Question

Constant-volume calorimeters are sometimes calibrated by running a combustion reaction of known ΔE and measuring the change in temperature. For example, the combustion energy of glucose is 15.57 kJ/g. When a 2.000 g sample of glucose burns in a constant volume calorimeter, the calorimeter temperature increases from 21.45 to 23.34°C. Find the total heat capacity of the calorimeter (in kJ/K).

Answer 1

Ans. Total amount of heat released from combustion of 2.000-gram glucose =

                        2 g x combustion energy of 1 glucose molecule

                        = 2 g x 15.57 kJ/ g

                        = 31.14 kJ

In calorimeter, the heat produced during combustion of sample (here, glucose) is absorbed by the calorimeter. As a result, the temperature of calorimeter system increases.

Now,

Heat gained by calorimeter system, q = c x dT     

     Where, c= heat capacity of calorimeter system

                     dT = increase in temperature of the system

                Or, 31.14 kJ = c x [(273.15+ 23.34 K) – (273.15 +21.45 K)] = c x 1.89 K                        

                Or, c = 31.14 J / 1.89 K = 16.476 kJ K^-1

Thus, heat capacity of the calorimeter = 16.476 kJ K^-1