/*! 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 30 Ever wonder why espresso costs m... [FREE SOLUTION] | 91Ó°ÊÓ

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Ever wonder why espresso costs much more per cup than regular drip coffee? Part of the reason is the expensive equipment needed to brew a proper espresso. A high-powered burr grinder first shears the coffee beans to a fine powder without producing too much heat. (Heating the coffee in the grinding stage prematurely releases the volatile oils that give espresso its rich flavor and aroma.) The ground coffee is put into a cylindrical container called a gruppa and tamped down firmly to provide an even flow of water through it. An electrically heated boiler inside the espresso machine maintains water in a reservoir at 1.4 bar and \(109^{\circ} \mathrm{C}\). An electric pump takes cold water at \(15^{\circ} \mathrm{C}\) and 1 bar, raises its pressure to slightly above 9 bar, and feeds it into a heating coil that passes through the reservoir. Heat transferred from the reservoir through the coil wall raises the water temperature to \(96^{\circ} \mathrm{C}\). The heated water flows into the top of the gruppa at \(96^{\circ} \mathrm{C}\) and 9 bar, passes slowly through the tightly packed ground beans, and dissolves the oils and some of the solids in the beans to become espresso, which decompresses to 1 atm as it exits the machine. The water temperature and uniform flow through the bed of packed coffee in the gruppa lead to the more intense flavor of espresso relative to normal drip coffee. Water drawn directly from the reservoir is expanded to atmospheric pressure where it forms steam, which is used to heat and froth milk for lattes and cappuccinos. (a) Sketch this process, using blocks to represent the pump, reservoir, and gruppa. Label all heat and work flows in the process, including electrical energy. (b) To make a 14-oz latte, you would steam 12 ounces of cold milk (3^'C) until it reaches 71 ^ C and pour it over 2 ounces of espresso. Assume that the steam cools but none of it condenses as it bubbles through the milk. For each latte made, the heating element that maintains the reservoir temperature must supply enough energy to heat the espresso water plus enough to heat the milk, plus additional energy. Assuming $$ \left(C_{p}\right)_{\operatorname{milk}}=3.93 \frac{\mathrm{J}}{\mathrm{g} \cdot^{\circ} \mathrm{C}}, \quad \mathrm{SG}_{\mathrm{milk}}=1.03 $$ calculate the quantity of electrical energy that must be provided to the heating element to accomplish those two functions. Why would more energy than what you calculate be required? (There are several reasons.) (c) Coffee beans contain a considerable amount of trapped carbon dioxide, not all of which is released when the beans are ground. When the hot pressurized water percolates through the ground beans, some of the carbon dioxide is absorbed in the liquid. When the liquid is then dispensed at atmospheric pressure, fine \(\mathrm{CO}_{2}\) bubbles come out of solution. In addition, one of the chemical compounds formed when the coffee beans are roasted and extracted into the espresso is melanoidin, a surfactant. Surfactant molecules are asymmetrical, with one end being hydrophilic (drawn to water) and the other end hydrophobic (repelled by water). When the bubbles (thin water films containing \(\mathrm{CO}_{2}\) ) pass through the espresso liquid, the hydrophilic ends of the melanoidin molecules attach to the bubbles and the dissolved bean oils in turn attach to the hydrophobic ends. The result is that the bubbles emerge coated with the oils to form the crema , the familiar reddish brown stable foam at the surface of good espresso. Speculate on why you don't see crema in normal drip coffee. (Hint: Henry's law should show up in your explanation.) Note: All soaps and shampoos contain at least one surfactant species. (A common one is sodium lauryl sulfate.) Its presence explains why if you have greasy hands, washing with plain water may leave the grease untouched but washing with soap removes the grease. (d) Explain in your own words (i) how espresso is made, (ii) why espresso has a more intense flavor than regular drip coffee, (iii) what the crema in espresso is, how it forms, and why it doesn't appear in regular drip coffee, and (iv) why washing with plain water does not remove grease but washing with soap does. (Note: Many people automatically assume that all chemical engineers are extraordinarily intelligent. If you can explain those four things, you can help perpetuate that belief.)

Short Answer

Expert verified
The process of espresso making entails steps that are carefully monitored for temperature and pressure, and involve more expensive equipment than regular drip coffee, leading to its cost. The electrical energy supplied for a 14-oz latte is higher than the energy required to heat water and milk due to inherent inefficiencies and losses. The intense flavor in espresso and the crema formation in it are attributed to the controlled conditions of brewing and the role of melanoidin which acts as a surfactant. Soaps remove grease effectively due to similar surfactant behavior.

Step by step solution

01

Diagram Creation

Draw a block diagram with blocks representing the pump, reservoir, and gruppa. Label the pump taking in cold water at \(15^{\circ} \mathrm{C}\) and 1 bar, raising its pressure to 9 bar, and feeding it into a heating coil. The reservoir maintained by an electrically heated boiler at \(109^{\circ} \mathrm{C}\) and 1.4 bar, the heated water flows into the top of the gruppa at \(96^{\circ} \mathrm{C}\) and 9 bar and becomes espresso upon exiting the machine. Illustrate the water drawn causing steam forming to heat the milk. Add labels for heat and work flows involved and specify the electrical energy.
02

Energy Calculation

Calculate the amount of electrical energy required to heat the water and the milk for a 14-oz latte. Heat provided to the milk can be calculated using the equation \(Q = m*C_{p}*\Delta T\), where \(m\) is the mass of the milk, \(C_{p}\) is the specific heat capacity of the milk, and \(\Delta T\) is the change in temperature of the milk. Do not forget to consider the specific gravity for the milk to get the mass for the volume given. The calculations should be done in suitable units. The electrical energy provided would be higher than this calculation due to inefficiencies, and losses.
03

Theoretical Considerations

Explain why espresso has a more intense flavor than regular drip coffee and why it has a crema. The mechanism of carbon dioxide absorption from coffee beans and its release into the liquid plays a major role as does the role of melanoidin acting as a surfactant. Point out that conditions like heat, pressure, and slow uniform flow are aiding the intense flavor in espresso. Also, relate the surfactant's role in forming the crema to its similar effect in soaps removing grease.
04

Explanation

In your own words, explain (i) the espresso making process, (ii) reasons for the intense flavor in espresso and regular drip coffee, (iii) crema formation in espresso and its absence in regular coffee, and (iv) the effectiveness of soaps over plain water in removing grease based on the theoretical considerations made in step 3.

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Key Concepts

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

Heat Transfer in Espresso Machine
When savoring an espresso, we rarely consider the sophisticated principles of heat transfer that go into brewing it to perfection. Well, the process is quite intricate and begins with preheated water being propelled through a heating coil in an espresso machine. Here, heat is transferred from the reservoir's hot water to the cold water in the coil. This is a classic example of \textbf{conduction} where thermal energy is passed through the material without the movement of the material itself – think of it like a baton being passed in a relay race.

Now, why does this matter? The controlled heat transfer ensures that the water reaches the optimal temperature of around \(96^{\circ} \mathrm{C}\) before mingling with the coffee grounds. This precise temperature is key to extracting rich flavors without burning the grounds or compromising the quality. Moreover, steam generated from heat transfer is also essential for creating that velvety froth in lattes and cappuccinos.
Specific Heat Capacity
Let's switch gears and talk about \textbf{specific heat capacity}, a term that sounds daunting but simply describes how much heat a substance can hold. Imagine it as the thermal 'endurance' of a substance – the higher the specific heat capacity, the more heat it can take before its temperature changes. This is crucial for our espresso calculations since it determines the amount of energy required to heat our espresso and milk to the desired temperatures.

In homework problem (b), we saw \(C_{p}\) representing specific heat capacity. For milk, this value is \(3.93 \frac{\mathrm{J}}{\mathrm{g} \cdot^{\circ} \mathrm{C}}\). To achieve that comforting sip of latte, your espresso machine has to take this into account to ensure enough energy is supplied to heat the milk from its starting chill of \(3^{\circ} \mathrm{C}\) to a cozy \(71^{\circ} \mathrm{C}\). This is calculated using the formula \(Q = m*C_{p}*\Delta T\), incorporating the specific gravity to relate the volume of milk to its mass and ensuring the energy supplied caters to these thermal requirements.
Surfactants in Chemical Processes
Surfactants might not ring a bell, but they're everywhere – from espresso to our daily shower. Think of surfactants as social molecules; one part loves water (hydrophilic), while the other would rather keep its distance (hydrophobic). This contradictory nature makes them extraordinary chemical process aids. In our beloved espresso, \textbf{melanoidin} is the surfactant superstar. It's busy during the brewing process, latching onto the small \(\mathrm{CO}_{2}\) bubbles with one end while its other end plays host to the delicious coffee oils.

Diving Deeper into Crema

The melanoidin not only helps to form the espresso's crema but also ensures the crema's stability. Why is this important? Well, that crema is more than a pretty layer; it's a flavor-packed emulsion that makes the espresso distinct from the regular drip coffee, which lacks this process and the conditions enabling the crema to form and persist. Surfactants in other contexts, like soaps, work in a similar way to wash away grease, making them indispensable in both our kitchen rituals and bathroom routines.
Henry's Law in Espresso Crema Formation
Let's unwrap some chemistry magic with \textbf{Henry's Law}, a gem that partly explains why espresso gets its signature crema. Henry's Law tells us about the solubility of gases: the higher the pressure, the more gas can be dissolved in a liquid. In the world of espresso, when the pressurized water interacts with the coffee grounds in the gruppa, it scoops up \(\mathrm{CO}_{2}\) trapped in the grounds. As the newly brewed espresso plummets back to atmospheric pressure, guess what happens? \(\mathrm{CO}_{2}\) bubbles escape, dancing their way to the top and voila – crema is born!

Why don't we see this frothy crema on regular coffee? The answer lies in the lack of that tightly packed coffee bed, high pressure, and the hot percolating process that are quintessential for espresso. This fizzy phenomenon doesn't occur in drip coffee, which is brewed without the high pressure that unleashes \(\mathrm{CO}_{2}\) with such abundance and helps create that rich, foamy layer on top of the espresso.

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Most popular questions from this chapter

A mixture of \(n\) -hexane vapor and air leaves a solvent recovery unit and flows through a \(70-\mathrm{cm}\) diameter duct at a velocity of \(3.00 \mathrm{m} / \mathrm{s}\). At a sampling point in the duct the temperature is \(40^{\circ} \mathrm{C}\), the pressure is \(850 \mathrm{mm}\) Hg, and the dew point of the sampled gas is \(25^{\circ} \mathrm{C}\). The gas is fed to a condenser in which it is cooled at constant pressure, condensing \(70 \%\) of the hexane in the feed. (a) Perform a degree-of-freedom analysis to show that enough information is available to calculate the required condenser outlet temperature \(\left(^{\circ} \mathrm{C}\right)\) and cooling rate \((\mathrm{kW})\) (b) Perform the calculations. (c) If the feed duct diameter were \(35 \mathrm{cm}\) for the same molar flow rate of the feed gas, what would be the average gas velocity (volumetric flow rate divided by cross-sectional area)? (d) Suppose you wanted to increase the percentage condensation of hexane for the same feed stream. Which three condenser operating variables might you change, and in which direction?

An aqueous slurry at \(30^{\circ} \mathrm{C}\) containing \(20.0 \mathrm{wt} \%\) solids is fed to an evaporator in which enough water is vaporized at 1 atm to produce a product slurry containing 35.0 wt\% solids. Heat is supplied to the evaporator by feeding saturated steam at 2.6 bar absolute into a coil immersed in the liquid. The steam condenses in the coil, and the slurry boils at the normal boiling point of pure water. The heat capacity of the solids may be taken to be half that of liquid water. (a) Calculate the required steam feed rate ( \(\mathrm{kg} / \mathrm{h}\) ) for a slurry feed rate of \(1.00 \times 10^{3} \mathrm{kg} / \mathrm{h}\). (b) Vapor recompression is often used in the operation of an evaporator. Suppose that the vapor (steam) generated in the evaporator described above is compressed to 2.6 bar and simultaneously heated to the saturation temperature at 2.6 bar, so that no condensation occurs. The compressed steam and additional saturated steam at 2.6 bar are then fed to the evaporator coil, in which isobaric condensation occurs. How much additional steam is required? (c) What more would you need to know to determine whether or not vapor recompression is economically advantageous in this process?

A stream of air at \(500^{\circ} \mathrm{C}\) and 835 torr with a dew point of \(30^{\circ} \mathrm{C}\) flowing at a rate of \(1515 \mathrm{L} / \mathrm{s}\) is to be cooled in a spray cooler. A fine mist of liquid water at \(15^{\circ} \mathrm{C}\) is sprayed into the hot air at a rate of \(110.0 \mathrm{g} / \mathrm{s}\) and evaporates completely. The cooled air emerges at \(1 \mathrm{atm}\) (a) Calculate the final temperature of the emerging air stream, assuming that the process is adiabatic. (Suggestion: Derive expressions for the enthalpies of dry air and water at the outlet air temperature, substitute them into the energy balance, and use a spreadsheet to solve the resulting fourth-order polynomial equation.) (b) At what rate (kW) is heat transferred from the hot air feed stream in the spray cooler? What becomes of this heat? (c) In a few sentences, explain how this process works in terms that a high school senior could understand. Incorporate the results of Parts (a) and (b) in your explanation.

The specific internal energy of formaldehyde (HCHO) vapor at 1 atm and moderate temperatures is given by the formula $$\hat{U}(\mathrm{J} / \mathrm{mol})=25.96 T+0.02134 T^{2}$$ where \(T\) is in \(^{\circ} \mathrm{C}\) (a) Calculate the specific internal energies of formaldehyde vapor at \(0^{\circ} \mathrm{C}\) and \(200^{\circ} \mathrm{C}\). What reference temperature was used to generate the given expression for \(\hat{U} ?\) (b) The value of \(\hat{U}\) calculated for \(200^{\circ} \mathrm{C}\) is not the true value of the specific internal energy of formaldehyde vapor at this condition. Why not? (Hint: Refer back to Section 7.5a.) Briefly state the physical significance of the calculated quantity. (c) Use the closed system energy balance to calculate the heat (J) required to raise the temperature of 3.0 mol HCHO at constant volume from 0^0 C to 200^'C. List all of your assumptions. (d) From the definition of heat capacity at constant volume, derive a formula for \(C_{v}(T)\left[\mathrm{J} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right]\) Then use this formula and Equation \(8.3-6\) to calculate the heat \((\) J) required to raise the temperature of 3.0 mol of HCHO(v) at constant volume from 0^ C to 200^'C. [You should get the same result you got in Part (c).]

(a) Determine the specific enthalpy ( \(\mathrm{kJ} / \mathrm{mol}\) ) of \(n\) -pentane vapor at \(200^{\circ} \mathrm{C}\) and 2.0 atm relative to n-pentane liquid at \(20^{\circ} \mathrm{C}\) and \(1.0 \mathrm{atm}\), assuming ideal-gas behavior for the vapor. Show clearly the process path you construct for this calculation and give the enthalpy changes for each step. State where you used the ideal-gas assumption.

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