How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 k to liquify the sample and lower the temperature to 335.0 k? the following physical data may be useful.

δhvap = 33.9 kj/mol
δhfus = 9.8 kj/mol
cliq = 1.73 j/g°c
cgas = 1.06 j/g°c
csol = 1.51 j/g°c
tmelting = 279.0 k
tboiling = 353.0 k

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

The expression used will be:

where,

= heat released by the reaction = ?

m = mass of benzene = 125 g

= specific heat of gaseous benzene =

= specific heat of liquid benzene =

= enthalpy change for vaporization =

Molar mass of benzene = 78.11 g/mole

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

Therefore, the energy removed must be, -67.7 kJ

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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.