An athlete is given 100 g of glucose (C₆H₁₂O₆) 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 ____________ 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
Molar mass of C, H O are 12, 1 and 16 g mol⁻¹.

Solution

Chemistry ma’am · 4 steps

Only the half he does not use has to be sweated away, so start from 900 kJ, not 1800 kJ.

1. Energy left over. The glucose gives 1800 kJ. Half of it, 900 kJ, goes into the sport. The other 900 kJ would be stored, so perspiration must remove it.
2. Moles of water. Each mole of water that evaporates carries away 45 kJ.

   n_(H₂O) = (900 kJ)/(45 kJ mol⁻¹) = 20 mol

3. Convert to mass. The molar mass of H₂O is 2 × 1 + 16 = 18 g mol⁻¹, so the mass is 20 × 18 = 360 g.
4. So. He must perspire 360 g of extra water.

The common slip: 720. Using the full 1800 kJ instead of the unused 900 kJ gives 40 mol, which is 720 g, not 360.

The idea to keep. In energy-balance questions, first find the energy that is actually left over. Then divide by the enthalpy per mole and convert moles to grams.

Answer from the NTA final answer key. Solution by PrepWiser.