The enthalpy of combustion of glucose is -2808 kJ/mol at 25degree C .How many grams of glucose do you need to consume(assume wt=62.5kg). (a) to climb a flight of stairs rising through 3M. (b) to climb a mountain of altitude 3000M? Assume that 25% of enthalpy can be converted to useful work.
Asked by m.nilu
23rd September 2018,
11:23 PM
Answered by Expert
Answer:
As given, 1 mole of glucose on combustion releases = -2808 kJ/mole at 25°C.
As we have asked, only 25% of enthalpy can be converted into work,
hence, For 1 mole of glucose = 
The molecular mass of sugar is 342 g hence it will provide -702 kJ energy to performed work.
As we know that work done is equal to the change potential energy
Work done = P. E. = mgh
Where m = mass, g = gravitational acceleration and h = height
weight = mg = 62.5 Kg
Now, let us calculate the energy consumed,
Case 1.: for h = 3 m
Work done = mgh = 62.5 X 3 = 187500 J = 187.5 kJ
Since on combustion of 342 gm of glucose 702 kJ energy released
Therefore, 187.5 kJ of energy released by = 
Case 2: For h= 3000 m
Work done = mgh = 62.5 X 3000 = 187.5 J = 0.187 kJ
Since on combustion of 342 gm of glucose 702 kJ energy released
Therefore, 0.187 kJ of energy released by = 
Answered by Expert
19th November 2018,
4:34 PM
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