Question

How many grams of sucrose would the hiker have to metabolize (quickly) to replace the heat of evaporating 1.00 L of water so that her temperature would not change? You can use the enthalpy of reaction at 25 ∘C; the sucrose (s) reacts with oxygen (g) to give carbon dioxide (g) and water (l).

Express your answer using three significant figures and include the appropriate units.

Answer 1

To replace the heat lost from evaporating 1.00 L of water, calculate the total energy needed for the evaporation at 37 °C using the enthalpy of vaporization and convert it to moles.

Step-by-step explanation:

The heat of evaporating 1.00 L of water can be replaced by metabolizing sucrose, a process that is exothermic and releases energy. First, calculate the amount of energy needed to evaporate the water. Using the given enthalpy of vaporization (ΔHvap) at 37 °C (normal body temperature) which is 43.46 kJ/mol, we can find how many moles of water there are in 1.00 L (which is approximately 1.00 kg or 1000 g) assuming the density of water is 1 g/cm³. Considering the molecular weight of water (H₂O) is about 18.01528 g/mol, we can convert 1000 g of water to moles and then multiply by the enthalpy of vaporization to find the total energy needed to evaporate the water.

Once we have the total energy in kJ, we can convert it to the amount of sucrose needed to provide that much energy using the energy released per gram of sucrose metabolized. As sucrose provides approximately 4 kcal/g (equal to 16.7 kJ/g), we can divide the total energy required by the energy provided per gram of sucrose. This will give us the weight in grams of sucrose required to replace the heat lost through the evaporation of 1.00 L of water.