Question

Many cereals are made with high moisture content so that the cereal can be formed into various shapes before it is dried. \(\mathrm{A}\) 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

To achieve a final cereal product containing \(20 \%\) water, \(380 \: kg/h\) of water must be evaporated per hour.

Step-by-step solution

Find the initial mass of water

We are given that the cereal product contains \(58 \%\) water by mass and the rate of production is \(1000 \: kg/h\). So, the initial mass of water in the product is $$ Initial \thinspace mass \thinspace of \thinspace water = (58 \% \thinspace of \thinspace 1000 \thinspace kg) \times \frac{1 \thinspace h}{1} $$ Calculating the value: $$ Initial \thinspace mass \thinspace of \thinspace water = 1000 \times \frac{58}{100} = 580 \: kg/h $$

Find the final mass of water when the product contains \(20 \%\) water

We are given that the final product should contain \(20 \%\) water. Since the initial mass of the product is \(1000 \: kg/h\), we can find the final mass of water in the product as $$ Final \thinspace mass \thinspace of \thinspace water = (20 \% \thinspace of \thinspace 1000 \thinspace kg) \times \frac{1 \thinspace h}{1} $$ Calculating the value: $$ Final \thinspace mass \thinspace of \thinspace water = 1000 \times \frac{20}{100} = 200 \: kg/h $$

Find the mass of water that must be evaporated per hour

Now, we have the initial mass of water in the product and the final mass of water in the product. The difference between these two values will give us the mass of water that must be evaporated per hour. So, we have $$ Mass \thinspace of \thinspace water \thinspace to \thinspace be \thinspace evaporated \thinspace per \thinspace hour = Initial \thinspace mass \thinspace of \thinspace water - Final \thinspace mass \thinspace of \thinspace water $$ Plugging in the values: $$ Mass \thinspace of \thinspace water \thinspace to \thinspace be \thinspace evaporated \thinspace per \thinspace hour = 580 \: kg/h - 200 \: kg/h = 380 \: kg/h $$ Thus, \(380 \: kg/h\) of water must be evaporated per hour so that the final product contains \(20 \%\) water.

Key concepts

Water Evaporation

In cereal production, water evaporation is a crucial process in which water is transformed from a liquid to a vapor and removed from the cereal mixture. This process is essential for achieving the desired moisture level in the final product. It helps in preserving the cereal, improving its texture, and enhancing its taste. When drying cereals, the main goal is to decrease the moisture content to a specific level that ensures freshness and longevity. In our case, we want to reduce the moisture from 58% to 20%. Understanding water evaporation helps in calculating how much water must be vaporized from a mixture to reach a desired consistency. The mass of water that needs to be evaporated can be computed by subtracting the mass of water in the final product from the mass of water in the initial product. After calculating, the result tells us how much water to remove to maintain the cereal's quality.

Moisture Content

Moisture content refers to the amount of water present in a cereal product by mass percentage. It is a critical factor in cereal manufacturing because it affects both the production process and the final product's quality. With precise control, manufacturers can ensure the cereal is neither too dry nor too wet. Calculating moisture content involves determining the percentage of water in a product compared to its total weight. For example, a cereal with 58% moisture content means that 58% of its total weight is water. Managing moisture content is essential, as it influences the texture, storage life, and appearance of the final cereal product. By understanding this concept, you see why it's necessary to manage and calculate moisture levels accurately during production.

Cereal Production

Cereal production involves multiple stages including mixing, shaping, and drying. Managing moisture throughout these stages is key to ensuring a product that meets quality standards. Initially, cereals often start with a high moisture content so they can be easily shaped and mixed. In our scenario, the cereal begins with 58% water. This is necessary for processing but must be adjusted through drying to achieve the preferred 20% moisture content. Understanding cereal production and how each stage influences the final product is important for maintaining quality control. The drying phase is particularly critical as it affects the water content and, subsequently, the storage duration and sensory properties of the cereal.

Unit Conversion

Unit conversion is a vital step in calculations related to cereal production, especially when dealing with mass and moisture percentages. In our example, interpreting percentages and converting them to practical figures is needed to determine the water content in a given mass of cereal. To convert a percentage to a mass, multiply the total mass by the percentage (in decimal form). For instance, converting 58% moisture in a 1000 kg cereal involves calculating: \[ \text{Mass of water} = 1000 \times \frac{58}{100} = 580 \text{ kg} \text{ of water}\]Conversion helps in precisely determining the quantity of each component in a mixture. It turns percentage values into usable data points that guide production steps and ensure the correct processing of cereal batches.