An athlete is given 100 g of glucose of energy equivalent to 1560 kJ. He utilizes 50 % of this

step 1 In the question, we are given that an athlete is given 100 gram of glucose, 100 gram of glucose.

An athlete is given 100 g of glucose of energy equivalent to...

An athlete is given 100 g of glucose of energy equivalent to $1560 \mathrm{~kJ}$. He utilizes $50 \%$ of this gained energy in the event. In order to avoid storage of energy in the body, calculate the mass of water he would need to perspire. Enthalpy of $\mathrm{H}_{2} \mathrm{O}$ for evaporation is $44 \mathrm{~kJ} \mathrm{~mol}^{-1}$. (a) $346 \mathrm{~g}$ (b) $316 \mathrm{~g}$ (c) $323 \mathrm{~g}$ (d) $319 \mathrm{~g}$

An athlete is given 100 g of glucose of energy equivalent to $1560 \mathrm{~kJ}$. He utilizes $50 \%$ of this gained energy in the event. In order to avoid storage of energy in the body, calculate the mass of water he would need
to perspire. Enthalpy of $\mathrm{H}_{2} \mathrm{O}$ for evaporation is $44 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
(a) $346 \mathrm{~g}$
(b) $316 \mathrm{~g}$
(c) $323 \mathrm{~g}$
(d) $319 \mathrm{~g}$

Key Concepts

Energy Content of Carbohydrates

This concept involves the measurement of the energy provided by carbohydrates when they are metabolized. It helps in understanding how much energy is available from a given mass of carbohydrate, and it is essential for calculating the energy yield from food sources based on their chemical composition.

Energy Utilization Efficiency

Energy utilization efficiency refers to the fraction of the total available energy that is effectively used by an organism during a specific process or activity. In many calculations, only a portion of the energy content is converted into useful work, with the rest potentially contributing to other processes such as heat loss or energy storage.

Latent Heat of Vaporization

The latent heat of vaporization is the energy required to convert a substance from its liquid phase to its gaseous phase without a change in temperature. This concept is crucial when calculating the amount of energy needed to evaporate a certain mass of water, as it directly relates the energy input to the mass of water that undergoes phase change.

Mole-to-Mass Conversion

Mole-to-mass conversion is the process of using the molar mass of a substance to convert from moles to grams. This fundamental concept in stoichiometry allows for the calculation of the actual mass corresponding to a given number of moles, which is necessary when connecting molecular energy changes to macroscopic masses in practical applications.

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