/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 94 The heat of combustion of ethano... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The heat of combustion of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(t),\) 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 mL)?

Short Answer

Expert verified
The caloric content due to the alcohol in 500 mL of beer is approximately 890 kJ.

Step by step solution

01

Determine mass of the beer

First, calculate the mass of the beer using its given density and volume. Since the density is \(1.0\, \text{g/mL}\) and the volume is \(500\, \text{mL}\), the mass of the beer is \(500\, \text{g}\).
02

Calculate mass of ethanol in beer

Next, find the mass of ethanol in the beer. Since the beer contains \(6.0\%\) ethanol by mass, the mass of ethanol is \(0.06 \times 500\, \text{g} = 30 \, \text{g}.\)
03

Determine moles of ethanol

We need to convert the mass of ethanol into moles. The molar mass of ethanol, \(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}\), is approximately \(46.08\, \text{g/mol}\). Thus, the number of moles is \(\frac{30\, \text{g}}{46.08\, \text{g/mol}} \approx 0.651\, \text{mol}.\)
04

Calculate caloric content from ethanol

Now, calculate the energy released by the ethanol. The heat of combustion of ethanol is \(-1367\, \text{kJ/mol}\). Therefore, the total energy released is \(0.651\, \text{mol} \times -1367\, \text{kJ/mol} \approx -890\, \text{kJ}.\) This negative sign indicates energy released.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Ethanol
Ethanol is a type of alcohol commonly found in alcoholic beverages like beer, wine, and spirits. It is also known as ethyl alcohol or grain alcohol. Ethanol is a volatile, flammable, and colorless liquid with a slightly sweet flavor. It has the molecular formula \(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}\).
  • Used not only in beverages but also as a fuel source and industrial solvent.
  • Ethanol in beverages is often identified by its volume percentage in products like beer, where 6% indicates that 6% of the beer's mass is ethanol.
  • It plays a crucial role in the caloric content of alcoholic drinks due to its energy-dense properties.
In our problem, the ethanol in beer contributes majorly to its caloric content, and knowing its percentage by mass helps in calculating its energy release through combustion. Ethanol's combustion releases heat, a process we tap into when discussing caloric content.
Exploring Caloric Content
Caloric content refers to the amount of energy provided by a food or drink to the body. It is usually measured in kilojoules (kJ) or calories (cal) and helps us understand how much energy we would obtain from consuming the beverage.
  • The term 'calorie' in food packaging and nutritional contexts often refers to kilocalories (kcal).
  • When ethanol burns, it releases energy, which can be quantitatively measured using its heat of combustion, an important aspect when calculating the caloric content of alcoholic drinks.
  • The concept helps us evaluate how much fuel (energy) ethanol provides compared to other nutrients or ingredients in food and beverages.
For our example, based on the heat of combustion, the ethanol in beer contributes significantly to its overall energy content, which is why alcoholic drinks with higher ethanol content also have higher caloric values.
Role of Density in Calculations
Density is a measure of how much mass exists in a given volume of a substance. It helps us convert between mass and volume, which is crucial when dealing with liquids like beer.
  • Expressed as mass per unit volume, such as grams per milliliter (g/mL).
  • For beer, given a density of \(1.0 \, \text{g/mL}\), we assume a straightforward conversion: 1 liter of beer weighs 1 kilogram.
  • Knowing the density allows us to calculate the mass of a given volume accurately, essential for further computations like finding the mass of ethanol.
In the context of our problem, the density was used to first derive the total mass of the beer from its volume. From there, calculating the mass of ethanol in it was made possible, laying the ground for finding the energy content based on the heat of combustion. Understanding density and its role in these calculations ensures that we arrive at precise and meaningful answers.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is blended with gasoline as an automobile fuel. (a) Write a balanced equation for the combustion of liquid ethanol in air. (b) Calculate the standard enthalpy change for the reaction, assuming \(\mathrm{H}_{2} \mathrm{O}(g)\) as a product. (c) Calculate the heat produced per liter of ethanol by combustion of ethanol under constant pressure. Ethanol has a density of \(0.789 \mathrm{~g} / \mathrm{mL}\). (d) Calculate the mass of \(\mathrm{CO}_{2}\) produced per \(\mathrm{kJ}\) of heat emitted.

(a) What is the electrostatic potential energy (in joules) between an electron and a proton that are separated by \(230 \mathrm{pm} ?\) (b) What is the change in potential energy if the distance separating the electron and proton is increased to \(1.0 \mathrm{nm} ?\) (c) Does the potential energy of the two particles increase or decrease when the distance is increased to \(1.0 \mathrm{nm} ?\)

It is estimated that the net amount of carbon dioxide fixed by photosynthesis on the landmass of Earth is \(5.5 \times 10^{16} \mathrm{~g} / \mathrm{yr}\) of \(\mathrm{CO}_{2}\). Assume that all this carbon is converted into glucose. (a) Calculate the energy stored by photosynthesis on land per year, in \(\mathrm{kJ} .\) (b) Calculate the average rate of conversion of solar energy into plant energy in megawatts, MW \((1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s})\). A large nuclear power plant produces about \(10^{3} \mathrm{MW}\). The energy of how many such nuclear power plants is equivalent to the solar energy conversion?

Consider the following hypothetical reactions: $$ \begin{array}{l} \mathrm{A} \longrightarrow \mathrm{B} \quad \Delta H_{I}=+60 \mathrm{k} \mathrm{J} \\ \mathrm{B} \longrightarrow \mathrm{C} \quad \Delta H_{I}=-90 \mathrm{k} \mathrm{J} \end{array} $$ (a) Use Hess's law to calculate the enthalpy change for the reaction \(\mathrm{A} \longrightarrow \mathrm{C}\) (b) Construct an enthalpy diagram for substances \(A, B\), and \(\mathrm{C},\) and show how Hess's law applies.

The specific heat of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(I),\) is \(2.22 \mathrm{~J} / \mathrm{g}-\mathrm{K} .(\mathrm{a}) \mathrm{How}\) many J of heat are needed to raise the temperature of \(80.0 \mathrm{~g}\) of octane from 10.0 to \(25.0^{\circ} \mathrm{C} ?\) (b) Which will require more heat, increasing the temperature of \(1 \mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}(I)\) by a certain amount or increasing the temperature of \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}(I)\) by the same amount?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.