In: Chemistry
How much energy, in kiloJoules, would be required to raise the temperature of 1.00 kilogram of liquid benzene (C6H6) from 20.0 dC to its boling point of 80.1dC and then to completely vaporize the benzene at this same temperature? The specific heat capacity of benzene is 1.74J g-1 0C-1. The molar enthalpy of vaporization of benzene is 1.929 kJ/mole.
1.00 kg = 1000 gm
Temperature change of water 200C to 80.10C T = 80.1 - 20 = 60.10C
specific heat capacity of benzene is 1.74J /g 0C
q = mass of benzene specific heat of benzene T
= 1000 1.74 60.1
q = 104574 J = 104.574 KJ
104.574 KJ energy would be required to raise the temperature of 1.00 kilogram of liquid benzene (C6H6) from 20.0 0C to its boling point of 80.10C.
molar mass of benzene = 78.11 gm/mol that mean 1 mole of benzene = 78.11 gm then 1000 gm of benzene =
1000/78.11 = 12.80 mol
1 mole of bezene require energy 1.929 KJ for vaporization then 12.80 mole require = 12.80 1.929 = 24.6912 KJ
completely vaporize the benzene at this same temperature require energy = 24.6912 KJ