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?

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: $$ Initial \thinspace mass \thinspace of \thinspace water = 1000 \times \frac{58}{100} = 580 \: kg/h $$

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: $$ Final \thinspace mass \thinspace of \thinspace water = 1000 \times \frac{20}{100} = 200 \: kg/h $$

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.