Question

The complete combustion of salicylic acid releases 21.90 kJ of energy per gram of salicylic acid. In a particular bomb calorimeter (initially at room temperature), the combustion of 0.1274 g of salicylic acid, in the presence of excess oxygen, causes the temperature of the calorimeter to rise by 2.76 °C. When a 0.2112-g sample of an unknown organic substance is similarly burned in the same calorimeter, the temperature rises by 4.24 °C. What is the energy of combustion per unit mass of the unknown substance?

Answer 1

Answer:Energy of combustion of unknown substance per gram is 13.27KJ/g

Solution:

1) According to question we will first calculate heat capacity of the bomb calorimeter,'C':

Energy of combustion salicylic acid per gram = 21.90 KJ/g = 21900 J/g  (1KJ=1000J)

Energy of combustion 0.1274 g of salicylic acid= (21900J/g)(0.1274 g)=2790.06 J

$\Delta T=2.76^oC=275.91 K(273.15 K=0^o Celcius)$

$q_{salicylic}=C\Delta T_{salicylic}$

$C=\frac{q_{salicylic}}{\Delta T_{salicylic}}=\frac{2790.60}{275.91K}=10.11J/k$

C= 10.11J/K

2) unknown substance of mass of 0.2112 g was burned in same calorimeter.

C= 10.11J/K (calculated above),$\Delta T_{unknown}=4.24^oC=277.4K$

$q_{unknown}=C\times \Delta T_{unknown}$

$q_{unknown}=10.11\times 277.39 K= 2804.41 J$

Energy of combustion of unknown substance per gram= $\frac{2804.41 J}{0.2112g}=13278.47 J/g$=13.27KJ/g