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Chapter 5: Problem 94

The heat of combustion of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l),\( is -1367 \)\mathrm{kJ} / \mathrm{mol}$. A bottle of stout (dark beer) contains up to \(6.0 \%\) ethanol by mass. Assuming the density of the beer to be \(1.0 \mathrm{~g} / \mathrm{mL},\) what is the caloric content due to the alcohol (ethanol) in a bottle of beer \((500 \mathrm{~mL})\) ?

Short Answer

Expert verified
The caloric content due to the ethanol in a \(500 \mathrm{~mL}\) bottle of beer is approximately -889.1 kJ.

Step by step solution

01

Find the mass of ethanol in the bottle of beer

First, we need to find the mass of the entire bottle of beer. To do this, we can use the density and volume given in the problem. Density of beer = \(1.0 \frac{g}{mL}\) Volume of the beer bottle = \(500 mL\) Mass of the beer \(=\) Density of beer \(\times\) Volume of the beer bottle Mass of the beer \(=1.0 \frac{g}{mL} \times 500 mL = 500 g\) Since the beer contains \(6.0\%\) ethanol by mass, we can find the mass of ethanol in the bottle: Mass of ethanol \(= 6.0 \% \times 500 g = 0.06 \times 500 g = 30 g\)
02

Convert the mass of ethanol to moles

To convert the mass of ethanol to moles, we need the molar mass of ethanol. Molar mass of ethanol \(\mathrm{C}_2\mathrm{H}_5\mathrm{OH} = 2\times 12.01 + 6\times 1.01 + 1\times 16.00 = 46.07 \frac{g}{mol}\) Moles of ethanol \(= \frac{Mass\;of\;Ethanol}{Molar\;Mass\;of\;Ethanol} = \frac{30 g}{46.07 \frac{g}{mol}}\) Moles of ethanol \(= 0.651 mol\)
03

Calculate the caloric content due to ethanol using the heat of combustion

Now that we have the moles of ethanol, we can calculate the caloric content using the heat of combustion. Heat of combustion of ethanol \(= -1367 \frac{kJ}{mol}\) (The negative sign indicates energy is released during combustion) Caloric content due to ethanol \(= -1367 \frac{kJ}{mol} \times 0.651 mol = -889.1 kJ\) The caloric content due to the ethanol in a bottle of beer is approximately -889.1 kJ.

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Most popular questions from this chapter

(a) A serving of a particular ready-to-serve brown \& wild rice meal contains \(4.5 \mathrm{~g}\) fat, \(42 \mathrm{~g}\) carbohydrate, and \(4.0 \mathrm{~g}\) protein. Estimate the number of calories in a serving. (b) According to its nutrition label, the same meal also contains $140 \mathrm{mg}$ of potassium ions. Do you think the potassium contributes to the caloric content of the food?

One of the best-selling light, or low-calorie, beers is \(4.2 \%\) alcohol by volume and a 355 -mL serving contains 110 Calories; remember: 1 Calorie $=1000 \mathrm{cal}=1 \mathrm{kcal} .$ To estimate the percentage of Calories that comes from the alcohol, consider the following questions. (a) Write a balanced chemical equation for the reaction of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH},$ with oxygen to make carbon dioxide and water. (b) Use enthalpies of formation in Appendix \(\mathrm{C}\) to determine \(\Delta H\) for this reaction. \((\mathbf{c})\) If \(4.2 \%\) of the total volume is ethanol and the density of ethanol is \(0.789 \mathrm{~g} / \mathrm{mL},\) what mass of ethanol does a \(355-\mathrm{mL}\) serving of light beer contain? (d) How many Calories are released by the metabolism of ethanol, the reaction from part (a)? (e) What percentage of the 110 Calories comes from the ethanol?

How much work (in J) is involved in a chemical reaction if the volume decreases from \(33.6 \mathrm{~L}\) to \(11.2 \mathrm{~L}\) against a constant pressure of \(90.5 \mathrm{kPa}\) ?

An aluminum can of a soft drink is placed in a freezer. Later, you find that the can is split open and its contents have frozen. Work was done on the can in splitting it open. Where did the energy for this work come from?

You may have noticed that when you compress the air in a bicycle pump, the body of the pump gets warmer. (a) Assuming the pump and the air in it comprise the system, what is the sign of \(w\) when you compress the air? (b) What is the sign of \(q\) for this process? (c) Based on your answers to parts (a) and (b), can you determine the sign of \(\Delta E\) for compressing the air in the pump? If not, what would you expect for the sign of \(\Delta E\) ? What is your reasoning? [Section 5.2]
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