A bomb calorimeter has a heat capacity of 675 J/°C and contains 925 g of water. If the combustion of 0.500 mole of a hydrocarbon increases the temperature of the calorimeter from 24.26°C to 53.88°C, determine the enthalpy change per mole of hydrocarbon.

Answer : The enthalpy change per mole of hydrocarbon is, 269 kJ/mole

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

where,

q = heat released by the reaction

= heat absorbed by the calorimeter

= heat absorbed by the water

= specific heat of calorimeter =

= specific heat of water =

= mass of water = 925 g

= change in temperature =

Now put all the given values in the above formula, we get:

Now we have to calculate the enthalpy change per mole of hydrocarbon.

where,

= enthalpy change = ?

q = heat released = 134.5 kJ

n = moles of hydrocarbon = 0.500 mol

Therefore, the enthalpy change per mole of hydrocarbon is, 269 kJ/mole

a boiling water reactor does not boil the water. a pressurized water reactor boils the water faster. a pressurized water reactor is heated directly by the core. a pressurized water reactor does not use fission.