Question

An athlete is given 100 g of glucose $\left(C_6{H}_{12}{O}_6\right)$ of energy equivalent to 1560 kJ. He utilises 50 percent of this gained energy in the event. In order to avoid storage of energy in the body, the weight of water he would need to perspire is (The enthalpy of evaporation of water is 44 kJ/mol.)

• A. 319 g
• B. 422 g
• C. 293 g
• D. 378 g

Answer 1

The correct option is A 319 g

Given energy to the athlete = 1560 kJ

Now, as given he ustilised 50% of that energy.so
Utilised energy $=\frac{1560\times 50}{100}=780kJ$

Now remaining energy will be used for the evaporation of water in form of perspiration.
So, energy to be used for perspiration= 1560 kJ - 780 kJ= 780 kJ

44 kJ energy utilised for evaporation of one mole of water $(i.e.18gof{H}_{2}O)$

$\therefore$ 780 kJ energy utilised for evaporation of $H_{2}O=\frac{780}{44}\times 18=319g$