Question

In: Chemistry

How much energy in kJ is required to fully vaporize (at temperature of boiling point =...

How much energy in kJ is required to fully vaporize (at temperature of boiling point = 2525C) a 270g sample of Al(s), initially at 25C?

(Enthalpy of vap = 300. kJ/mol, enthalpy of fusion = 11. kJ/mol)

(specific heat of aluminum for both solid and liquid form = 0.9 J/(K * g))

(temperature of melting point = 625C, temp of boiling point = 2525C)

Expert Solution

no of moles of Al = W/G.M.Wt

= 270/27 = 10moles

q1 = mCT

= 270*0.9*(625-25)

= 145800J = 145.800KJ

q2 = no of moles*enthalpy of fusion

=10*11 = 110Kj

q3 = mCT

= 270*0.9*(2525-625)

= 461700J = 461.7KJ

q4 = no of moles * Enthalpy of vap

= 10*300 = 3000KJ

Total energy = q1+q2+q3+q4

= 145.8+110+461.7+3000 = 3717.5KJ

queen_honey_blossom answered 1 year ago

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