What is the enthalpy of pure, liquid aluminum at 1000 K? (Use pure, solid alumin

Question

What is the enthalpy of pure, liquid aluminum at 1000 K? (Use pure, solid aluminum at 298 K as the reference state.) An electric resistance furnace is used to melt pure aluminum at the rate of 100kgh-1. The furnace is fed with solid aluminum at 298 K. The liquid aluminum leaves the furnace at 1000 K. What is the minimum electric power of the furnace (kW)? Data for aluminum: atomic weight 27g mol-1. heat capacity of solid (Cp,mol 26 Jol1K-1; heat capacity of liquid (Cp.mol) 29Jmol K- melting point Tm-932 K: heat of fusion AHm) = 10 700 J mol-

Explanation / Answer

Moles of Aluminum = mass/molecular weight

= (100 kg/h x 1000 g/kg) / (27g/mol)

= 3703.704 mol/h

Enthalpy required to heat the solid aluminum from 298 K to 932 K

H1 = moles x Cps x (T2-T1)

= 3703.704 mol/h x 26 J/mol·K x (932 - 298)K

= 61051851.85 J/h

Phase change at 932 K from solid to liquid

H2 = moles x heat of fusion

= 3703.704 mol/h x 10700 J/mol

= 39629629.63 J/h

Enthalpy required to heat the liquid aluminum from 932 K to 1000 K (pure liquid)

H3 = moles x Cpl x (T2-T1)

= 3703.704 mol/h x 29 J/mol·K x (1000 - 932)K

= 7303704.288 J/h

Enthalpy of pure liquid at 1000 K

H3 = 7303704.288 J/h x 1h/3600s x 1kJ/1000 J

= 2.03 kJ/s

Total enthalpy required = H1 + H2 + H3

= 61051851.85 + 39629629.63 + 7303704.288

= 107985185.768 J/h x 1h/3600s x 1kJ/1000 J

= 29.995 kW

Minimum electric power required = 29.995 kW or 30 kW

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