Question

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 0}C^-1. The molar enthalpy of vaporization of benzene is 1.929 kJ/mole.

Answer 1

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