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Chemical Thermodynamics: Chemistry | JEE Main

An athlete is given

100\mathrm{~g}

of glucose (

\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6

) for energy. This is equivalent to

1800\mathrm{~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

\underline{\quad\quad\quad}\mathrm{g}

(Nearest integer) Assume that there is no other way of consuming stored energy. Given : The enthalpy of evaporation of water is

45\mathrm{~kJ}\mathrm{~mol}^{-1}

Molar mass of

\mathrm{C}, \mathrm{H}\&\mathrm{O}

are

12, 1

and

16\mathrm{~g}\mathrm{~mol}^{-1}

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Correct answer

The athlete must perspire 360 g of water to dissipate the remaining 50% of unused energy.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Energy from $100\text{ g}$ glucose $= 1800\text{ kJ}$ . Fraction utilized in sports $= 50\%$ . Remaining energy to be dissipated $= 50\%$ . $\Delta H_{\text{vap}}(\text{H}_2\text{O}) = 45\text{ kJ mol}^{-1}$ . Molar masses: $\text{C}=12, \text{H}=1, \text{O}=16\text{ g mol}^{-1}$ .

02find

Find the mass of water (in g) that must evaporate as perspiration to dissipate the unutilized energy.

03strategise

Calculate the unutilized energy: $E_{\text{excess}} = 1800 \times 0.50 = 900\text{ kJ}$ . Then find moles of water needed to absorb this heat via evaporation: $n_{\text{H}_2\text{O}} = \frac{E_{\text{excess}}}{\Delta H_{\text{vap}}}$ . Finally, convert moles of water to mass: $m = n_{\text{H}_2\text{O}} \times M_{\text{H}_2\text{O}}$ , where $M_{\text{H}_2\text{O}} = 2(1) + 16 = 18\text{ g mol}^{-1}$ .

04execute

Molar mass of water $= 18\text{ g mol}^{-1}$ . Unutilized energy $= 1800 \times 0.5 = 900\text{ kJ}$ . Moles of water evaporated $= \frac{900}{45} = 20\text{ mol}$ . Mass of water evaporated $= 20 \times 18 = 360\text{ g}$ .

✓verify

Check dimensions: $(\text{kJ}) / (\text{kJ mol}^{-1}) \times (\text{g mol}^{-1}) = \text{g}$ . Perspiring $360\text{ g}$ of water during an intense athletic event is physiologically sensible.

Hints that build this answer step by step

How much energy must be dissipated through perspiration to avoid storage?
900 kJ
How many moles of water need to evaporate to dissipate 900 kJ, given an enthalpy of evaporation of 45 kJ/mol?
20 mol
What is the mass of 20 mol of water (H2O)?
360 g

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why was the mass of glucose (100 g) given if we did not use its molar mass (180 g/mol)?

The problem directly specifies that 'This [100 g] is equivalent to 1800 kJ of energy'. The 100 g value is redundant context; the available thermal energy is directly 1800 kJ.

Answer

The athlete must perspire 360 g of water to dissipate the remaining 50% of unused energy.

Why each option works or fails

Step-by-step solution

given: Energy from $100\text{ g}$ glucose $= 1800\text{ kJ}$ . Fraction utilized in sports $= 50\%$ . Remaining energy to be dissipated $= 50\%$ . $\Delta H_{\text{vap}}(\text{H}_2\text{O}) = 45\text{ kJ mol}^{-1}$ . Molar masses: $\text{C}=12, \text{H}=1, \text{O}=16\text{ g mol}^{-1}$ .
find: Find the mass of water (in g) that must evaporate as perspiration to dissipate the unutilized energy.
strategise: Calculate the unutilized energy: $E_{\text{excess}} = 1800 \times 0.50 = 900\text{ kJ}$ . Then find moles of water needed to absorb this heat via evaporation: $n_{\text{H}_2\text{O}} = \frac{E_{\text{excess}}}{\Delta H_{\text{vap}}}$ . Finally, convert moles of water to mass: $m = n_{\text{H}_2\text{O}} \times M_{\text{H}_2\text{O}}$ , where $M_{\text{H}_2\text{O}} = 2(1) + 16 = 18\text{ g mol}^{-1}$ .
execute: Molar mass of water $= 18\text{ g mol}^{-1}$ . Unutilized energy $= 1800 \times 0.5 = 900\text{ kJ}$ . Moles of water evaporated $= \frac{900}{45} = 20\text{ mol}$ . Mass of water evaporated $= 20 \times 18 = 360\text{ g}$ .
verify: Check dimensions: $(\text{kJ}) / (\text{kJ mol}^{-1}) \times (\text{g mol}^{-1}) = \text{g}$ . Perspiring $E_{\text{excess}} = 1800 \times 0.50 = 900\text{ kJ}$ of water during an intense athletic event is physiologically sensible.