Question

An athelete 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, Determine the weight of water he would need to perspire. (The enthalpy of evaporation of water is 44 kJ/mole.)

• A. 319 gm
• B. 323 gm
• C. 342 gm
• D. 312

Answer 1

The correct option is A 319 gm

Net amount of energy given to athlete $=1560kJ$

$1560\times 50/100$ Energy lost in an event $=780kJ$

Energy left out $=1560–780kJ=780kJ$

Now, consider the evaporation of water $H_{2}O\left(l\right)\to H_{2}O\left(g\right)$; $\Delta H=44kJmol{e}^{–1}$

Thus, for consumption of $44kJ$ of energy the amount of water evaporated $=1mole=18g$

For consumption of $780kJ$ of energy the amount of water to be evaporated $\frac{18\times 780}{44}=319\cdot 09g$