An athelete is given 100 g glucose of energy equivalenet to 1560 kJ. He utilizes 50% of this gained energy in an event. In order to avoid storage of energy in body, calculate the weight of water he need to perspire. Enthalpy of `H_(2)O` for evaporation is 44 kJ `"mole"^(-1)`.

Explanation

Amount of energy given to athlete = 1560 kJ
Energy lost in the event `= (1560 xx50)/(100) = 780 kJ`
Energy stored in the body = 1560 - 780 = 780 kJ
For evaporation of water,
`H_(2)O(l) to H_(2)O(g) DeltaH = 44 kJ "mol"^(-1)`
For consumption of 44 kJ of energy, water evaporated = 18 g
For consumption of 780 kJ of energy, water evaporated `= 18/44 xx 780 = 319.1g`.