Please help with the second question: For the above current at 3.29 V, how much

Question

Please help with the second question:

For the above current at 3.29 V, how much electrical energy (in kw-hr) is consumed in the production of 1 kg of aluminum? (The electrical energy or work =qV and 1 kW-hr is 3.6 x106 J.)

The electrolysis of molten Al203 at 980 °C is used to produce metallic aluminum. A current of 529 A is used in the electrolysis cell. What is the rate of formation of aluminum, in kilograms per hour? Answer: 0.178 For the above current at 3.29 V, how much electrical energy (in kw-hr) is consumed in the production of 1 kg of aluminum? (The electrical energy or work =q V and 1 kW- hr is 3.6 x106 J.)

Explanation / Answer

P = VI

if V = 0.178 and I = 529 A

then

P = 0.178 *529 = 94.162 J/s = 94.162 W

now..

find total electrical energy in Kw-Hr fo 1 kg of aluminium

1 kg = 1000 g

mol of Al+3 = mass/MW = 1000/26.98 = 37.06 mol

mol of e- per Al+3 = 37.06*3 = 111.18 mol of e-

charge:

1 mol of e- = 96500 C

111.18 mol of e- > X

X = 111.18*96500 = 10728870 C

time required = charge/current = 10728870/529 = 369961.0 seconds

Total Energy = Ework * time = 94.162 *369961.0 = 34836267.6 J

change to kW-h

3.6*10^6 J = 1 kwh

34836267.6 J = x

x = 34836267.6/(3.6*10^6) = 9.6767 KwH

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