Question

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

• A. $319g$
• B. $638g$
• C. $14040g$
• D. $35.45g$

Answer 1

Answer: (A)

$319g$

100 g of glucose $=1560kJ$

Energy utilized in body $=\frac{50}{100}\times 1560=780KJ$

Energy left un-utilized in body $=1560–780=780kJ$

Energy to be given out $=1560–780=780kJ$

Enthalpy of evaporation of water $=44kJ/mole=\frac{44kJ}{18g}$ of water [1 mole H2O = 18g water]

Hence amount water to be perspired to avoid storage of energy $=\frac{18}{44}\times 780=319.1g$

Hence option A is correct.