Logo
Jeeify
NVAEasyJEE 2023Internal Energy & Enthalpy

JEE Chemistry 2023 Question with Solution

An athlete is given 100 g100 \, \text{g} of glucose (C6H12O6\mathrm{C_6H_{12}O_6}) for energy, which is equivalent to 1800 kJ1800 \, \text{kJ} of energy. If 50%50\% of this energy is utilized for activities, the weight of extra water needed to perspire is _____ g\text{g}. (Nearest integer)

Given: Enthalpy of evaporation of water = 45 kJ/mol45 \, \text{kJ/mol}; molar masses: C=12  g/mol\mathrm{C} = 12 \; \text{g/mol}, H=1  g/mol\mathrm{H} = 1 \; \text{g/mol}, O=16  g/mol\mathrm{O} = 16 \; \text{g/mol}.

Answer

Correct answer:360

Step-by-step solution

Standard Method

Given: Total energy from glucose is 1800 kJ1800 \, \text{kJ} and 50%50\% of this energy is utilized for activities. Enthalpy of evaporation of water is 45 kJ/mol45 \, \text{kJ/mol}.

Find: The mass of extra water needed to perspire.

The solution states that the correct answer is 360360.

Energy to be dissipated through perspiration:

50100×1800=900 kJ\frac{50}{100} \times 1800 = 900 \, \text{kJ}

Now use enthalpy of vaporization of water:

900=nH2O×45900 = n_{\mathrm{H_2O}} \times 45 nH2O=90045=20  moln_{\mathrm{H_2O}} = \frac{900}{45} = 20 \; \text{mol}

Mass of water required:

WH2O=20×18=360 gW_{\mathrm{H_2O}} = 20 \times 18 = 360 \, \text{g}

Therefore, the extra water needed to perspire is 360 g360 \, \text{g}.

Energy Dissipation Approach

Given: Half of the available energy must be dissipated, and ΔHvap=45 kJ/mol\Delta H_{\text{vap}} = 45 \, \text{kJ/mol}.

Find: The corresponding mass of water evaporated.

The extracted working gives:

  1. Energy to be dissipated:
Energy=50100×1800=900 kJ\text{Energy} = \frac{50}{100} \times 1800 = 900 \, \text{kJ}
  1. Moles of water required:
Moles of water=EnergyΔHvap=90045=20 mol\text{Moles of water} = \frac{\text{Energy}}{\Delta H_{\text{vap}}} = \frac{900}{45} = 20 \, \text{mol}
  1. Mass of water required:
Mass=Moles×Molar mass of water=20×18=360 g\text{Mass} = \text{Moles} \times \text{Molar mass of water} = 20 \times 18 = 360 \, \text{g}

Hence, the required extra water is 360 g360 \, \text{g}.

Common mistakes

  • Using the full 1800 kJ1800 \, \text{kJ} instead of only 50%50\% of it. This is wrong because only the remaining energy must be dissipated through perspiration. First calculate 900 kJ900 \, \text{kJ}, then proceed.

  • Treating 45 kJ/mol45 \, \text{kJ/mol} as energy for 1 g1 \, \text{g} of water instead of 1 mol1 \, \text{mol} of water. This gives an incorrect mass directly. Convert energy to moles first, then moles to grams using molar mass.

  • Using the given atomic molar masses to recompute glucose-related moles unnecessarily. This is wrong because the question already gives the total usable energy as 1800 kJ1800 \, \text{kJ} for 100 g100 \, \text{g} glucose. The calculation only needs the fraction of energy dissipated and the vaporization enthalpy of water.

Practice more Internal Energy & Enthalpy questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Internal Energy & Enthalpy

Related questions