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Question: An athlete is given 180 g of glucose \[\left( {{{\text{C}}_6}{{\text{H}}_{12}}{{\text{O}}_6}} \right...

An athlete is given 180 g of glucose (C6H12O6)\left( {{{\text{C}}_6}{{\text{H}}_{12}}{{\text{O}}_6}} \right). He utilises 50%50\% of the energy due to internal combustion in the body, in order to avoid storage of energy in the body. Calculate the mass of water he would need to perspire. Given enthalpy of combustion of glucose is −2800 KJ mol−1 - 2800{\text{ KJ mo}}{{\text{l}}^{ - 1}} and enthalpy of evaporation of water is 44 KJ mol−1{\text{44 KJ mo}}{{\text{l}}^{ - 1}}

Explanation

Solution

Using the molar mass of the glucose we will get the energy that is actually consumed by the glucose. The enthalpy of evaporation of water is given and hence using the mass of water and consumed energy by the athlete we will calculate the mass of water needed.

Complete step by step solution:
- The molecular formula for glucose is C6H12O6{{\text{C}}_6}{{\text{H}}_{12}}{{\text{O}}_6}. The Mass of 1 mole of glucose is 12×6+12×1+6×16=18012 \times 6 + 12 \times 1 + 6 \times 16 = 180
- The combustion of 1 mole of glucose is −2800 KJ mol−1 - 2800{\text{ KJ mo}}{{\text{l}}^{ - 1}}. We can also say that combustion of 180 g of glucose is −2800 KJ mol−1 - 2800{\text{ KJ mo}}{{\text{l}}^{ - 1}}.
- It is given to us that the athlete consumes only fifty percent of glucose. Hence the enthalpy of combustion will be:
50100×−2800 KJ mol−1=1400 KJ mol−1\dfrac{{50}}{{100}} \times - 2800{\text{ KJ mo}}{{\text{l}}^{ - 1}} = 1400{\text{ KJ mo}}{{\text{l}}^{ - 1}}
For 1 mole the enthalpy of evaporation is 44 KJ mol−1{\text{44 KJ mo}}{{\text{l}}^{ - 1}}. The molar mass of water is 18. Number of moles is the ratio of mass divided by the molar mass. Hence the mass of water needed for perspiration is:
1844×1400=572.7 g\dfrac{{18}}{{44}} \times 1400 = 572.7{\text{ g}}

Hence, the mass of water is 572.7 g572.7{\text{ g}}.

Note: Enthalpy of combustion is defined as the amount of energy release when a substance burns completely in the presence of oxygen. The enthalpy of combustion is always negative. Enthalpy of evaporation is the amount of energy released or consumed when 1 mole of substance changes into gaseous phase from liquid phase. Generally the heat is consumed during evaporation and hence the enthalpy of evaporation is positive generally. Glucose is monosaccharide. Glucose is converted to energy in the form of ATP that is adenosine triphosphate along with carbon dioxide and water. Perspiration is the process of loss of water from the body.