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nydimaria [60]
3 years ago
5

In a distillation operation, the heat added to the boiler a) increases with increase in reflux ratio b) decreases with increase

in reflux ratio c) remains unaltered with change in reflux ratio d) None of the above
Chemistry
1 answer:
Leni [432]3 years ago
7 0

Answer:

a) increases with increase in reflux ratio

Explanation:

If we increase the reflux ratio, which can de defined as : L/D = Condesate back to column /distillate discarded from column, it means that we increase the amount of condensate back to column. So the reboiler load will increase and be maximum in case of Total reflux ratio. And as vapor load will be increased, diameter of column will be increased too.  

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A 23.5g aluminum block is warmed to 65.9°C and plunged into an insulated beaker containing 55.0g water initially at 22.3°C. The
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Answer:

25.97oC

Explanation:

Heat lost by aluminum = heat gained by water

M(Al) x C(Al) x [ Temp(Al) – Temp(Al+H2O) ] = M(H2O) x C(H2O) x [ Temp(Al+H2O) – Temp(H2O) ]

Where M(Al) = 23.5g, C(Al) = specific heat capacity of aluminum = 0.900J/goC, Temp(Al) = 65.9oC, Temp(Al+H2O)= temperature of water and aluminum at equilibrium = ?, M(H2O) = 55.0g, C(H2O)= specific heat capacity of liquid water = 4.186J/goC

Let Temp(Al+H2O) = X

23.5 x 0.900 x (65.9-X) = 55.0 x 4.186 x (X-22.3)

21.15(65.9-X) = 230.23(X-22.3)

1393.785 - 21.15X = 230.23X – 5134.129

230.23X + 21.15X = 1393.785 + 5134.129

251.38X = 6527.909

X = 6527.909/251.38

X = 25.97oC

So, the final temperature of the water and aluminum is = 25.97oC

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