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Chapter 7: Problem 107

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

Expert verified
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

01

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.
02

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.
03

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).
04

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).
05

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.

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Most popular questions from this chapter

The following substances undergo complete combustion in a bomb calorimeter. The calorimeter assembly has a heat capacity of \(5.136 \mathrm{kJ} /^{\circ} \mathrm{C} .\) In each case, what is the final temperature if the initial water temperature is \(22.43^{\circ} \mathrm{C} ?\) \(\begin{array}{lllll}\text { (a) } 0.3268 & \text { g caffeine, } & \mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{N}_{4} & \text { (heat of }\end{array}\) combustion \(=-1014.2 \mathrm{kcal} / \mathrm{mol} \text { caffeine })\) (b) \(1.35 \mathrm{mL}\) of methyl ethyl ketone, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}(1)\) \(d=0.805 \mathrm{g} / \mathrm{mL}\) (heat of combustion \(=-2444 \mathrm{kJ} / \mathrm{mol}\) methyl ethyl ketone).

The heat of neutralization of \(\mathrm{HCl}(\text { aq) by } \mathrm{NaOH}(\mathrm{aq})\) is \(-55.84 \mathrm{kJ} / \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) produced. If \(50.00 \mathrm{mL}\) of \(1.05 \mathrm{M}\) \(\mathrm{NaOH}\) is added to \(25.00 \mathrm{mL}\) of \(1.86 \mathrm{M} \mathrm{HCl}\), with both solutions originally at \(24.72^{\circ} \mathrm{C},\) what will be the final solution temperature? (Assume that no heat is lost to the surrounding air and that the solution produced in the neutralization reaction has a density of \(1.02 \mathrm{g} / \mathrm{mL}\) and a specific heat of \(3.98 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\).

Under the entry \(\mathrm{H}_{2} \mathrm{SO}_{4},\) a reference source lists many values for the standard enthalpy of formation. For example, for pure \(\mathrm{H}_{2} \mathrm{SO}_{4}(1), \Delta H_{\mathrm{f}}^{\circ}=-814.0 \mathrm{kJ} / \mathrm{mol}\) for a solution with \(1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) \(-841.8 ;\) with \(10 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-880.5 ;\) with \(50 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) \(-886.8 ;\) with \(100 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-887.7 ;\) with \(500 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) \(-890.5 ;\) with \(1000 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-892.3 ;\) with \(10,000 \mathrm{mol}\) \(\mathrm{H}_{2} \mathrm{O},-900.8 ;\) and with \(100,000 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-907.3\) (a) Explain why these values are not all the same. (b) The value of \(\Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\right]\) in an infinitely dilute solution is \(-909.3 \mathrm{kJ} / \mathrm{mol} .\) What data from this chapter can you cite to confirm this value? Explain. (c) If \(500.0 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is prepared from pure \(\mathrm{H}_{2} \mathrm{SO}_{4}(1),\) what is the approximate change in temperature that should be observed? Assume that the \(\mathrm{H}_{2} \mathrm{SO}_{4}(1)\) and \(\mathrm{H}_{2} \mathrm{O}(1)\) are at the same temperature initially and that the specific heat of the \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is about \(4.2 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\).

You are planning a lecture demonstration to illustrate an endothermic process. You want to lower the temperature of \(1400 \mathrm{mL}\) water in an insulated container from 25 to \(10^{\circ} \mathrm{C} .\) Approximately what mass of \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})\) should you dissolve in the water to achieve this result? The heat of solution of \(\mathrm{NH}_{4} \mathrm{Cl}\) is \(+14.7 \mathrm{kJ} / \mathrm{mol} \mathrm{NH}_{4} \mathrm{Cl}\).

We can determine the purity of solid materials by using calorimetry. A gold ring (for pure gold, specific heat \(=0.1291 \mathrm{Jg}^{-1} \mathrm{K}^{-1}\) ) with mass of \(10.5 \mathrm{g}\) is heated to \(78.3^{\circ} \mathrm{C}\) and immersed in \(50.0 \mathrm{g}\) of \(23.7^{\circ} \mathrm{C}\) water in a constant-pressure calorimeter. The final temperature of the water is \(31.0^{\circ} \mathrm{C}\). Is this a pure sample of gold?
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