JEE Main 2023 · 25 Jan · Shift 1 · numerical answer

JEE Main 2023 Chemistry question on Chemical Thermodynamics

An athlete is given 100 g of glucose (C6H12O6\text{C}_6\text{H}_{12}\text{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^{-1}

Molar mass of C, H & O are 12, 1 and 16 g mol−1^{-1}.
Show answer and solution

Answer: 360

Solution

The athlete uses 50% of 1800 kJ = 900 kJ for sports. The remaining 50% = 900 kJ must be dissipated via perspiration to avoid energy storage.

Moles of water needed to evaporate = 900 kJ ÷ 45 kJ/mol = 20 mol.

Mass of water = 20 × 18 g/mol = 360 g.

[JEE Main 2023 · 25 Jan · Shift 1]

Practise Chemical Thermodynamics the way it is asked

A timed set of real JEE Main Chemical Thermodynamics questions, graded as the exam grades. No sign-up.

Start a Chemical Thermodynamics test

More: all JEE Main Chemical Thermodynamics questions · JEE Main 2023 · Chemistry chapter list