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 \(1000 .\) kg/h. What mass of water must be evaporated per hour if the final product contains only \(20 . \%\) water?

We are given that the cereal product contains 58% water by mass and is produced at a rate of 1000 kg/h. To find the mass of water and the mass of dry cereal in the initial product, we will use these percentages and the total product mass. The mass of water in the initial product is: \(Mass_{water} = Mass_{total} \times \frac{58}{100}\) The mass of dry cereal (not including the water) in the initial product is: \(Mass_{dry\_cereal} = Mass_{total} - Mass_{water}\)

Now that we have the mass of water and mass of dry cereal in the initial product, we can calculate the mass of water that needs to be evaporated to reach the desired 20% final mass percentage of water. We can find the final total mass by using the following equation based on the final mass percentage of water: \(Mass_{final\_total} = \frac{Mass_{dry\_cereal}}{1 - \frac{20}{100}}\) Now we can find the mass of water in the final product: \(Mass_{final\_water} = Mass_{final\_total} - Mass_{dry\_cereal}\) Finally, we can find the mass of water that must be evaporated per hour by subtracting the mass of water in the final product from the mass of water in the initial product: \(Mass_{evaporated\_water\_per\_hour} = Mass_{water} - Mass_{final\_water}\) Now, we just need to plug in the numbers and calculate:

Initial mass of water: \(Mass_{water} = 1000 \times \frac{58}{100} = 580 kg/h\) Initial mass of dry cereal (not including water): \(Mass_{dry\_cereal} = 1000 - 580 = 420 kg/h\) Final total mass: \(Mass_{final\_total} = \frac{420}{1 - \frac{20}{100}} = \frac{420}{0.8} = 525 kg/h\) Final mass of water: \(Mass_{final\_water} = 525 - 420 = 105 kg/h\) Mass of water that must be evaporated per hour: \(Mass_{evaporated\_water\_per\_hour} = 580 - 105 = 475 kg/h\) So, 475 kg of water must be evaporated per hour to achieve a final product containing only 20% water.