Join WhatsApp Icon JEE WhatsApp Group
Question 60

An athlete is given 100 g of glucose (C$$_6$$H$$_{12}$$O$$_6$$) for energy. This is equivalent to 1800 kJ of energy. The 50% of this energy gained is utilized by the athlete for sports activities at the event. In order to avoid storage of energy, the weight of extra water he would need to perspire is _____ g (Nearest integer) Assume that there is no other way of consuming stored energy.
Given: The enthalpy of evaporation of water is 45 kJ mol$$^{-1}$$
Molar mass of C, H & O are 12.1 and 16 g mol$$^{-1}$$.


Correct Answer: 360

Given: 100 g of glucose provides 1800 kJ of energy. 50% is used for sports, so 50% (= 900 kJ) must be dissipated through perspiration.

Enthalpy of evaporation of water = 45 kJ/mol.

Moles of water to evaporate = $$\frac{900}{45} = 20$$ mol

Mass of water = $$20 \times 18 = 360$$ g

The answer is $$360$$ g.

Get AI Help

Create a FREE account and get:

  • Free JEE Mains Previous Papers PDF
  • Take JEE Mains paper tests

50,000+ JEE Students Trusted Our Score Calculator

Predict your JEE Main percentile, rank & performance in seconds

Ask AI

Ask our AI anything

AI can make mistakes. Please verify important information.