calorimeter containing $250\;g$ water. The temperature rises by $1.48^oC$. Calculate the specific heat capacity of a calorimeter, knowing that the heat of combustion (= lost during the combustion of 1 mole) of benzoic acid $C_6H_5COOH$ equals $771.2\frac{kcal}{mole}$
The system: $0.325\;g$ benzoic acid and the oxygen The surroundings: The calorimeter and the water.
$Q$ $=$ $-4184\cdot m_{water}\cdot \Delta \theta$ $ -$ $ C \cdot \Delta \theta\;(1)$
$771.2kcal$ are $771.2\cdot 1000\cdot 4.184\;J$ $Q$ $=$ $-771.2\cdot 1000\cdot 4.184\cdot n_{acid}$ $=$ $-\frac{771,2\cdot 1000\cdot 4,184\cdot 0,325}{7\cdot 12+6\cdot 1+2\cdot 16 }$ $\approx$ $ -8593\;J$ $Q$ = heat received, therefore the - sign!
$(1)$ $-8593$ $=$ $-4184\cdot 0.25\cdot 1.48$ $ - $ $C \cdot 1.48$ $C$ $=$ $4760 \frac{J}{K}$