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Thermodynamics question

2023 · 25 Jan · Shift 1 · Q19
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  5. /2023 · 25 Jan · Shift 1 · Q19

Thermodynamics question

2023 · 25 Jan · Shift 1 · Q19

JEE MainChemistryThermodynamicsNumerical+4 / −1
An athlete is given 100 g of glucose (C 6_66​ H 12_{12}12​ O 6_66​) for energy. This is equivalent to 1800kJ 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{\hspace{2cm}}​ 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}−1 Molar mass of C, H & O are 12, 1 and 16 g mol −1^{-1}−1.
Numerical answer
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Correct answer: 360

  1. Total energy obtained from glucose

    Given, 100 g100\,\text{g}100g glucose provides 1800 kJ1800\,\text{kJ}1800kJ energy.

  2. Energy utilized for sports activity

    Only 50%50\%50% of this energy is used: Eused=0.5×1800=900 kJE_{\text{used}} = 0.5 \times 1800 = 900\,\text{kJ}Eused​=0.5×1800=900kJ

  3. Excess energy to be removed

    The remaining 50%50\%50% is not used for work, so to avoid storage it must be lost via perspiration: Eexcess=1800−900=900 kJE_{\text{excess}} = 1800 - 900 = 900\,\text{kJ}Eexcess​=1800−900=900kJ

  4. Use latent heat of evaporation of water

    Enthalpy of evaporation of water = 45 kJ mol−145\,\text{kJ mol}^{-1}45kJ mol−1

    Moles of water needed to evaporate to remove 900 kJ900\,\text{kJ}900kJ: n=90045=20 moln = \frac{900}{45} = 20\,\text{mol}n=45900​=20mol

  5. Convert moles of water to mass

    Molar mass of water: M(H2O)=2(1)+16=18 g mol−1M(\text{H}_2\text{O}) = 2(1) + 16 = 18\,\text{g mol}^{-1}M(H2​O)=2(1)+16=18g mol−1

    Therefore, mass of water evaporated: m=20×18=360 gm = 20 \times 18 = 360\,\text{g}m=20×18=360g

  6. Final Answer

    The extra water he must perspire is: 360 g\boxed{360\,\text{g}}360g​

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