Question

The metabolism of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) yields \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) as products. Heat released in the process is converted to useful work with about \(70 \%\) efficiency. Calculate the mass of glucose metabolized by a \(58.0 \mathrm{kg}\) person in climbing a mountain with an elevation gain of \(1450 \mathrm{m}\). Assume that the work performed in the climb is about four times that required to simply lift \(58.0 \mathrm{kg}\) by \(1450 \mathrm{m} \cdot\left(\Delta H_{\mathrm{f}}^{2} \text { of } \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s}) \text { is }-1273.3 \mathrm{kJ} / \mathrm{mol} .\right)\)

Short answer

To calculate the final result, execute the instructions in the mentioned sequence. The answer will be in grams, the metabolized glucose mass required for climbing the mountain.

Step-by-step solution

Calculate the basic work required to lift the person

Start by calculating the work done (or energy needed) without considering any inefficiencies, using the gravitational potential energy formula: \(Work = mass \times g \times height\), where mass = 58.0 Kg, \(g = 9.8 m/s^{2}\) (approximate gravitational acceleration), and height = 1450 m. Work out the result.

Calculate actual work or energy required

The problem notes that the actual work done in climbing the mountain is about four times that required to simply lift the person. Thus, work was underestimated in Step 1, and should be multiplied by four to get to the actual number. Hence, the actual required energy equals four times the result from step 1.

Account for inefficiency in energy conversion

The efficiency of energy conversion is said to be 70%. This means that for every 100 parts of energy obtained from glucose, only 70 parts are converted into useful work- the rest is lost as heat. Therefore, you need to calculate how much energy we would need to derive from glucose in order to have enough after the 30% loss. This is done by dividing the required energy from step 2 by 0.7 (corresponding to 70% efficiency).

Calculate moles of glucose needed

Next, calculate the number of glucose moles required to provide that energy. The molar enthalpy of formation for glucose is -1273.3 kJ/mol, but we are interested in the absolute value of this energy (as we are considering energy release), so we take 1273.3 kJ/mol. The number of moles equals the energy needed (from step 3) divided by the energy released per mole of glucose (1273.3 kJ/mol).

Calculate mass of glucose metabolized

Finally, convert the moles of glucose required into mass. The molar mass of glucose is 180.156 g/mol (12*12 + 1*22 + 16*6). Calculate the result by multiplying the number of moles (from step 4) by the molar mass of glucose.

Key concepts

Energy Efficiency in Glucose Metabolism

In the context of glucose metabolism, energy efficiency refers to how effectively our bodies convert the energy stored in glucose into usable work. The process is not 100% efficient. This is because not all the energy from glucose is transformed into work; some portion is always lost, primarily as heat.
The metabolism of glucose has an efficiency of 70%. This means that for every 100 units of energy that glucose can potentially provide, only 70 units are actually converted into energy for activities like climbing, while 30 units are lost.
When calculating the actual energy needed for activities, you must account for this inefficiency. In the exercise, this is done by dividing the total energy by the efficiency percentage (0.7 in this case). This adjustment ensures that the calculated energy truly reflects the amount of glucose required to perform the work.

Enthalpy of Formation

Enthalpy of formation is a crucial concept when discussing energy changes in chemical reactions. It represents the change in energy when one mole of a substance is formed from its elements in their standard states.
In glucose metabolism, the molar enthalpy of formation is given as -1273.3 kJ/mol. Here, the negative sign indicates that energy is released during the formation, signifying an exothermic reaction. However, for the purpose of calculating the metabolic energy of glucose, the absolute value is used: 1273.3 kJ/mol.
This value is critical as it allows you to compute how much energy is released per mole of glucose during metabolism. This provides a basis for determining how many moles of glucose are necessary to perform metabolic work, considering the body's energy efficiency.

Gravitational Potential Energy and Work

Gravitational potential energy is the energy an object possesses because of its position relative to the earth's surface. It is calculated using the formula: \[ \text{Potential Energy} = m \cdot g \cdot h \]where \(m\) is the mass, \(g\) is the gravitational acceleration (approximately 9.8 m/s² on the earth's surface), and \(h\) is the height.
In the exercise, this energy calculation is the first step in determining the work required to lift a person to a certain height, which in this case is 1450 meters. While this gives the basic potential energy, the problem specifies that actual work done is four times this amount due to additional energy requirements during climbing.
Thus, calculating gravitational potential energy provides a baseline from which total energy expenditure can be evaluated and adjusted for real-world inefficiencies and unexpected energy losses, such as heat dissipation and biomechanical factors.

Molar Mass Calculation

Calculating the molar mass is a critical step in transforming the amount of a substance from moles to grams, allowing for practical measurement in laboratories or metabolic analyses.
For glucose, the chemical formula is \(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\). To find its molar mass, you need to sum the atomic masses of all the atoms present in one molecule of glucose:

• Carbon (C): 12 atoms × 6 = 72
• Hydrogen (H): 1 atom × 12 = 12
• Oxygen (O): 16 atoms × 6 = 96

Adding these together results in a molar mass of 180.156 g/mol for glucose.
This calculation is a key step in converting the calculated moles of glucose (determined by the energy requirement) into an actual mass, providing a tangible amount of glucose needed for metabolic processes.