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Many cereals are made with high moisture content so that the cereal can be formed into various shapes before it is dried. A cereal product containing \(58 \% \mathrm{H}_{2} \mathrm{O}\) by mass is produced at the rate of \(1000 . \mathrm{kg} / \mathrm{h} .\) What mass of water must be evaporated per hour if the final product contains only \(20 . \%\) water?

Short Answer

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#tag_title#Step 2: Calculate the final mass of water#tag_content#Now, let's find the mass of water in the final cereal product. The mass percentage is given as 20%. We will multiply the mass percentage by the total mass of the cereal product: Mass of water finally = (Percentage of water finally) x (Total mass of cereal product) \(Mass\,of\,water\,finally = 0.20 \times 1000\, kg\) #tag_title#Step 3: Calculate the mass of water evaporated#tag_content#To find the mass of water evaporated per hour, we subtract the final mass of water from the initial mass of water: Mass of water evaporated = Mass of water initially - Mass of water finally #tag_title#Answer#tag_content#\(Mass\,of\,water\,evaporated = (0.58 \times 1000\, kg) - (0.20 \times 1000\, kg) = 380\, kg/h\)

Step by step solution

01

Calculate the initial mass of water

First, let's find the mass of water in the initial cereal product. The mass percentage is given as 58%. To find the mass of water, we will multiply the mass percentage by the total mass of the cereal product: Mass of water initially = (Percentage of water initially) x (Total mass of cereal product) \(Mass\,of\,water\,initially = 0.58 \times 1000\, kg\)

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Key Concepts

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

Water Evaporation
Water evaporation is a key process in drying cereals to reduce their moisture content. It involves converting liquid water into vapor and removing it from the cereal. This process is essential to achieve the desired moisture level in the final product. Removing excess water helps stabilize the cereal, preventing spoilage and allowing it to maintain its shape and texture during storage. When you have a cereal with high initial moisture content, like 58% by mass, it's crucial to reduce it efficiently. For example, if a cereal is produced at a rate of 1000 kg per hour initially containing 58% water, it means that 580 kg of the product is water. In the drying process, water evaporates and is removed until the cereal contains only the desired 20% moisture. To achieve this, you'll need to calculate how much of the initial water needs to be evaporated to leave behind only the desired percentage of moisture.
Mass Percentage
Mass percentage is an important concept when dealing with mixtures, like cereals, where you need to know how much of a given substance, such as water, is present. It is expressed as a percentage of the total mass of the mixture. To calculate mass percentage, divide the mass of the substance by the total mass of the mixture, and then multiply the result by 100. For instance, if a cereal contains 58% water by mass, this means that in every 100 kg of cereal, there are 58 kg of water. Understanding this concept helps us quantify the amount of water that must be evaporated to reach a specific dryness, like lowering the water content to 20%. This concept also helps in adjusting processes to ensure consistent quality in cereal production, considering factors like shelf life and consumer preferences for texture.
Initial Mass Calculation
The initial mass calculation is the first step in determining how much water needs to be removed from a product to achieve a desired moisture level. To perform this calculation, you need to know the total mass of the product and the initial mass percentage of the water it contains. Here's how you can calculate it for the cereal mentioned:
  • The total mass of the cereal product is 1000 kg.
  • The initial water content is 58% by mass.
  • The mass of water initially present is calculated as: 0.58 × 1000 kg = 580 kg.
These calculations are essential in planning the water evaporation process. Understanding the initial mass of water allows you to quantify how much needs to be evaporated to achieve a final moisture content, such as 20%. Such initial assessments ensure that the drying process is efficient and that the cereal reaches the correct moisture level for safe storage. Initial mass calculation gives you a clear starting point for achieving your goals, ensuring that each cereal batch is processed correctly and uniformly.

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Most popular questions from this chapter

A \(0.4230-\mathrm{g}\) sample of impure sodium nitrate was heated, converting all the sodium nitrate to \(0.2864 \mathrm{~g}\) of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the original sample.

Chloral hydrate \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3} \mathrm{O}_{2}\right)\) is a drug formerly used as a sedative and hypnotic. It is the compound used to make "Mickey Finns" in detective stories. a. Calculate the molar mass of chloral hydrate. b. What amount (moles) of \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3} \mathrm{O}_{2}\) molecules are in \(500.0 \mathrm{~g}\) chloral hydrate? c. What is the mass in grams of \(2.0 \times 10^{-2}\) mol chloral hydrate? d. What number of chlorine atoms are in \(5.0 \mathrm{~g}\) chloral hydrate? e. What mass of chloral hydrate would contain \(1.0 \mathrm{~g} \mathrm{Cl} ?\) \(\mathbf{f}\). What is the mass of exactly 500 molecules of chloral hydrate?

Balance each of the following chemical equations. a. \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{KOH}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) b. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) c. \(\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{PCl}_{5}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{HCl}(g)\) e. \(\mathrm{CaO}(s)+\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}_{2}(g)\) f. \(\operatorname{MoS}_{2}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{MoO}_{3}(s)+\mathrm{SO}_{2}(g)\) g. \(\mathrm{FeCO}_{3}(s)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{HCO}_{3}\right)_{2}(a q)\)

A compound that contains only carbon, hydrogen, and oxygen is \(48.64 \% \mathrm{C}\) and \(8.16 \% \mathrm{H}\) by mass. What is the empirical formula of this substance?

In using a mass spectrometer, a chemist sees a peak at a mass of \(30.0106\). Of the choices \({ }^{12} \mathrm{C}_{2}{ }^{1} \mathrm{H}_{6},{ }^{12} \mathrm{C}^{1} \mathrm{H}_{2}{ }^{16} \mathrm{O}\), and \({ }^{14} \mathrm{~N}^{16} \mathrm{O}\), which is responsible for this peak? Pertinent masses are \({ }^{1} \mathrm{H}, 1.007825\); \({ }^{16} \mathrm{O}, 15.994915 ;\) and \({ }^{14} \mathrm{~N}, 14.003074 .\)

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